# A Proof of The Erdos-Turan Conjecture on Asymptotic Additive Bases

Konstantinos Smpokos

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# A Proof of The Erdos-Turan Conjecture on Asymptotic Additive Bases

Konstantinos Smpokos

<figure><img src="image_1.png" /></figure>

To cite this version:

Konstantinos Smpokos. A Proof of The Erdos-Turan Conjecture on Asymptotic Additive Bases. 2024. [hal-04611663](https://hal.science/hal-04611663v1)

HAL Id: hal-04611663

https://hal.science/hal-04611663v1

Preprint submitted on 13 Jun 2024

**HAL** is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire **HAL**, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d'enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

<figure><img src="image_2.png" /></figure>

Distributed under a Creative Commons Attribution 4.0 International License
A PROOF OF THE ERDOS-TURAN CONJECTURE ON ASYMPTOTIC
ADDITIVE BASES

KONSTANTINOS SMPOKOS

ABSTRACT. In this article we present a full proof of the Erdos-Turan conjecture on asymptotic additive bases. The conjecture states that if a set $B$ of non-negative integers is an asymptotic base of natural numbers of order two and we denote the sequence which counts the number of ways that a natural number $N$ can be written as sum of two elements of the set $B$ as $R_B(N)$, then we have that $\limsup_{N\to\infty} R_B(N) = \infty$.

# 1. INTRODUCTION

By [2], in 1937 I. M. Vinogradov succeeded in solving the original Goldbach problem for all odd $N$ which are sufficiently large. Before him, G. H. Hardy and J. E. Littlewood made a serious attack by means of a then new circle method. In the article [1] of Erdos and Turan in 1941 the Erdos-Turan conjecture is stated. It states that for every set $B$ of non-negative integers which is an asymptotic base of the natural numbers of order two, the sequence which counts the number of ways that a natural number $N$ is written as a sum of two elements of the set $B$, is unbounded.

In this paper we prove Erdos-Turan conjecture. Here, we denote as $R_B(N)$ the sequence which counts the number of ways that a natural number $N$ is written as a sum of two elements of the set $B$. We use only the starting point of the circle method and using analytical tools we present a full proof of this conjecture.

<footnote n="2020">2020 Mathematics Subject Classification. 11P82.</footnote>

<footnote>Key words and phrases. Additive Number Theory, Erdos-Turan Conjecture on Additive Bases, Asymp-
totic Bases of Order Two.</footnote>
# A PROOF OF THE ERDOS-TURAN CONJECTURE ON ASYMPTOTIC ADDITIVE BASES

KONSTANTINOS SMPOKOS

ABSTRACT. In this article we present a full proof of the Erdos-Turan conjecture on asymptotic additive bases. The conjecture states that if a set $B$ of non-negative integers is an asymptotic base of natural numbers of order two and we denote the sequence which counts the number of ways that a natural number $N$ can be written as sum of two elements of the set $B$ as $R_B(N)$, then we have that $\limsup_{N\to\infty} R_B(N) = \infty$.

## 1. INTRODUCTION

By [2], in 1937 I. M. Vinogradov succeeded in solving the original Goldbach problem for all odd $N$ which are sufficiently large. Before him, G. H. Hardy and J. E. Littlewood made a serious attack by means of a then new circle method. In the article [1] of Erdos and Turan in 1941 the Erdos-Turan conjecture is stated. It states that for every set $B$ of non-negative integers which is an asymptotic base of the natural numbers of order two, the sequence which counts the number of ways that a natural number $N$ is written as a sum of two elements of the set $B$, is unbounded.

In this paper we prove Erdos-Turan conjecture. Here, we denote as $R_B(N)$ the sequence which counts the number of ways that a natural number $N$ is written as a sum of two elements of the set $B$. We use only the starting point of the circle method and using analytical tools we present a full proof of this conjecture.

<footnote n="2020">2020 Mathematics Subject Classification. 11P82.</footnote>

<footnote>Key words and phrases. Additive Number Theory, Erdos-Turan Conjecture on Additive Bases, Asymptotic Bases of Order Two.</footnote>
## 2. PRELIMINARY RESULTS

Write $R_B(N) = \sum_{(n,m) \in B^2} 1$. Write also $B(t) = \sum_{n \in B, n \le t} 1$.

**Lemma 2.1.** Let $B$ be an infinite set of natural numbers. Then, For every $\rho \in (0, 1)$, we have

$$R_B(N) = \frac{1}{2\pi i} \oint_{|z|=\rho} \frac{\log^2 z}{z^{N+1}} \int_1^\infty \int_1^\infty B(t)B(s)z^{t+s}dt ds dz$$

*Proof.* Take the complex analytic function $f(z) = \sum_{n \in B} z^n$ for $|z| < 1$. It is easy to see that the powerseries has radius of convergence 1. Then, we have

$$f^2(z) = \sum_{n \in B} \sum_{m \in B} z^{n+m} = \sum_{k=1}^{\infty} \left( \sum_{\substack{(n,m) \in B^2 \\ n+m=k}} 1 \right) z^k = \sum_{k=1}^{\infty} R_B(k) z^k$$

By the Cauchy integral formula, we get

$$R_B(N) = \frac{1}{2\pi i} \oint_{|z|=\rho} \frac{f^2(z)}{z^{N+1}} dz$$

This holds for every $\rho \in (0, 1)$. Let a complex number $z$ eith $|z| = \rho \in (0, 1)$.

Then $z = \rho e^{i\theta}$, where $\theta \in (-\pi, \pi]$. Let $f_k(z) = \sum_{n=1}^k \chi_B(n)z^n$. Here, the function $\chi_B$ is the characteristic function of the set $B$.

We use Abel's summation formula which can be found in [3]. We have,

$$
\begin{align*}
f_k(z) &= \left(\sum_{n=1}^{k} \chi_B(n)\right) \rho^k e^{ik\theta} - \int_1^k \sum_{n \le t} \chi_B(n) (\rho^t e^{i\theta t})' dt = \\
&= B(k) \rho^k e^{ik\theta} - \int_1^k B(t) (\rho^t e^{i\theta t})' dt = \\
&= B(k) \rho^k e^{ik\theta} - \int_1^k B(t) (\rho^t \log \rho e^{i\theta t} + \rho^t e^{i\theta t} i\theta) dt = \\
&= B(k) \rho^k e^{ik\theta} - \int_1^k B(t) \rho^t e^{i\theta t} \log(\rho e^{i\theta}) dt = \\
&= B(k) z^k - \log z \int_1^k B(t) z^t dt
\end{align*}
$$
## 2. PRELIMINARY RESULTS

Write $R_B(N) = \sum_{(n,m) \in B^2} 1$. Write also $B(t) = \sum_{n \in B, n \le t} 1$.

**Lemma 2.1.** Let $B$ be an infinite set of natural numbers. Then, For every $\rho \in (0, 1)$, we have

$$R_B(N) = \frac{1}{2\pi i} \oint_{|z|=\rho} \frac{\log^2 z}{z^{N+1}} \int_1^\infty \int_1^\infty B(t)B(s)z^{t+s}dt ds dz$$

*Proof.* Take the complex analytic function $f(z) = \sum_{n \in B} z^n$ for $|z| < 1$. It is easy to see that the powerseries has radius of convergence 1. Then, we have

$$f^2(z) = \sum_{n \in B} \sum_{m \in B} z^{n+m} = \sum_{k=1}^{\infty} \left( \sum_{\substack{(n,m) \in B^2 \\ n+m=k}} 1 \right) z^k = \sum_{k=1}^{\infty} R_B(k) z^k$$

By the Cauchy integral formula, we get

$$R_B(N) = \frac{1}{2\pi i} \oint_{|z|=\rho} \frac{f^2(z)}{z^{N+1}} dz$$

This holds for every $\rho \in (0, 1)$. Let a complex number $z$ eith $|z| = \rho \in (0, 1)$.

Then $z = \rho e^{i\theta}$, where $\theta \in (-\pi, \pi]$. Let $f_k(z) = \sum_{n=1}^k \chi_B(n)z^n$. Here, the function $\chi_B$ is the characteristic function of the set $B$.

We use Abel's summation formula which can be found in [3]. We have,

$$
\begin{align*}
f_k(z) &= \left(\sum_{n=1}^{k} \chi_B(n)\right) \rho^k e^{ik\theta} - \int_1^k \sum_{n \le t} \chi_B(n) (\rho^t e^{i\theta t})' dt = \\
&= B(k) \rho^k e^{ik\theta} - \int_1^k B(t) (\rho^t e^{i\theta t})' dt = \\
&= B(k) \rho^k e^{ik\theta} - \int_1^k B(t) (\rho^t \log \rho e^{i\theta t} + \rho^t e^{i\theta t} i\theta) dt = \\
&= B(k) \rho^k e^{ik\theta} - \int_1^k B(t) \rho^t e^{i\theta t} \log(\rho e^{i\theta}) dt = \\
&= B(k) z^k - \log z \int_1^k B(t) z^t dt
\end{align*}
$$
Here, we use the branch of logarithm that is real on every positive real number.

We adopt the convention that $B(t) = 0$ for every $t < 1$.

Therefore, using the fact that $\lim_{k \to \infty} B(k)z^k = 0$ (since $|z| < 1$ and $0 \le B(k) \le k$), we get

$$f(z) = \lim_{k \to \infty} f_k(z) = -\log z \int_{1}^{\infty} B(t)z^t dt$$

Therefore,

$$f^2(z) = \log^2 z \int_{1}^{\infty} \int_{1}^{\infty} B(t)B(s)z^{t+s} dt ds$$

Thus, we have

$$R_B(N) = \frac{1}{2\pi i} \oint_{|z|=\rho} \frac{\log^2 z}{z^{N+1}} \int_1^\infty \int_1^\infty B(t)B(s)z^{t+s} dt ds \, dz$$

This completes the proof. $\square$

**Proposition 2.2.** Let

$$\Lambda(x) = -\frac{4\pi^2 \cos(x\pi)}{x} - \frac{4\pi(\cos(\pi x) - \sin(\pi x))}{x^2} + \frac{4\sin(\pi x)}{x^3}$$

Then, we have

$$R_B(N) = \frac{1}{2\sqrt{2\pi}} \int_{2-N}^{\infty} \int_{t+s=\lambda, t,s \ge 1-\frac{N}{2}} B\left(t+\frac{N}{2}\right) B\left(s+\frac{N}{2}\right) dH^1(t,s) e^{-\pi\lambda} \Lambda(\lambda) d\lambda$$

Here $H^1$ stands for the one dimensional Hausdorff measure in $\mathbb{R}^2$.

*Proof.* From the previous Lemma, we have that

$$R_B(N) = \frac{1}{2\pi i} \oint_{|z|=\rho} \frac{\log^2 z}{z^{N+1}} \int_1^\infty \int_1^\infty B(t)B(s)z^{t+s} dt ds \, dz$$

Therefore, we get

$$\begin{align*}
R_B(N) &= \frac{1}{2\pi i} \int_{-\pi}^{\pi} \frac{\log^2(\rho e^{i\theta}) (\int_1^{\infty} B(t) \rho^t e^{it\theta} dt)^2 i \rho e^{i\theta}}{\rho^{N+1} e^{i(N+1)\theta}} d\theta \\
&= \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{(\log \rho + i\theta)^2 (\int_1^{\infty} B(t) \rho^t e^{it\theta} dt)^2}{\rho^N e^{iN\theta}} d\theta \\
&= \frac{1}{2\pi \rho^N} \int_{-\pi}^{\pi} (\log \rho + i\theta)^2 (\int_1^{\infty} B(t) \rho^t e^{i(t-\frac{N}{2})\theta} dt)^2 d\theta \\
&= \frac{1}{2\pi} \int_{-\pi}^{\pi} (\log \rho + i\theta)^2 (\int_{1-\frac{N}{2}}^{\infty} B(t+\frac{N}{2}) \rho^t e^{it\theta} dt)^2 d\theta
\end{align*}$$
Here, we use the branch of logarithm that is real on every positive real number.

