# Extended Prudnikov generating function involving Euler's polynomial

Robert Reynolds

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Robert Reynolds. Extended Prudnikov generating function involving Euler's polynomial. 2023. hal-04152722

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# Extended Prudnikov generating function involving Euler's polynomial

Robert Reynolds

**ABSTRACT.** This research paper utilizes a technique called contour integral method to obtain and assess the infinite series of the Euler polynomial using the Hurwitz Zeta function. The paper provides equations for various types of infinite series of the Euler polynomial using the Riemann Zeta function and important mathematical constants, such as Catalan's constant.

## 1. Introduction

Infinite sums of higher transcendental functions refer to the summation of series involving functions that are considered to be more complex and transcendental in nature. These functions typically cannot be expressed in terms of elementary functions like polynomials, exponential functions, logarithmic functions, and trigonometric functions. Examples of higher transcendental functions include the Riemann Zeta function, Hurwitz-zeta function, Bessel functions, hypergeometric functions, and elliptic functions, among others. These functions often arise in various areas of mathematics, physics, and engineering, and their infinite sums play a significant role in understanding and solving mathematical problems. Analyzing and evaluating infinite sums of higher transcendental functions can be a challenging task due to their intricate properties and behaviour. Mathematicians and researchers employ a variety of techniques, such as contour integration, series transformations, special identities, and numerical methods, to study and compute these infinite sums. Investigating the convergence, divergence, and analytical properties of infinite sums involving higher transcendental functions contribute to advancing our understanding of these complex functions and their applications in diverse fields. Current literature on topics related to generating functions and Euler's polynomial are in [1, 2, 3, 4]. Euler polynomials also appear in statistical physics as well as in semi-classical approximations to quantum probability distributions Ballentine and McRae [5] and evaluations of these polynomials at rational arguments was published by Cvijović et al. [6]. Fourier series expansions for Euler's polynomial is listed in section (24.8) in [7].

<footnote n="2020">2020 *Mathematics Subject Classification*. Primary 30E20, 33-01, 33-03, 33-04.

*Key words and phrases*. Euler's polynomial, contour integration, Catalan's constant, gamma function.</footnote>
We proceed by using the contour integral method [8] applied to equation (6.3.4.3) in [9] to yield the contour integral representation given by:

$$ (1.1) \quad \frac{1}{2\pi i} \int_C \sum_{p=0}^{\infty} \frac{(-1)^p a^w E_{2p+1}(x) w^{2p-k}}{(2p+1)!} dw \\ = \frac{1}{2\pi i} \int_C a^w w^{-k-1} \sec\left(\frac{w}{2}\right) \sin\left(w\left(x - \frac{1}{2}\right)\right) dw $$

where $a, x, k \in \mathbb{C}$, $Re(w(1/2 - x)) > 0$. Using equation (1.1) the main Theorem to be derived and evaluated is given by

$$ (1.2) \quad \sum_{p=0}^{\infty} \frac{(-1)^p \left(\frac{i}{\pi a}\right)^{2p} (1-k)_{2p} E_{1+2p}(x)}{\Gamma(2+2p)} \\ = -\frac{a\pi}{k} \left(\frac{2}{a\pi}\right)^k \left( \zeta\left(-k, \frac{1}{2}(1+a\pi-x)\right) - \zeta\left(-k, \frac{1}{2}(2+a\pi-x)\right) \right. \\ \left. -\zeta\left(-k, \frac{1}{2}(a\pi+x)\right) + \zeta