We adopt the convention that $B(t) = 0$ for every $t < 1$.

Therefore, using the fact that $\lim_{k \to \infty} B(k)z^k = 0$ (since $|z| < 1$ and $0 \le B(k) \le k$), we get

$$f(z) = \lim_{k \to \infty} f_k(z) = -\log z \int_1^\infty B(t)z^t dt$$

Therefore,

$$f^2(z) = \log^2 z \int_1^\infty \int_1^\infty B(t)B(s)z^{t+s} dt ds$$

Thus, we have

$$R_B(N) = \frac{1}{2\pi i} \oint_{|z|=\rho} \frac{\log^2 z}{z^{N+1}} \int_1^\infty \int_1^\infty B(t)B(s)z^{t+s} dt ds \, dz$$

This completes the proof. $\square$

**Proposition 2.2.** Let

$$\Lambda(x) = -\frac{4\pi^2 \cos(x\pi)}{x} - \frac{4\pi(\cos(\pi x) - \sin(\pi x))}{x^2} + \frac{4\sin(\pi x)}{x^3}$$

Then, we have

$$R_B(N) = \frac{1}{2\sqrt{2\pi}} \int_{2-N}^{\infty} \int_{t+s=\lambda, t,s \ge 1-\frac{N}{2}} B(t+\frac{N}{2})B(s+\frac{N}{2}) dH^1(t,s) e^{-\pi\lambda} \Lambda(\lambda) d\lambda$$

Here $H^1$ stands for the one dimensional Hausdorff measure in $\mathbb{R}^2$.

*Proof.* From the previous Lemma, we have that

$$R_B(N) = \frac{1}{2\pi i} \oint_{|z|=\rho} \frac{\log^2 z}{z^{N+1}} \int_1^\infty \int_1^\infty B(t)B(s)z^{t+s} dt ds \, dz$$

Therefore, we get

$$\begin{align*}
R_B(N) &= \frac{1}{2\pi i} \int_{-\pi}^{\pi} \frac{\log^2(\rho e^{i\theta}) (\int_1^{\infty} B(t) \rho^t e^{it\theta} dt)^2 i \rho e^{i\theta}}{\rho^{N+1} e^{i(N+1)\theta}} d\theta \\
&= \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{(\log \rho + i\theta)^2 (\int_1^{\infty} B(t) \rho^t e^{it\theta} dt)^2}{\rho^N e^{iN\theta}} d\theta \\
&= \frac{1}{2\pi \rho^N} \int_{-\pi}^{\pi} (\log \rho + i\theta)^2 (\int_1^{\infty} B(t) \rho^t e^{i(t-\frac{N}{2})\theta} dt)^2 d\theta \\
&= \frac{1}{2\pi} \int_{-\pi}^{\pi} (\log \rho + i\theta)^2 (\int_{1-\frac{N}{2}}^{\infty} B(t+\frac{N}{2}) \rho^t e^{it\theta} dt)^2 d\theta
\end{align*}$$
$$ = \frac{1}{2\pi} \int_{-\pi}^{\pi} (\log \rho + i\theta)^2 \int_{[1-\frac{N}{2}, \infty)^2} B(t+\frac{N}{2})B(s+\frac{N}{2})\rho^{t+s} e^{i(t+s)\theta} d(t,s)d\theta $$

Now, since the function $(\log \rho + i\theta)^2 B(t + \frac{N}{2})B(s + \frac{N}{2})\rho^{t+s} e^{i(t+s)\theta}$ is integrable for $(t, s, \theta) \in [1 - \frac{N}{2}, \infty)^2 \times [-\pi, \pi]$, by Fubini Theorem we have

$$ R_B(N) = \frac{1}{2\pi} \int_{[1-\frac{N}{2}, \infty)^2} B(t+\frac{N}{2})B(s+\frac{N}{2})\rho^{t+s} \int_{-\pi}^{\pi} (\log \rho + i\theta)^2 e^{i(t+s)\theta} d\theta d(t,s) $$

This holds for every $\rho \in (0, 1)$. We set $\rho = e^{-\pi}$ and we have

$$ R_B(N) = \frac{1}{2\pi} \int_{[1-\frac{N}{2}, \infty)^2} B(t+\frac{N}{2})B(s+\frac{N}{2})e^{-\pi(t+s)} \int_{-\pi}^{\pi} (-\pi + i\theta)^2 e^{i(t+s)\theta} d\theta d(t,s) $$

Now, we have

$$ \begin{aligned} \mathrm{Re}(\int_{-\pi}^{\pi} (-\pi + i\theta)^2 e^{i(t+s)\theta} d\theta) &= \int_{-\pi}^{\pi} \mathrm{Re}((-\pi + i\theta)^2 e^{i(t+s)\theta}) d\theta \\ &= \int_{-\pi}^{\pi} (\pi^2 - \theta^2) \cos((t+s)\theta) + 2\pi\theta \sin((t+s)\theta) d\theta \end{aligned} $$

Therefore, we get

$$ R_B(N) = \frac{1}{2\pi} \int_{[1-\frac{N}{2}, \infty)^2} B(t+\frac{N}{2})B(s+\frac{N}{2})e^{-\pi(t+s)} \int_{-\pi}^{\pi} (\pi^2 - \theta^2) \cos((t+s)\theta) + 2\pi\theta \sin((t+s)\theta) d\theta d(t,s) $$

Also, we have

$$ \int_{-\pi}^{\pi} \pi^2 \cos((t+s)\theta) d\theta = \frac{\pi^2}{t+s} (\sin(\pi(t+s)) - \sin(-\pi(t+s))) = 2\pi^2 \frac{\sin(\pi(t+s))}{t+s} $$

By integration by parts we get

$$ \begin{aligned} \int_{-\pi}^{\pi} \theta^2 \cos((t+s)\theta) d\theta &= 2\pi^2 \frac{\sin((t+s)\pi)}{t+s} - \frac{2}{t+s} \int_{-\pi}^{\pi} \theta \sin((t+s)\theta) d\theta \\ &= 2\pi^2 \frac{\sin((t+s)\pi)}{t+s} + \frac{2}{t+s} \left(2\pi \frac{\cos(\pi(t+s))}{t+s} - \frac{1}{t+s} \int_{-\pi}^{\pi} \cos((t+s)\theta) d\theta \right) \\ &= 2\pi^2 \frac{\sin((t+s)\pi)}{t+s} + \frac{4\pi}{(t+s)^2} \cos(\pi(t+s)) - \frac{4}{(t+s)^3} \sin((t+s)\pi) \end{aligned} $$

Therefore, we have

$$ \int_{-\pi}^{\pi} (\pi^2 - \theta^2) \cos((t+s)\theta) d\theta = -\frac{4\pi}{(t+s)^2} \cos(\pi(t+s)) + \frac{4}{(t+s)^3} \sin((t+s)\pi) $$
$$= \frac{1}{2\pi} \int_{-\pi}^{\pi} (\log \rho + i\theta)^2 \int_{[1-\frac{N}{2}, \infty)^2} B(t+\frac{N}{2})B(s+\frac{N}{2})\rho^{t+s} e^{i(t+s)\theta} d(t,s)d\theta$$

Now, since the function $(\log \rho + i\theta)^2 B(t + \frac{N}{2})B(s + \frac{N}{2})\rho^{t+s} e^{i(t+s)\theta}$ is integrable for $(t, s, \theta) \in [1 - \frac{N}{2}, \infty)^2 \times [-\pi, \pi]$, by Fubini Theorem we have

$$R_B(N) = \frac{1}{2\pi} \int_{[1-\frac{N}{2}, \infty)^2} B(t+\frac{N}{2})B(s+\frac{N}{2})\rho^{t+s} \int_{-\pi}^{\pi} (\log \rho + i\theta)^2 e^{i(t+s)\theta} d\theta d(t,s)$$

This holds for every $\rho \in (0, 1)$. We set $\rho = e^{-\pi}$ and we have

$$R_B(N) = \frac{1}{2\pi} \int_{[1-\frac{N}{2}, \infty)^2} B(t+\frac{N}{2})B(s+\frac{N}{2})e^{-\pi(t+s)} \int_{-\pi}^{\pi} (-\pi + i\theta)^2 e^{i(t+s)\theta} d\theta d(t,s)$$

Now, we have

$$\begin{aligned} \operatorname{Re}(\int_{-\pi}^{\pi} (-\pi + i\theta)^2 e^{i(t+s)\theta} d\theta) &= \int_{-\pi}^{\pi} \operatorname{Re}((-\pi + i\theta)^2 e^{i(t+s)\theta}) d\theta \\ &= \int_{-\pi}^{\pi} (\pi^2 - \theta^2) \cos((t+s)\theta) + 2\pi\theta \sin((t+s)\theta) d\theta \end{aligned}$$

Therefore, we get

$$R_B(N) = \frac{1}{2\pi} \int_{[1-\frac{N}{2}, \infty)^2} B(t+\frac{N}{2})B(s+\frac{N}{2})e^{-\pi(t+s)} \int_{-\pi}^{\pi} (\pi^2 - \theta^2) \cos((t+s)\theta) + 2\pi\theta \sin((t+s)\theta) d\theta d(t,s)$$

Also, we have

$$\int_{-\pi}^{\pi} \pi^2 \cos((t+s)\theta) d\theta = \frac{\pi^2}{t+s} (\sin(\pi(t+s)) - \sin(-\pi(t+s))) = 2\pi^2 \frac{\sin(\pi(t+s))}{t+s}$$

By integration by parts we get

$$\begin{aligned} \int_{-\pi}^{\pi} \theta^2 \cos((t+s)\theta) d\theta &= 2\pi^2 \frac{\sin((t+s)\pi)}{t+s} - \frac{2}{t+s} \int_{-\pi}^{\pi} \theta \sin((t+s)\theta) d\theta \\ &= 2\pi^2 \frac{\sin((t+s)\pi)}{t+s} + \frac{2}{t+s} \left(2\pi \frac{\cos(\pi(t+s))}{t+s} - \frac{1}{t+s} \int_{-\pi}^{\pi} \cos((t+s)\theta) d\theta \right) \\ &= 2\pi^2 \frac{\sin((t+s)\pi)}{t+s} + \frac{4\pi}{(t+s)^2} \cos(\pi(t+s)) - \frac{4}{(t+s)^3} \sin((t+s)\pi) \end{aligned}$$

Therefore, we have

$$\int_{-\pi}^{\pi} (\pi^2 - \theta^2) \cos((t+s)\theta) d\theta = -\frac{4\pi}{(t+s)^2} \cos(\pi(t+s)) + \frac{4}{(t+s)^3} \sin((t+s)\pi)$$
Also, easily we can see that

$$ \int_{-\pi}^{\pi} \theta \sin((t+s)\theta) d\theta = \frac{2}{(t+s)^2} \sin((t+s)\pi) - 2\pi \frac{\cos((t+s)\pi)}{t+s} $$

Therefore, we see that

$$ R_B(N) = \frac{1}{2\pi} \int_{[1-\frac{N}{2}, \infty)^2} B(t+\frac{N}{2})B(s+\frac{N}{2})e^{-\pi(t+s)}\Lambda(t+s)d(t,s) $$

By integration on level sets(which can be found in [5]), we can see that

$$ R_B(N) = \frac{1}{2\sqrt{2\pi}} \int_{2-N}^{\infty} \int_{t+s=\lambda, t,s \ge 1-\frac{N}{2}} B(t+\frac{N}{2})B(s+\frac{N}{2})dH^1(t,s)e^{-\pi\lambda}\Lambda(\lambda)d\lambda $$

This completes the proof. □

Let

$$ B^*(\lambda, N) = \int_{t+s=\lambda, t,s \ge 1-\frac{N}{2}} B(t+\frac{N}{2})B(s+\frac{N}{2})dH^1(t,s) $$

We have

$$ B^*(\lambda, N) = \sqrt{2} \int_{1}^{\lambda+N-1} B(t)B(\lambda - t + N)dt = B^*(\lambda + N) $$

for the function $B^*(x) = \int_{1}^{x-1} B(t)B(x-t)dt$. We continue with the following Lemma.

**Lemma 2.3.** Assume that $c_1\sqrt{x} \le B(x) \le c_2\sqrt{x}$ for $x \ge M_0$, where $M_0$ is the least element of the set B. Then, there exist $w_1, w_2 > 0$ such that

$$ w_1x^2 \le B^*(x) \le w_2x^2 $$

for $x > 2M_0$.

*Proof.* We have

$$ \begin{aligned} \int_{M_0}^{x-M_0} \sqrt{t\sqrt{x-t}} dt &= \int_{M_0}^{x-M_0} \sqrt{\frac{x^2}{4} - \left(t - \frac{x}{2}\right)^2} dt = \\ &c^2 \frac{x}{2} \int_{M_0}^{x-M_0} \sqrt{1 - \left(\frac{2t}{x} - 1\right)^2} dt = c^2 \frac{x^2}{4} \int_{\frac{2M_0}{x}-1}^{1-\frac{2M_0}{x}} \sqrt{1-t^2} dt = \\ &\frac{c^2 x^2}{8} \left( \left(1-\frac{2M_0}{x}\right) \sqrt{\frac{4M_0}{x} - \frac{4M_0}{x^2}} + \arcsin\left(1-\frac{2M_0}{x}\right) \right) - c^2 \frac{x^2}{8} \left( \left(\frac{2M_0}{x}-1\right) \sqrt{\frac{4M_0}{x} - \frac{4M_0}{x^2}} + \arcsin\left(\frac{2M_0}{x}-1\right) \right) = \\ &c^2 \frac{x^2}{4} \arcsin\left(1-\frac{2M_0}{x}\right) + O\left(x^{\frac{3}{2}}\right) \end{aligned} $$
Also, easily we can see that

$$ \int_{-\pi}^{\pi} \theta \sin((t+s)\theta) d\theta = \frac{2}{(t+s)^2} \sin((t+s)\pi) - 2\pi \frac{\cos((t+s)\pi)}{t+s} $$

Therefore, we see that

$$ R_B(N) = \frac{1}{2\pi} \int_{[1-\frac{N}{2}, \infty)^2} B(t+\frac{N}{2})B(s+\frac{N}{2})e^{-\pi(t+s)}\Lambda(t+s)d(t,s) $$

By integration on level sets(which can be found in [5]), we can see that

$$ R_B(N) = \frac{1}{2\sqrt{2\pi}} \int_{2-N}^{\infty} \int_{t+s=\lambda, t,s \ge 1-\frac{N}{2}} B(t+\frac{N}{2})B(s+\frac{N}{2})dH^1(t,s)e^{-\pi\lambda}\Lambda(\lambda)d\lambda $$

This completes the proof.

<figure><img src="image_3.png" /></figure>

Let

$$ B^*(\lambda, N) = \int_{t+s=\lambda, t,s \ge 1-\frac{N}{2}} B(t+\frac{N}{2})B(s+\frac{N}{2})dH^1(t,s) $$

We have

$$ B^*(\lambda, N) = \sqrt{2} \int_{1}^{\lambda+N-1} B(t)B(\lambda - t + N)dt = B^*(\lambda + N) $$

for the function $B^*(x) = \int_{1}^{x-1} B(t)B(x-t)dt$. We continue with the following Lemma.

**Lemma 2.3.** Assume that $c_1\sqrt{x} \le B(x) \le c_2\sqrt{x}$ for $x \ge M_0$, where $M_0$ is the least element of the set B. Then, there exist $w_1, w_2 > 0$ such that

$$ w_1x^2 \le B^*(x) \le w_2x^2 $$

for $x > 2M_0$.

*Proof.* We have

$$ \begin{aligned} \int_{M_0}^{x-M_0} \sqrt{t\sqrt{x-t}} dt &= \int_{M_0}^{x-M_0} \sqrt{\frac{x^2}{4} - (t-\frac{x}{2})^2} dt = \\ c^2 \frac{x}{2} \int_{M_0}^{x-M_0} \sqrt{1-(\frac{2t}{x}-1)^2} dt &= c^2 \frac{x^2}{4} \int_{\frac{2M_0}{x}-1}^{1-\frac{2M_0}{x}} \sqrt{1-t^2} dt = \\ \frac{c^2 x^2}{8} &((1-\frac{2M_0}{x})\sqrt{\frac{4M_0}{x} - \frac{4M_0}{x^2}}) + \arcsin(1-\frac{2M_0}{x}) - c^2 \frac{x^2}{8}((\frac{2M_0}{x}-1)\sqrt{\frac{4M_0}{x} - \frac{4M_0}{x^2}}) + \arcsin(\frac{2M_0}{x}-1) = \\ c^2 \frac{x^2}{4} \arcsin(1-\frac{2M_0}{x}) + O(x^{\frac{3}{2}}) \end{aligned} $$
By assumption we can see easily that there exist $w_1, w_2 > 0$ such that

$$w_1x^2 \leq B^*(x) \leq w_2x^2$$

for $x > 2M_0$. This completes the proof.

<figure><img src="image_3.png" /></figure>

### 3. RESULTS TOWARDS THE FINAL PROOF

We have

$$R_B(N) = \frac{1}{2\pi} \int_{-\infty}^{\infty} B^*(\lambda + N) e^{-\pi\lambda} \Lambda(\lambda) d\lambda$$

By changing variables, we get

$$R_B(N) = \frac{1}{2\pi} \int_{-\infty}^{\infty} B^*(\lambda) e^{-\pi(\lambda-N)} \Lambda(\lambda-N) d\lambda = e^{\pi N} \frac{1}{2\pi} \int_{0}^{\infty} B^*(\lambda) e^{-\pi\lambda} \Lambda(\lambda - N) d\lambda$$

$$\Lambda(\lambda - N) = (-1)^N \left( -\frac{4\pi^2 \cos(\lambda\pi)}{\lambda - N} - \frac{4\pi(\cos(\pi\lambda) - \sin(\pi\lambda))}{(\lambda - N)^2} + \frac{4\sin(\pi\lambda)}{(\lambda - N)^3} \right)$$

Let

$$\Lambda^*(\lambda, N) = -\frac{4\pi^2 \cos(\lambda\pi)}{\lambda - N} - \frac{4\pi(\cos(\pi\lambda) - \sin(\pi\lambda))}{(\lambda - N)^2} + \frac{4\sin(\pi\lambda)}{(\lambda - N)^3}$$

Then we get,

$$R_B(N) = \frac{(-1)^N}{2\pi} \int_0^\infty B^*(\lambda) e^{-\pi(\lambda-N)} \Lambda^*(\lambda, N) d\lambda$$

Therefore, if we denote by

$$W(N) = \frac{1}{2\pi} \int_{0}^{\infty} B^{*}(\lambda) e^{-\pi(\lambda-N)} \Lambda^{*}(\lambda, N) d\lambda$$

we have $R_B(N) = (-1)^N W(N)$. Since, $R_B(N) \ge 0$ for every $N$, we can see that $W(N) \ge 0$ if $N$ is even and $W(N) \le 0$ if $N$ is odd.

**Proposition 3.1.** There exists a constant $C > 0$ such that for every natural number $N$, we have

$$W(N) = \frac{1}{2\pi} \int_{(\lambda-N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-N)} |\Lambda^*(\lambda, N)| d\lambda - \frac{1}{2\pi} \int_{(\lambda-N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-N)} |\Lambda^*(\lambda, N)| d\lambda \\ + O\left( \int_{|\lambda-N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-N)} \min\left(1, \frac{1}{(\lambda-N)^2}\right) d\lambda \right)$$
By assumption we can see easily that there exist $w_1, w_2 > 0$ such that

$$w_1x^2 \leq B^*(x) \leq w_2x^2$$

for $x > 2M_0$. This completes the proof. □

### 3. RESULTS TOWARDS THE FINAL PROOF

We have

$$R_B(N) = \frac{1}{2\pi} \int_{-\infty}^{\infty} B^*(\lambda + N)e^{-\pi\lambda}\Lambda(\lambda)d\lambda$$

By changing variables, we get

$$R_B(N) = \frac{1}{2\pi} \int_{-\infty}^{\infty} B^*(\lambda)e^{-\pi(\lambda-N)}\Lambda(\lambda-N)d\lambda = e^{\pi N} \frac{1}{2\pi} \int_{0}^{\infty} B^*(\lambda)e^{-\pi\lambda}\Lambda(\lambda-N)d\lambda \\ \Lambda(\lambda - N) = (-1)^N \left( -\frac{4\pi^2 \cos(\lambda\pi)}{\lambda - N} - \frac{4\pi(\cos(\pi\lambda) - \sin(\pi\lambda))}{(\lambda - N)^2} + \frac{4\sin(\pi\lambda)}{(\lambda - N)^3} \right)$$

Let

$$\Lambda^*(\lambda, N) = -\frac{4\pi^2 \cos(\lambda\pi)}{\lambda - N} - \frac{4\pi(\cos(\pi\lambda) - \sin(\pi\lambda))}{(\lambda - N)^2} + \frac{4\sin(\pi\lambda)}{(\lambda - N)^3}$$

Then we get,

$$R_B(N) = \frac{(-1)^N}{2\pi} \int_0^\infty B^*(\lambda)e^{-\pi(\lambda-N)}\Lambda^*(\lambda, N)d\lambda$$

Therefore, if we denote by

$$W(N) = \frac{1}{2\pi} \int_{0}^{\infty} B^{*}(\lambda)e^{-\pi(\lambda-N)}\Lambda^{*}(\lambda, N)d\lambda$$

we have $R_B(N) = (-1)^N W(N)$. Since, $R_B(N) \ge 0$ for every $N$, we can see that $W(N) \ge 0$ if $N$ is even and $W(N) \le 0$ if $N$ is odd.

**Proposition 3.1.** There exists a constant $C > 0$ such that for every natural number $N$, we have

$$W(N) = \frac{1}{2\pi} \int_{(\lambda-N)\cos(\pi\lambda) < 0} B^*(\lambda)e^{-\pi(\lambda-N)}|\Lambda^*(\lambda, N)|d\lambda - \frac{1}{2\pi} \int_{(\lambda-N)\cos(\pi\lambda) > 0} B^*(\lambda)e^{-\pi(\lambda-N)}|\Lambda^*(\lambda, N)|d\lambda \\ +O\left(\int_{|\lambda-N|\cos(\pi\lambda)| \le C} B^*(\lambda)e^{-\pi(\lambda-N)}\min\left(1, \frac{1}{(\lambda-N)^2}\right)d\lambda\right)$$
Proof. We have,

$$
\begin{align*}
W(N) &= \frac{1}{2\pi} \int_0^\infty B^*(\lambda) e^{-\pi(\lambda-N)} \Lambda^*(\lambda, N) d\lambda \\
&= \frac{1}{2\pi} \int_{(\lambda-N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-N)} \Lambda^*(\lambda, N) d\lambda + \frac{1}{2\pi} \int_{(\lambda-N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-N)} \Lambda^*(\lambda, N) d\lambda \\
&= \frac{1}{2\pi} \int_{(\lambda-N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-N)} |\Lambda^*(\lambda, N)| d\lambda - \frac{1}{2\pi} \int_{(\lambda-N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-N)} |\Lambda^*(\lambda, N)| d\lambda \\
&\quad + \frac{1}{\pi} \int_{(\lambda-N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-N)} \max(-\Lambda^*(\lambda, N), 0) d\lambda \\
&\quad + \frac{1}{\pi} \int_{(\lambda-N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-N)} \max(\Lambda^*(\lambda, N), 0) d\lambda
\end{align*}
$$

We have

$$
\Lambda^*(\lambda, N) = - \frac{4\pi^2 \cos(\lambda\pi)}{\lambda - N} - \frac{4\pi(\cos(\lambda\pi) - \sin(\lambda\pi))}{(\lambda - N)^2} + \frac{4\sin(\lambda\pi)}{(\lambda - N)^3}
$$

Assume that $\Lambda^*(\lambda, N) \le 0$ and $(\lambda - N) \cos(\pi\lambda) < 0$. Then, we get

$$
|\lambda - N|4\pi^2|\cos(\pi\lambda)| \le 4\pi(\cos(\pi\lambda) - \sin(\pi\lambda)) - \frac{4\sin(\lambda\pi)}{\lambda - N}
$$

Similarly, if $\Lambda^*(\lambda, N) \ge 0$ and $(\lambda - N) \cos(\pi\lambda) > 0$, we get that

$$
|\lambda - N|4\pi^2|\cos(\pi\lambda)| \le -4\pi(\cos(\pi\lambda) - \sin(\pi\lambda)) + \frac{4\sin(\lambda\pi)}{\lambda - N}
$$

Since, the function $4\pi(\cos(\pi\lambda) - \sin(\pi\lambda))$ is bounded and the function $\frac{4\sin(\lambda\pi)}{\lambda}$ is also bounded and using the fact that

$$
\frac{4 \sin(\lambda \pi)}{\lambda - N} = (-1)^N \frac{4 \sin((\lambda - N) \pi)}{\lambda - N}
$$

, we have that there exists a constant $C > 0$ independent of $N$, such that in both cases

$$
|\lambda - N| |\cos(\pi\lambda)| \leq C
$$

Therefore, we have

$$
\int_{(\lambda-N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-N)} \max(-\Lambda^*(\lambda, N), 0) d\lambda = O\left( \int_{|\lambda-N| |\cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-N)} |\Lambda^*(\lambda, N)| d\lambda \right)
$$

and

$$
\int_{(\lambda-N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-N)} \max(\Lambda^*(\lambda, N), 0) d\lambda = O\left( \int_{|\lambda-N| |\cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-N)} |\Lambda^*(\lambda, N)| d\lambda \right)
$$
Proof. We have,

$$
\begin{align*}
W(N) &= \frac{1}{2\pi} \int_0^\infty B^*(\lambda) e^{-\pi(\lambda-N)} \Lambda^*(\lambda, N) d\lambda \\
&= \frac{1}{2\pi} \int_{(\lambda-N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-N)} \Lambda^*(\lambda, N) d\lambda + \frac{1}{2\pi} \int_{(\lambda-N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-N)} \Lambda^*(\lambda, N) d\lambda \\
&= \frac{1}{2\pi} \int_{(\lambda-N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-N)} |\Lambda^*(\lambda, N)| d\lambda - \frac{1}{2\pi} \int_{(\lambda-N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-N)} |\Lambda^*(\lambda, N)| d\lambda \\
&\quad + \frac{1}{\pi} \int_{(\lambda-N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-N)} \max(-\Lambda^*(\lambda, N), 0) d\lambda \\
&\quad + \frac{1}{\pi} \int_{(\lambda-N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-N)} \max(\Lambda^*(\lambda, N), 0) d\lambda
\end{align*}
$$

We have

$$
\Lambda^*(\lambda, N) = - \frac{4\pi^2 \cos(\lambda\pi)}{\lambda - N} - \frac{4\pi(\cos(\lambda\pi) - \sin(\lambda\pi))}{(\lambda - N)^2} + \frac{4\sin(\lambda\pi)}{(\lambda - N)^3}
$$

Assume that $\Lambda^*(\lambda, N) \le 0$ and $(\lambda - N) \cos(\pi\lambda) < 0$. Then, we get

$$
|\lambda - N|4\pi^2|\cos(\pi\lambda)| \le 4\pi(\cos(\pi\lambda) - \sin(\pi\lambda)) - \frac{4\sin(\lambda\pi)}{\lambda - N}
$$

Similarly, if $\Lambda^*(\lambda, N) \ge 0$ and $(\lambda - N) \cos(\pi\lambda) > 0$, we get that

$$
|\lambda - N|4\pi^2|\cos(\pi\lambda)| \le -4\pi(\cos(\pi\lambda) - \sin(\pi\lambda)) + \frac{4\sin(\lambda\pi)}{\lambda - N}
$$

Since, the function $4\pi(\cos(\pi\lambda) - \sin(\pi\lambda))$ is bounded and the function $\frac{4\sin(\lambda\pi)}{\lambda}$ is also bounded and using the fact that

$$
\frac{4 \sin(\lambda \pi)}{\lambda - N} = (-1)^N \frac{4 \sin((\lambda - N) \pi)}{\lambda - N}
$$

, we have that there exists a constant $C > 0$ independent of $N$, such that in both cases

$$
|\lambda - N| |\cos(\pi\lambda)| \le C
$$

Therefore, we have

$$
\int_{(\lambda-N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-N)} \max(-\Lambda^*(\lambda, N), 0) d\lambda = O\left( \int_{|\lambda-N| \cos(\pi\lambda) \le C} B^*(\lambda) e^{-\pi(\lambda-N)} |\Lambda^*(\lambda, N)| d\lambda \right)
$$

and

$$
\int_{(\lambda-N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-N)} \max(\Lambda^*(\lambda, N), 0) d\lambda = O\left( \int_{|\lambda-N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-N)} |\Lambda^*(\lambda, N)| d\lambda \right)
$$
Now, since $\lim_{\lambda \to 0} \Lambda(\lambda)$ exists and it is a real number and also $\lim_{|\lambda| \to +\infty} \Lambda(\lambda) = 0$, we get by continuity, that the function $\Lambda(\lambda)$ is bounded.

Now, using the fact that $\Lambda^*(\lambda, N) = (-1)^N \Lambda(\lambda - N)$, we have that

$$|\Lambda^*(\lambda, N)| = O(1)$$

Also, if we have that $|\lambda - N| |\cos(\pi\lambda)| \le C$, we get that

$$|\Lambda^*(\lambda, N)| = O\left(\frac{1}{(\lambda - N)^2} + \frac{1}{|\lambda - N|^3}\right)$$

Therefore, we get that for $\lambda$ such that $|\lambda - N| |\cos(\pi\lambda)| \le C$, we have

$$|\Lambda^*(\lambda, N)| = O(\min(1, \frac{1}{(\lambda - N)^2}))$$

together with an absolute implied constant.

Therefore, we get

$$\int_{(\lambda-N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-N)} \max(-\Lambda^*(\lambda, N), 0) d\lambda = O\left(\int_{|\lambda-N| |\cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-N)} \min\left(1, \frac{1}{(\lambda-N)^2}\right) d\lambda\right)$$

and

$$\int_{(\lambda-N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-N)} \max(\Lambda^*(\lambda, N), 0) d\lambda = O\left(\int_{|\lambda-N| |\cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-N)} \min\left(1, \frac{1}{(\lambda-N)^2}\right) d\lambda\right)$$

This completes the proof. □

We continue with the following Proposition.

**Proposition 3.2.** We have,

$$\frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} \Lambda^*(\lambda, 2N+1) d\lambda - \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} \Lambda^*(\lambda, 2N) d\lambda = -\frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)| d\lambda +$$
Now, since $\lim_{\lambda \to 0} \Lambda(\lambda)$ exists and it is a real number and also $\lim_{|\lambda| \to +\infty} \Lambda(\lambda) = 0$, we get by continuity, that the function $\Lambda(\lambda)$ is bounded.

Now, using the fact that $\Lambda^*(\lambda, N) = (-1)^N \Lambda(\lambda - N)$, we have that

$$|\Lambda^*(\lambda, N)| = O(1)$$

Also, if we have that $|\lambda - N| |\cos(\pi\lambda)| \le C$, we get that

$$|\Lambda^*(\lambda, N)| = O\left(\frac{1}{(\lambda - N)^2} + \frac{1}{|\lambda - N|^3}\right)$$

Therefore, we get that for $\lambda$ such that $|\lambda - N| |\cos(\pi\lambda)| \le C$, we have

$$|\Lambda^*(\lambda, N)| = O(\min(1, \frac{1}{(\lambda - N)^2}))$$

together with an absolute implied constant.

Therefore, we get

$$\int_{(\lambda-N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-N)} \max(-\Lambda^*(\lambda, N), 0) d\lambda = O\left(\int_{|\lambda-N| |\cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-N)} \min\left(1, \frac{1}{(\lambda-N)^2}\right) d\lambda\right)$$

and

$$\int_{(\lambda-N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-N)} \max(\Lambda^*(\lambda, N), 0) d\lambda = O\left(\int_{|\lambda-N| |\cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-N)} \min\left(1, \frac{1}{(\lambda-N)^2}\right) d\lambda\right)$$

This completes the proof. □

We continue with the following Proposition.

**Proposition 3.2.** We have,

$$\frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} \Lambda^*(\lambda, 2N+1) d\lambda - \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} \Lambda^*(\lambda, 2N) d\lambda = -\frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)| d\lambda +$$
$$O\left( \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda)e^{-\pi(\lambda-2N)}\min\left(1, \frac{1}{(\lambda-2N)^2}d\lambda\right) + O\left(\int_{2N-1}^{\infty} B^*(\lambda)e^{-\pi(\lambda-2N)}d\lambda\right)$$

*Proof.* By Proposition 3.1, we get

$$\begin{aligned} & \frac{1}{2\pi} \int_{(\lambda-2N-1)\cos(\pi\lambda)<0} B^*(\lambda)e^{-\pi(\lambda-2N)}\Lambda^*(\lambda, 2N+1)d\lambda - \\ & \qquad \frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0} B^*(\lambda)e^{-\pi(\lambda-2N)}\Lambda^*(\lambda, 2N)d\lambda = \\ & \frac{1}{2\pi} \int_{(\lambda-2N-1)\cos(\pi\lambda)<0} B^*(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^*(\lambda, 2N+1)|d\lambda - \\ & \qquad \frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0} B^*(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^*(\lambda, 2N)|d\lambda + \\ & +O\left(\int_{|\lambda-2N-1||\cos(\pi\lambda)|\le C} B^*(\lambda)e^{-\pi(\lambda-2N)}\min\left(1, \frac{1}{(\lambda-2N-1)^2}d\lambda\right)\right) + \\ & O\left(\int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda)e^{-\pi(\lambda-2N)}\min\left(1, \frac{1}{(\lambda-2N)^2}d\lambda\right)\right) \end{aligned}$$

We have,

$$\begin{aligned} & \frac{1}{2\pi} \int_{(\lambda-2N-1)\cos(\pi\lambda)<0} B^*(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^*(\lambda, 2N+1)|d\lambda - \\ & \qquad \frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0} B^*(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^*(\lambda, 2N)|d\lambda = \\ & \frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0, \cos(\pi\lambda)>0} B^*(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|d\lambda + \\ & O\left(\int_{2N}^{\infty} B^*(\lambda)e^{-\pi(\lambda-2N)}d\lambda\right) \end{aligned}$$

This is true because, if $\cos(\pi\lambda) < 0$, then forces $\lambda > 2N$ in both of the terms and also if $(\lambda - 2N) \cos(\pi\lambda) \ge 0$ in the first term and $\cos(\pi\lambda) > 0$, then surely $\lambda \ge 2N$.

Now,

$$\begin{aligned} & \frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0, \cos(\pi\lambda)>0} B^*(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|d\lambda = \\ & -\frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0, \cos(\pi\lambda)>0} B^*(\lambda)e^{-\pi(\lambda-2N)}||\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)||d\lambda+ \end{aligned}$$
$$O\left(\int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda)e^{-\pi(\lambda-2N)}\min\left(1, \frac{1}{(\lambda-2N)^2}d\lambda\right) + O\left(\int_{2N-1}^{\infty} B^*(\lambda)e^{-\pi(\lambda-2N)}d\lambda\right)$$

*Proof.* By Proposition 3.1, we get

$$\begin{aligned} & \frac{1}{2\pi} \int_{(\lambda-2N-1)\cos(\pi\lambda)<0} B^*(\lambda)e^{-\pi(\lambda-2N)}\Lambda^*(\lambda, 2N+1)d\lambda - \\ & \qquad \frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0} B^*(\lambda)e^{-\pi(\lambda-2N)}\Lambda^*(\lambda, 2N)d\lambda = \\ & \frac{1}{2\pi} \int_{(\lambda-2N-1)\cos(\pi\lambda)<0} B^*(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^*(\lambda, 2N+1)|d\lambda - \\ & \qquad \frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0} B^*(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^*(\lambda, 2N)|d\lambda + \\ & +O\left(\int_{|\lambda-2N-1||\cos(\pi\lambda)|\le C} B^*(\lambda)e^{-\pi(\lambda-2N)}\min\left(1, \frac{1}{(\lambda-2N-1)^2}d\lambda\right)\right) + \\ & O\left(\int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda)e^{-\pi(\lambda-2N)}\min\left(1, \frac{1}{(\lambda-2N)^2}d\lambda\right)\right) \end{aligned}$$

We have,

$$\begin{aligned} & \frac{1}{2\pi} \int_{(\lambda-2N-1)\cos(\pi\lambda)<0} B^*(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^*(\lambda, 2N+1)|d\lambda - \\ & \qquad \frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0} B^*(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^*(\lambda, 2N)|d\lambda = \\ & \frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0, \cos(\pi\lambda)>0} B^*(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|d\lambda + \\ & O\left(\int_{2N}^{\infty} B^*(\lambda)e^{-\pi(\lambda-2N)}d\lambda\right) \end{aligned}$$

This is true because, if $\cos(\pi\lambda) < 0$, then forces $\lambda > 2N$ in both of the terms and also if $(\lambda - 2N) \cos(\pi\lambda) \ge 0$ in the first term and $\cos(\pi\lambda) > 0$, then surely $\lambda \ge 2N$.

Now,

$$\begin{aligned} & \frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0, \cos(\pi\lambda)>0} B^*(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|d\lambda = \\ & -\frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0, \cos(\pi\lambda)>0} B^*(\lambda)e^{-\pi(\lambda-2N)}||\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)||d\lambda+ \end{aligned}$$
$$
\begin{align*}
& \frac{1}{\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} \max(|\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|, 0) d\lambda = \\
& \quad -\frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|| d\lambda + \\
& \frac{1}{\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0, |\Lambda(\lambda, 2N+1)| \ge |\Lambda(\lambda, 2N)|} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|| d\lambda = \\
& \quad -\frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|| d\lambda + \\
& \frac{1}{\pi} \int_{\lambda < 2N-1, \cos(\pi\lambda) > 0, \Lambda(\lambda, 2N-1) \ge \Lambda(\lambda, 2N) \ge 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1) - \Lambda^*(\lambda, 2N)| d\lambda + \\
& \quad O\left(\int_{|\lambda-2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda\right) \\
& \quad + O\left(\int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda\right)
\end{align*}
$$

But if we have $\lambda < 2N - 1$ and $\cos(\pi\lambda) > 0$ and $\Lambda(\lambda, 2N + 1) \ge \Lambda(\lambda, 2N)$, then

$$
\begin{aligned}
& -\frac{4\pi^2 \cos(\lambda\pi)}{\lambda - 2N - 1} - \frac{4\pi(\cos(\pi\lambda) - \sin(\pi\lambda))}{(\lambda - 2N - 1)^2} + \frac{4\sin(\pi\lambda)}{(\lambda - 2N - 1)^3} \ge \\
& \quad -\frac{4\pi^2 \cos(\lambda\pi)}{\lambda - 2N} - \frac{4\pi(\cos(\pi\lambda) - \sin(\pi\lambda))}{(\lambda - 2N)^2} + \frac{4\sin(\pi\lambda)}{(\lambda - 2N)^3}
\end{aligned}
$$

Therefore,

$$
\begin{aligned}
& -4\pi |\cos(\lambda\pi)| \left( \frac{\pi}{\lambda - 2N - 1} - \frac{\pi}{\lambda - 2N} + \frac{1}{(\lambda - 2N - 1)^2} - \frac{1}{(\lambda - 2N)^2} \right) \ge \\
& -4\pi \sin(\lambda\pi) \left( \frac{1}{(\lambda - 2N - 1)^2} - \frac{1}{(\lambda - 2N)^2} + \frac{1}{\pi(\lambda - 2N - 1)^3} - \frac{1}{(\lambda - 2N)^3} \right)
\end{aligned}
$$

which, gives that

$$
\begin{aligned}
& |\cos(\lambda\pi)|\left(\left(\frac{\pi}{(\lambda - 2N - 1)(\lambda - 2N)} + \frac{2\lambda - 4N - 1}{(\lambda - 2N - 1)^2(\lambda - 2N)^2}\right) \le \\
& \sin(\lambda\pi)\left(\frac{2\lambda - 4N - 1}{(\lambda - 2N - 1)^2(\lambda - 2N)^2} + \frac{3(\lambda - 2N - 1)^2 + 3(\lambda - 2N - 1) + 1}{\pi(\lambda - 2N - 1)^3(\lambda - 2N)^3}\right)
\end{aligned}
$$

Therefore, we get

$$
|\cos(\lambda\pi)||\lambda - 2N - 1||\lambda - 2N| \le C_1(2\lambda - 4N - 1) + C_2 \frac{3(\lambda - 2N - 1)^2 + 3(\lambda - 2N - 1) + 1}{(\lambda - 2N - 1)(\lambda - 2N)}
$$
$$
\begin{align*}
& \frac{1}{\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} \max(|\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|, 0) d\lambda = \\
& \quad -\frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|| d\lambda + \\
& \frac{1}{\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0, |\Lambda(\lambda, 2N+1)| \ge |\Lambda(\lambda, 2N)|} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|| d\lambda = \\
& \quad -\frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|| d\lambda + \\
& \frac{1}{\pi} \int_{\lambda < 2N-1, \cos(\pi\lambda) > 0, \Lambda(\lambda, 2N-1) \ge \Lambda(\lambda, 2N) \ge 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1) - \Lambda^*(\lambda, 2N)| d\lambda + \\
& \quad O\left( \int_{|\lambda-2N| \cos(\pi\lambda) \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \\
& \quad + O\left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)
\end{align*}
$$

But if we have $\lambda < 2N - 1$ and $\cos(\pi\lambda) > 0$ and $\Lambda(\lambda, 2N + 1) \ge \Lambda(\lambda, 2N)$, then

$$
\begin{aligned}
& -\frac{4\pi^2 \cos(\lambda\pi)}{\lambda - 2N - 1} - \frac{4\pi(\cos(\pi\lambda) - \sin(\pi\lambda))}{(\lambda - 2N - 1)^2} + \frac{4\sin(\pi\lambda)}{(\lambda - 2N - 1)^3} \ge \\
& \quad -\frac{4\pi^2 \cos(\lambda\pi)}{\lambda - 2N} - \frac{4\pi(\cos(\pi\lambda) - \sin(\pi\lambda))}{(\lambda - 2N)^2} + \frac{4\sin(\pi\lambda)}{(\lambda - 2N)^3}
\end{aligned}
$$

Therefore,

$$
\begin{aligned}
& -4\pi |\cos(\lambda\pi)| \left( \frac{\pi}{\lambda - 2N - 1} - \frac{\pi}{\lambda - 2N} + \frac{1}{(\lambda - 2N - 1)^2} - \frac{1}{(\lambda - 2N)^2} \right) \ge \\
& -4\pi \sin(\lambda\pi) \left( \frac{1}{(\lambda - 2N - 1)^2} - \frac{1}{(\lambda - 2N)^2} + \frac{1}{\pi(\lambda - 2N - 1)^3} - \frac{1}{(\lambda - 2N)^3} \right)
\end{aligned}
$$

which, gives that

$$
\begin{aligned}
& |\cos(\lambda\pi)|\left(\frac{\pi}{(\lambda - 2N - 1)(\lambda - 2N)} + \frac{2\lambda - 4N - 1}{(\lambda - 2N - 1)^2(\lambda - 2N)^2}\right) \le \\
& \sin(\lambda\pi)\left(\frac{2\lambda - 4N - 1}{(\lambda - 2N - 1)^2(\lambda - 2N)^2} + \frac{3(\lambda - 2N - 1)^2 + 3(\lambda - 2N - 1) + 1}{\pi(\lambda - 2N - 1)^3(\lambda - 2N)^3}\right)
\end{aligned}
$$

Therefore, we get

$$
|\cos(\lambda\pi)||\lambda - 2N - 1||\lambda - 2N| \le C_1(2\lambda - 4N - 1) + C_2 \frac{3(\lambda - 2N - 1)^2 + 3(\lambda - 2N - 1) + 1}{(\lambda - 2N - 1)(\lambda - 2N)}
$$
, which gives that

$$|\cos(\lambda\pi)||\lambda - 2N| \le C_1 \frac{2\lambda - 4N - 1}{\lambda - 2N - 1} + C_2 \frac{3}{\lambda - 2N} + C_2 \frac{3}{(\lambda - 2N)(\lambda - 2N - 1)} + C_2 \frac{1}{(\lambda - 2N)(\lambda - 2N - 1)^2}$$

Using the fact that $\lambda < 2N - 1$, we can easily see that there exists $C > 0$ independent of $N$ and $\lambda$ such that

$$|\cos(\lambda\pi)||\lambda - 2N| \le C$$

Therefore, we can easily see that

$$\frac{1}{\pi} \int_{\lambda < 2N - 1, \cos(\pi\lambda) > 0, \Lambda(\lambda, 2N - 1) \ge \Lambda(\lambda, 2N) \ge 0} B^*(\lambda) e^{-\pi(\lambda - 2N)} |\Lambda^*(\lambda, 2N + 1) - \Lambda^*(\lambda, 2N)| d\lambda = \\ O\left( \int_{|\lambda - 2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda - 2N)} \min\left(1, \frac{1}{(\lambda - 2N)^2}\right) d\lambda \right) + O\left( \int_{2N - 1}^{\infty} B^*(\lambda) e^{-\pi(\lambda - 2N)} d\lambda \right)$$

This completes the proof. $\square$

**Proposition 3.3.** We have that there exists $C > 0$ such that for every natural number $N$

$$\frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|| d\lambda \le \\ W(2N) - e^{\pi}W(2N+1) + O\left( \int_{|\lambda-2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda + \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)$$

*Proof.* By Proposition 3.2, we get

$$-\frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|| d\lambda = \\ \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} \Lambda^*(\lambda, 2N+1) d\lambda - \\ \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} \Lambda^*(\lambda, 2N) d\lambda + \\ O\left( \int_{|\lambda-2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \\ + O\left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)$$

Also, we have

$$e^{\pi}W(2N + 1) - W(2N) = \\ \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda -$$
, which gives that

$$|\cos(\lambda\pi)||\lambda - 2N| \le C_1 \frac{2\lambda - 4N - 1}{\lambda - 2N - 1} + C_2 \frac{3}{\lambda - 2N} + C_2 \frac{3}{(\lambda - 2N)(\lambda - 2N - 1)} + C_2 \frac{1}{(\lambda - 2N)(\lambda - 2N - 1)^2}$$

Using the fact that $\lambda < 2N - 1$, we can easily see that there exists $C > 0$ independent of $N$ and $\lambda$ such that

$$|\cos(\lambda\pi)||\lambda - 2N| \le C$$

Therefore, we can easily see that

$$\frac{1}{\pi} \int_{\lambda < 2N - 1, \cos(\pi\lambda) > 0, \Lambda(\lambda, 2N - 1) \ge \Lambda(\lambda, 2N) \ge 0} B^*(\lambda) e^{-\pi(\lambda - 2N)} |\Lambda^*(\lambda, 2N + 1) - \Lambda^*(\lambda, 2N)| d\lambda = \\ O\left( \int_{|\lambda - 2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda - 2N)} \min\left(1, \frac{1}{(\lambda - 2N)^2}\right) d\lambda \right) + O\left( \int_{2N - 1}^{\infty} B^*(\lambda) e^{-\pi(\lambda - 2N)} d\lambda \right)$$

This completes the proof. $\square$

**Proposition 3.3.** We have that there exists $C > 0$ such that for every natural number $N$

$$\frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|| d\lambda \le \\ W(2N) - e^{\pi}W(2N+1) + O\left( \int_{|\lambda-2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda + \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)$$

*Proof.* By Proposition 3.2, we get

$$-\frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)|| d\lambda = \\ \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} \Lambda^*(\lambda, 2N+1) d\lambda - \\ \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} \Lambda^*(\lambda, 2N) d\lambda + \\ O\left( \int_{|\lambda-2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \\ + O\left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)$$

Also, we have

$$e^{\pi}W(2N + 1) - W(2N) = \\ \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda -$$
$$
\begin{aligned}
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda + \\
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda + \\
& O\left( \int_{|\lambda-2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) + \\
& O\left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)
\end{aligned}
$$

Therefore, we get

$$
\begin{aligned}
& -\frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} \left( |\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)| \right) d\lambda = \\
& e^{\pi W(2N+1)} - W(2N) + \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda + \\
& O\left( \int_{|\lambda-2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) + \\
& +O\left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)
\end{aligned}
$$

It suffices to show that

$$
\begin{aligned}
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda \ge \\
& 2e^{\pi W(2N+1)} - 2W(2N) \\
& +O\left( \int_{|\lambda-2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \\
& +O\left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)
\end{aligned}
$$

,which is true if and only if

$$
\frac{1}{\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda -
$$
$$
\begin{aligned}
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda + \\
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda + \\
& O\left( \int_{|\lambda-2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) + \\
& O\left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)
\end{aligned}
$$

Therefore, we get

$$
\begin{aligned}
& -\frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} \left( |\Lambda^*(\lambda, 2N+1)| - |\Lambda^*(\lambda, 2N)| \right) d\lambda = \\
& e^{\pi W(2N+1)} - W(2N) + \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda + \\
& O\left( \int_{|\lambda-2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) + \\
& +O\left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)
\end{aligned}
$$

It suffices to show that

$$
\begin{aligned}
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda \ge \\
& 2e^{\pi W(2N+1)} - 2W(2N) \\
& +O\left( \int_{|\lambda-2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \\
& +O\left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)
\end{aligned}
$$

,which is true if and only if

$$
\frac{1}{\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda -
$$
$$ 
\begin{aligned}
& \frac{1}{\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda + \\
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda \le \\
& O\left( \int_{|\lambda-2N| |\cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) + \\
& \qquad + O\left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)
\end{aligned}
$$

, which is true if and only if

$$ 
\begin{aligned}
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda \ge \\
& \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda + \\
& \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda + \\
& O\left( \int_{|\lambda-2N| |\cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) + \\
& \qquad + O\left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)
\end{aligned}
$$

This always holds, firstly because $W(2N) \ge 0$, therefore always

$$ 
\begin{aligned}
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda \ge \\
& O\left( \int_{|\lambda-2N| |\cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) +
\end{aligned}
$$
$$
\begin{aligned}
& \frac{1}{\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda + \\
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda \le \\
& O\left( \int_{|\lambda-2N| |\cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) + \\
& \qquad + O\left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)
\end{aligned}
$$

, which is true if and only if

$$
\begin{aligned}
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda \ge \\
& \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda + \\
& \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N-1) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N+1)| d\lambda + \\
& O\left( \int_{|\lambda-2N| |\cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) + \\
& \qquad + O\left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda \right)
\end{aligned}
$$

This always holds, firstly because $W(2N) \ge 0$, therefore always

$$
\begin{aligned}
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda - \\
& \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} |\Lambda^*(\lambda, 2N)| d\lambda \ge \\
& O\left( \int_{|\lambda-2N| |\cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) +
\end{aligned}
$$
$$+O\left(\int_{2N-1}^{\infty} B^{*}(\lambda)e^{-\pi(\lambda-2N)}d\lambda\right)$$

Secondly, we have $e^{\pi W(2N + 1)} \le 0$. Therefore

$$\frac{1}{2\pi} \int_{(\lambda-2N-1)\cos(\pi\lambda)<0} B^{*}(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^{*}(\lambda, 2N+1)|d\lambda-$$

$$\frac{1}{2\pi} \int_{(\lambda-2N-1)\cos(\pi\lambda)>0} B^{*}(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^{*}(\lambda, 2N+1)|d\lambda \le$$

$$O\left(\int_{|\lambda-2N|\cos(\pi\lambda)|\le C} B^{*}(\lambda)e^{-\pi(\lambda-2N)}\min\left(1, \frac{1}{(\lambda-2N)^2}\right)d\lambda\right) + \\ +O\left(\int_{2N-1}^{\infty} B^{*}(\lambda)e^{-\pi(\lambda-2N)}d\lambda\right)$$

and finally because as we showed in the previous Proposition we have

$$\frac{1}{2\pi} \int_{(\lambda-2N-1)\cos(\pi\lambda)<0} B^{*}(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^{*}(\lambda, 2N+1)|d\lambda-$$

$$\frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0} B^{*}(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^{*}(\lambda, 2N)|d\lambda \le$$

$$O\left(\int_{|\lambda-2N|\cos(\pi\lambda)|\le C} B^{*}(\lambda)e^{-\pi(\lambda-2N)}\min\left(1, \frac{1}{(\lambda-2N)^2}\right)d\lambda\right) + \\ +O\left(\int_{2N-1}^{\infty} B^{*}(\lambda)e^{-\pi(\lambda-2N)}d\lambda\right)$$

This completes the proof. □

### 4. PROOF OF THE ERDOS-TURAN CONJECTURE

In this section we will present a full proof of the Erdos-Turin Conjecture on Additive Bases.

This conjecture states that if for a set $B$ of natural numbers we have that $R_B(N) > 0$ for $N$ sufficiently large, then we must have that $\limsup_{N\to\infty} R_B(N) = \infty$.

From now on we will write

$$L(\lambda) = |\Lambda(\lambda - 1)| - |\Lambda(\lambda)|$$
$$+O\left(\int_{2N-1}^{\infty} B^{*}(\lambda)e^{-\pi(\lambda-2N)}d\lambda\right)$$

Secondly, we have $e^{\pi W(2N + 1)} \le 0$. Therefore

$$\frac{1}{2\pi} \int_{(\lambda-2N-1)\cos(\pi\lambda)<0} B^{*}(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^{*}(\lambda, 2N+1)|d\lambda-$$

$$\frac{1}{2\pi} \int_{(\lambda-2N-1)\cos(\pi\lambda)>0} B^{*}(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^{*}(\lambda, 2N+1)|d\lambda \le$$

$$O\left(\int_{|\lambda-2N|\cos(\pi\lambda)|\le C} B^{*}(\lambda)e^{-\pi(\lambda-2N)}\min\left(1, \frac{1}{(\lambda-2N)^2}\right)d\lambda\right) + \\ +O\left(\int_{2N-1}^{\infty} B^{*}(\lambda)e^{-\pi(\lambda-2N)}d\lambda\right)$$

and finally because as we showed in the previous Proposition we have

$$\frac{1}{2\pi} \int_{(\lambda-2N-1)\cos(\pi\lambda)<0} B^{*}(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^{*}(\lambda, 2N+1)|d\lambda-$$

$$\frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda)<0} B^{*}(\lambda)e^{-\pi(\lambda-2N)}|\Lambda^{*}(\lambda, 2N)|d\lambda \le$$

$$O\left(\int_{|\lambda-2N|\cos(\pi\lambda)|\le C} B^{*}(\lambda)e^{-\pi(\lambda-2N)}\min\left(1, \frac{1}{(\lambda-2N)^2}\right)d\lambda\right) + \\ +O\left(\int_{2N-1}^{\infty} B^{*}(\lambda)e^{-\pi(\lambda-2N)}d\lambda\right)$$

This completes the proof. □

### 4. PROOF OF THE ERDOS-TURAN CONJECTURE

In this section we will present a full proof of the Erdos-Turin Conjecture on Additive Bases.

This conjecture states that if for a set $B$ of natural numbers we have that $R_B(N) > 0$ for $N$ sufficiently large, then we must have that $\limsup_{N\to\infty} R_B(N) = \infty$.

From now on we will write

$$L(\lambda) = |\Lambda(\lambda - 1)| - |\Lambda(\lambda)|$$
Now, for every asymptotic additive base of the positive integrers $B$, we have $B(\lambda) \ge c\sqrt{\lambda}$ for every $\lambda > M_0$, where $M_0$ is the least element of $B$, where $c > 0$ is a constant, by [4].
Also, for every $B$ with $R_B(N) = O(1)$, we get that

$$ \{(a, b) \in B^2 : a, b \le \lambda\} \subseteq \cup_{k \le 2\lambda, k \in \mathbb{N}} \{(a, b) \in B^2 : a + b = k\} $$

,which gives

$$ B(\lambda)^2 \le \sum_{k \le 2\lambda} R_B(k) \le v\lambda $$

for some constant $v > 0$. This shows that

$$ B(\lambda) \le c\sqrt{\lambda} $$

for some constant $c > 0$.

Therefore, by Lemma 2.3, we get $w_1\lambda^2 \le B^*(\lambda) \le w_2\lambda^2$, for some constants $w_1, w_2 > 0$ and every $\lambda > 2M_0$.

**Proposition 4.1.** Assume that for a set $B$ of natural numbers we have

$$ \frac{N^2 \int_{\lambda < 2N, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi\lambda} |L(\lambda - 2N)| d\lambda}{e^{\pi N} \left( \int_{|\lambda - 2N| \cos(\pi\lambda)| \le C} B^*(\lambda) \min\left(1, \frac{1}{(\lambda - 2N)^2}\right) e^{-\pi\lambda} d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\frac{\pi}{2}\lambda} d\lambda \right)} \to \infty $$

as $N \to \infty$. Then, $\limsup_{N \to \infty} R_B(N) = \infty$.

*Proof.* Assume for the sake of contradiction that $R_B(N) = O(1)$.

Therefore by Proposition 3.3, we get that

$$ \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)|| - ||\Lambda^*(\lambda, 2N)|| d\lambda \le W(2N) - e^{\pi W(2N+1)} + O(1) $$

$$ O\left( \int_{|\lambda - 2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda - 2N)} \min\left(1, \frac{1}{(\lambda - 2N)^2}\right) d\lambda + \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda - 2N)} d\lambda \right) $$

Therefore, we get that

$$ \frac{1}{2\pi} \int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)|| - ||\Lambda^*(\lambda, 2N)|| d\lambda \le W(2N) - e^{\pi W(2N-1)} + O(1) $$
Now, for every asymptotic additive base of the positive integrers $B$, we have $B(\lambda) \ge c\sqrt{\lambda}$ for every $\lambda > M_0$, where $M_0$ is the least element of $B$, where $c > 0$ is a constant, by [4].

Also, for every $B$ with $R_B(N) = O(1)$, we get that

$$ \{(a, b) \in B^2 : a, b \le \lambda\} \subseteq \cup_{k \le 2\lambda, k \in \mathbb{N}} \{(a, b) \in B^2 : a + b = k\} $$

,which gives

$$ B(\lambda)^2 \le \sum_{k \le 2\lambda} R_B(k) \le v\lambda $$

for some constant $v > 0$. This shows that

$$ B(\lambda) \le c\sqrt{\lambda} $$

for some constant $c > 0$.

Therefore, by Lemma 2.3, we get $w_1\lambda^2 \le B^*(\lambda) \le w_2\lambda^2$, for some constants $w_1, w_2 > 0$ and every $\lambda > 2M_0$.

**Proposition 4.1.** Assume that for a set $B$ of natural numbers we have

$$ \frac{N^2 \int_{\lambda < 2N, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi\lambda} |L(\lambda - 2N)| d\lambda}{e^{\pi N} \left( \int_{|\lambda - 2N| \cos(\pi\lambda)| \le C} B^*(\lambda) \min\left(1, \frac{1}{(\lambda - 2N)^2}\right) e^{-\pi\lambda} d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\frac{\pi}{2}\lambda} d\lambda \right)} \to \infty $$

as $N \to \infty$. Then, $\limsup_{N \to \infty} R_B(N) = \infty$.

*Proof.* Assume for the sake of contradiction that $R_B(N) = O(1)$.

Therefore by Proposition 3.3, we get that

$$ \frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)|| - ||\Lambda^*(\lambda, 2N)|| d\lambda \le W(2N) - e^{\pi W(2N+1)} + O(1) $$

$$ O\left( \int_{|\lambda - 2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda - 2N)} \min\left(1, \frac{1}{(\lambda - 2N)^2}\right) d\lambda + \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda - 2N)} d\lambda \right) $$

Therefore, we get that

$$ \frac{1}{2\pi} \int_{(\lambda-2N)\cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)|| - ||\Lambda^*(\lambda, 2N)|| d\lambda \le W(2N) - e^{\pi W(2N-1)} + O(1) $$
$$O\left(\int_{|\lambda-2N| \cos(\pi\lambda) \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda + \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda\right)$$

By assumption, we get

$$W(N) - e^{\pi}W(2N-1) = O(1)$$

Therefore, we can write the following

$$\int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)|| - ||\Lambda^*(\lambda, 2N)|| d\lambda \le O(1) + O\left(\int_{|\lambda-2N| \cos(\pi\lambda)| \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda + \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda\right)$$

Changing variables we have that

$$0 \le \int_{\lambda<0, \cos(\pi\lambda)>0} B^*(\lambda + 2N) e^{-\pi\lambda} |L(\lambda)| d\lambda \le O(1) + O\left(\int_{|\lambda| \cos(\pi\lambda)| \le C} B^*(\lambda + 2N) e^{-\pi\lambda} \min\left(1, \frac{1}{\lambda^2}\right) d\lambda + \int_{-1}^{\infty} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda\right)$$

The inequality $B^*(\lambda + 2N) \ge (\lambda + 2N)^2 \chi_{(2M_0, \infty)}(\lambda + 2N)$ also shows that

$$\int_{-1}^{\infty} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda \ge uN^2$$

and

$$\int_{|\lambda| \cos(\pi\lambda)| \le C} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda \ge u' N^2$$

for some constants $u, u' > 0$ independent of $N$ and $N$ sufficiently large.

Therefore, dividing this asymptotic double inequality by

$$(\int_{|\lambda| \cos(\pi\lambda)| \le C} B^*(\lambda + 2N) e^{-\pi\lambda} \min\left(1, \frac{1}{\lambda^2}\right) d\lambda) (\int_{-1}^{\infty} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda) \frac{1}{N^2}$$

we get

$$\frac{N^2 \int_{\lambda<0, \cos(\pi\lambda)>0} B^*(\lambda + 2N) e^{-\pi\lambda} |L(\lambda)| d\lambda}{(\int_{|\lambda| \cos(\pi\lambda)| \le C} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda) (\int_{-1}^{\infty} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda)} = O(1)$$

Also, we have

$$(\int_{|\lambda| \cos(\pi\lambda)| \le C} B^*(\lambda + 2N) e^{-\pi\lambda} \min\left(1, \frac{1}{\lambda^2}\right) d\lambda) (\int_{-1}^{\infty} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda) =$$
$$O\left(\int_{|\lambda-2N| \cos(\pi\lambda) \le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda + \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda\right)$$

By assumption, we get

$$W(N) - e^{\pi}W(2N-1) = O(1)$$

Therefore, we can write the following

$$\int_{(\lambda-2N) \cos(\pi\lambda) < 0, \cos(\pi\lambda) > 0} B^*(\lambda) e^{-\pi(\lambda-2N)} ||\Lambda^*(\lambda, 2N+1)|| - ||\Lambda^*(\lambda, 2N)|| d\lambda \le O(1) + O\left(\int_{|\lambda-2N| \cos(\pi\lambda)|\le C} B^*(\lambda) e^{-\pi(\lambda-2N)} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda + \int_{2N-1}^{\infty} B^*(\lambda) e^{-\pi(\lambda-2N)} d\lambda\right)$$

Changing variables we have that

$$0 \le \int_{\lambda<0, \cos(\pi\lambda)>0} B^*(\lambda + 2N) e^{-\pi\lambda} |L(\lambda)| d\lambda \le O(1) + O\left(\int_{|\lambda| \cos(\pi\lambda)|\le C} B^*(\lambda + 2N) e^{-\pi\lambda} \min\left(1, \frac{1}{\lambda^2}\right) d\lambda + \int_{-1}^{\infty} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda\right)$$

The inequality $B^*(\lambda + 2N) \ge (\lambda + 2N)^2 \chi_{(2M_0, \infty)}(\lambda + 2N)$ also shows that

$$\int_{-1}^{\infty} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda \ge uN^2$$

and

$$\int_{|\lambda| \cos(\pi\lambda)|\le C} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda \ge u' N^2$$

for some constants $u, u' > 0$ independent of $N$ and $N$ sufficiently large.

Therefore, dividing this asymptotic double inequality by

$$(\int_{|\lambda| \cos(\pi\lambda)|\le C} B^*(\lambda + 2N) e^{-\pi\lambda} \min\left(1, \frac{1}{\lambda^2}\right) d\lambda) \left(\int_{-1}^{\infty} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda\right) \frac{1}{N^2}$$

we get

$$\frac{N^2 \int_{\lambda<0, \cos(\pi\lambda)>0} B^*(\lambda + 2N) e^{-\pi\lambda} |L(\lambda)| d\lambda}{(\int_{|\lambda| \cos(\pi\lambda)|\le C} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda) (\int_{-1}^{\infty} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda)} = O(1)$$

Also, we have

$$(\int_{|\lambda| \cos(\pi\lambda)|\le C} B^*(\lambda + 2N) e^{-\pi\lambda} \min\left(1, \frac{1}{\lambda^2}\right) d\lambda) \left(\int_{-1}^{\infty} B^*(\lambda + 2N) e^{-\pi\lambda} d\lambda\right) =$$
$$e^{4\pi N} \left( \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda)e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda)e^{-\pi\lambda} d\lambda \right) \le \\ C' e^{3\pi N} \left( \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} (B^*(\lambda))e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda)e^{-\frac{\pi}{2}\lambda} d\lambda \right)$$

for some constant $C' > 0$. We have

$$\int_{\lambda<2N, \cos(\pi\lambda)>0} B^*(\lambda)e^{-\pi\lambda}|L(\lambda-2N)|d\lambda = e^{-2\pi N} \int_{\lambda<0, \cos(\pi\lambda)>0} B^*(\lambda+2N)e^{-\pi\lambda}|L(\lambda)|d\lambda$$

By assumption we have that for every $W > 0$

$$\int_{\lambda<2N, \cos(\pi\lambda)>0} B^*(\lambda)e^{-\pi\lambda}|L(\lambda-2N)|d\lambda \\ \ge W \frac{e^{\pi N}}{N^2} \left( \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda)e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda)e^{-\frac{\pi}{2}\lambda} d\lambda \right)$$

for $N$ large enough.

Therefore, we get that for every $W > 0$

$$\int_{\lambda<0, \cos(\pi\lambda)>0} B^*(\lambda+2N)e^{-\pi\lambda}|L(\lambda)| \ge \frac{W}{N^2} \left( \int_{|\lambda||\cos(\pi\lambda)|\le C} B^*(\lambda+2N)e^{-\pi\lambda} \min\left(1, \frac{1}{\lambda^2}\right) d\lambda \right) \left( \int_{-1}^{\infty} B^*(\lambda+2N)e^{-\pi\lambda} d\lambda \right)$$

for $N$ large enough. This contradicts the fact that

$$\frac{N^2 \int_{\lambda<0, \cos(\pi\lambda)>0} B^*(\lambda+2N)e^{-\pi\lambda}|L(\lambda)|d\lambda}{(\int_{|\lambda||\cos(\pi\lambda)|\le C} B^*(\lambda+2N)e^{-\pi\lambda} \min\left(1, \frac{1}{\lambda^2}\right) d\lambda) (\int_{-1}^{\infty} B^*(\lambda+2N)e^{-\pi\lambda} d\lambda)} = O(1)$$

This completes the proof. $\square$

**Theorem 4.2.** Erdos Turan conjecture is true.

*Proof.* We have by Fatou's Lemma

$$\liminf_{N \to \infty} \int_{\lambda < 2N, \cos(\pi\lambda) > 0} N^2 B^*(\lambda) e^{-\pi\lambda} |L(\lambda - 2N)| d\lambda \ge \\ \int_{\cos(\pi\lambda) > 0} \liminf_{N \to \infty} (B^*(\lambda) N^2 |L(\lambda - 2N)| e^{-\pi\lambda} \chi_{(M_0, 2N)}(\lambda)) d\lambda$$

Also, by a computation on Mathematica we get

$$\lim_{N \to \infty} N^2 |L(\lambda - N)| = \frac{\pi^2}{\sqrt{2}} \sqrt{1 + \cos(2\pi\lambda)}$$
$$e^{4\pi N} \left( \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda)e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda)e^{-\pi\lambda} d\lambda \right) \le \\ C' e^{3\pi N} \left( \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} (B^*(\lambda))e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda)e^{-\frac{\pi}{2}\lambda} d\lambda \right)$$

for some constant $C' > 0$. We have

$$\int_{\lambda<2N, \cos(\pi\lambda)>0} B^*(\lambda)e^{-\pi\lambda}|L(\lambda-2N)|d\lambda = e^{-2\pi N} \int_{\lambda<0, \cos(\pi\lambda)>0} B^*(\lambda+2N)e^{-\pi\lambda}|L(\lambda)|d\lambda$$

By assumption we have that for every $W > 0$

$$\int_{\lambda<2N, \cos(\pi\lambda)>0} B^*(\lambda)e^{-\pi\lambda}|L(\lambda-2N)|d\lambda \\ \ge W \frac{e^{\pi N}}{N^2} \left( \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda)e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda)e^{-\frac{\pi}{2}\lambda} d\lambda \right)$$

for $N$ large enough.

Therefore, we get that for every $W > 0$

$$\int_{\lambda<0, \cos(\pi\lambda)>0} B^*(\lambda+2N)e^{-\pi\lambda}|L(\lambda)| \ge \frac{W}{N^2} \left( \int_{|\lambda||\cos(\pi\lambda)|\le C} B^*(\lambda+2N)e^{-\pi\lambda} \min\left(1, \frac{1}{\lambda^2}\right) d\lambda \right) \left( \int_{-1}^{\infty} B^*(\lambda+2N)e^{-\pi\lambda} d\lambda \right)$$

for $N$ large enough. This contradicts the fact that

$$\frac{N^2 \int_{\lambda<0, \cos(\pi\lambda)>0} B^*(\lambda+2N)e^{-\pi\lambda}|L(\lambda)|d\lambda}{(\int_{|\lambda||\cos(\pi\lambda)|\le C} B^*(\lambda+2N)e^{-\pi\lambda} \min\left(1, \frac{1}{\lambda^2}\right) d\lambda) (\int_{-1}^{\infty} B^*(\lambda+2N)e^{-\pi\lambda} d\lambda)} = O(1)$$

This completes the proof. $\square$

**Theorem 4.2.** Erdos Turan conjecture is true.

*Proof.* We have by Fatou's Lemma

$$\liminf_{N \to \infty} \int_{\lambda < 2N, \cos(\pi\lambda) > 0} N^2 B^*(\lambda) e^{-\pi\lambda} |L(\lambda - 2N)| d\lambda \ge \\ \int_{\cos(\pi\lambda) > 0} \liminf_{N \to \infty} (B^*(\lambda) N^2 |L(\lambda - 2N)| e^{-\pi\lambda} \chi_{(M_0, 2N)}(\lambda)) d\lambda$$

Also, by a computation on Mathematica we get

$$\lim_{N \to \infty} N^2 |L(\lambda - N)| = \frac{\pi^2}{\sqrt{2}} \sqrt{1 + \cos(2\pi\lambda)}$$
Therefore, it is easy to see that

$$ \int_{\lambda<2N, \cos(\pi\lambda)>0} B^*(\lambda) e^{-\pi\lambda} |L(\lambda - 2N)| d\lambda \geq \frac{w}{N^2} $$

for some constant $w > 0$ independent of $N$. Then, we have

$$ \frac{N^2 \int_{\lambda<2N, \cos(\pi\lambda)>0} B^*(\lambda) e^{-\pi\lambda} |L(\lambda - 2N)| d\lambda}{e^{\pi N} \left( \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda) e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\frac{\pi}{2}\lambda} d\lambda \right)} \ge \frac{w}{e^{\pi N} \left( \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda) e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\frac{\pi}{2}\lambda} d\lambda \right)} $$

Since $B^*(\lambda) \le C\lambda^2$, we get

$$ \int_{2N-1}^{\infty} B^*(\lambda) e^{-\frac{\pi}{2}\lambda} d\lambda \le C \int_{2N-1}^{\infty} \lambda^2 e^{-\frac{\pi}{2}\lambda} d\lambda = O(N^2 e^{-\pi N}) $$

Therefore, we get

$$ \frac{N^2 \int_{\lambda<2N, \cos(\pi\lambda)>0} B^*(\lambda) e^{-\pi\lambda} |L(\lambda - 2N)| d\lambda}{e^{\pi N} \left( \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda) e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\frac{\pi}{2}\lambda} d\lambda \right)} \ge \frac{w}{N^2 \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda) e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda} \ge \frac{w}{N^2 \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} \lambda^2 e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda} $$

Surely, we have

$$ N^2 \int_{N}^{\infty} \lambda^2 e^{-\pi\lambda} d\lambda \to 0 $$

as $N \to \infty$. Now if $\lambda \in (1, N)$, then $|\lambda - 2N| \ge N$ Therefore, we can write

$$ \frac{w}{N^2 \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} \lambda^2 e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda} \ge \frac{w}{N^2 \int_{|\cos(\pi\lambda)|\le \frac{C}{N}, 1<\lambda<N} \lambda^2 e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda + o(1)} $$

To prove the Theorem it suffices to show that

$$ N^2 \int_{|\cos(\pi\lambda)|\le \frac{C}{N}, 1<\lambda<N} \lambda^2 e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda = o(1) $$
Therefore, it is easy to see that

$$ \int_{\lambda<2N, \cos(\pi\lambda)>0} B^*(\lambda) e^{-\pi\lambda} |L(\lambda - 2N)| d\lambda \geq \frac{w}{N^2} $$

for some constant $w > 0$ independent of $N$. Then, we have

$$ \frac{N^2 \int_{\lambda<2N, \cos(\pi\lambda)>0} B^*(\lambda) e^{-\pi\lambda} |L(\lambda - 2N)| d\lambda}{e^{\pi N} \left( \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda) e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\frac{\pi}{2}\lambda} d\lambda \right)} \ge \frac{w}{e^{\pi N} \left( \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda) e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\frac{\pi}{2}\lambda} d\lambda \right)} $$

Since $B^*(\lambda) \le C\lambda^2$, we get

$$ \int_{2N-1}^{\infty} B^*(\lambda) e^{-\frac{\pi}{2}\lambda} d\lambda \le C \int_{2N-1}^{\infty} \lambda^2 e^{-\frac{\pi}{2}\lambda} d\lambda = O(N^2 e^{-\pi N}) $$

Therefore, we get

$$ \frac{N^2 \int_{\lambda<2N, \cos(\pi\lambda)>0} B^*(\lambda) e^{-\pi\lambda} |L(\lambda - 2N)| d\lambda}{e^{\pi N} \left( \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda) e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda \right) \left( \int_{2N-1}^{\infty} B^*(\lambda) e^{-\frac{\pi}{2}\lambda} d\lambda \right)} \ge \frac{w}{N^2 \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} B^*(\lambda) e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda} \ge \frac{w}{N^2 \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} \lambda^2 e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda} $$

Surely, we have

$$ N^2 \int_{N}^{\infty} \lambda^2 e^{-\pi\lambda} d\lambda \to 0 $$

as $N \to \infty$. Now if $\lambda \in (1, N)$, then $|\lambda - 2N| \ge N$ Therefore, we can write

$$ \frac{w}{N^2 \int_{|\lambda-2N||\cos(\pi\lambda)|\le C} \lambda^2 e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda} \ge \frac{w}{N^2 \int_{|\cos(\pi\lambda)|\le \frac{C}{N}, 1<\lambda<N} \lambda^2 e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda + o(1)} $$

To prove the Theorem it suffices to show that

$$ N^2 \int_{|\cos(\pi\lambda)|\le \frac{C}{N}, 1<\lambda<N} \lambda^2 e^{-\pi\lambda} \min\left(1, \frac{1}{(\lambda-2N)^2}\right) d\lambda = o(1) $$
Since, $\lambda \in (1, N)$, then $|\lambda - 2N| \ge N$.

Therefore,

$$ \min(1, \frac{1}{(\lambda - 2N)^2}) \le \frac{1}{N^2} $$

Therefore,

$$ N^2 \int_{|\cos(\pi\lambda)| \le \frac{C}{N}, 1 < \lambda < N} \lambda^2 e^{-\pi\lambda} \min(1, \frac{1}{(\lambda - 2N)^2}) d\lambda \le $$

$$ \int_{|\cos(\pi\lambda)| \le \frac{C}{N}, 1 < \lambda < N} \lambda^2 e^{-\pi\lambda} d\lambda = \int_1^\infty \lambda^2 e^{-\pi\lambda} \chi_{A_N}(\lambda) d\lambda $$

Here $A_n = \{\lambda \in (1, N) : |\cos(\pi\lambda)| \le \frac{C}{N}\}$. Obviously, we have that

$$ \lim_{N \to \infty} \chi_{A_N}(\lambda) = 0 $$

for every $\lambda \in (1, \infty)$ such that $\cos(\pi\lambda) \ne 0$.

Additionally the Lebesgue measure of the set $\{\lambda \in (1, \infty) : \cos(\pi\lambda) = 0\}$ is equal to 0.

Thus, by dominated convergence theorem, we conclude that

$$ \int_{|\cos(\pi\lambda)| \le \frac{C}{N}, 1 < \lambda < N} \lambda^2 e^{-\pi\lambda} d\lambda = o(1) $$

This completes the proof. □

## REFERENCES

[1] ON A PROBLEM OF SIDON IN ADDITIVE NUMBER THEORY AND ON SOME RELATED PROBLEMS, P. Erdos, P. Turan, *Journal of the London Mathematical Society* Vol.16, 1941

[2] ANALYTIC NUMBER THEORY, Henryk Iwaniec, Emmanuel Kowalski, *American Mathematical Society* Colloquium Publications

[3] INTRODUCTION TO ANALYTIC NUMBER THEORY, Tom M. Apostol, Springer

[4] ADDITIVE NUMBER THEORY- THE CLASSICAL BASES, Melvyn B. Nathanson, Springer

[5] MEASURE THEORY AND FINE PROPERTIES OF FUNCTIONS, Lawrence C. Evans, Ronald F. Gariepy, CRC Press
Since, $\lambda \in (1, N)$, then $|\lambda - 2N| \ge N$.

Therefore,

$$ \min(1, \frac{1}{(\lambda - 2N)^2}) \le \frac{1}{N^2} $$

Therefore,

$$ \begin{aligned} & N^2 \int_{|\cos(\pi\lambda)| \le \frac{C}{N}, 1 < \lambda < N} \lambda^2 e^{-\pi\lambda} \min(1, \frac{1}{(\lambda - 2N)^2}) d\lambda \le \\ & \int_{|\cos(\pi\lambda)| \le \frac{C}{N}, 1 < \lambda < N} \lambda^2 e^{-\pi\lambda} d\lambda = \int_1^\infty \lambda^2 e^{-\pi\lambda} \chi_{A_N}(\lambda) d\lambda \end{aligned} $$

Here $A_n = \{\lambda \in (1, N) : |\cos(\pi\lambda)| \le \frac{C}{N}\}$. Obviously, we have that

$$ \lim_{N \to \infty} \chi_{A_N}(\lambda) = 0 $$

for every $\lambda \in (1, \infty)$ such that $\cos(\pi\lambda) \ne 0$.

Additionally the Lebesgue measure of the set $\{\lambda \in (1, \infty) : \cos(\pi\lambda) = 0\}$ is equal to 0.

Thus, by dominated convergence theorem, we conclude that

$$ \int_{|\cos(\pi\lambda)| \le \frac{C}{N}, 1 < \lambda < N} \lambda^2 e^{-\pi\lambda} d\lambda = o(1) $$

This completes the proof. □

## REFERENCES

[1] ON A PROBLEM OF SIDON IN ADDITIVE NUMBER THEORY AND ON SOME RELATED PROBLEMS, P. Erdos, P. Turan, *Journal of the London Mathematical Society* Vol.16, 1941

[2] ANALYTIC NUMBER THEORY, Henryk Iwaniec, Emmanuel Kowalski, *American Mathematical Society Colloquium Publications*

[3] INTRODUCTION TO ANALYTIC NUMBER THEORY, Tom M. Apostol, Springer

[4] ADDITIVE NUMBER THEORY- THE CLASSICAL BASES, Melvyn B. Nathanson, Springer

[5] MEASURE THEORY AND FINE PROPERTIES OF FUNCTIONS, Lawrence C. Evans, Ronald F. Gariepy, CRC Press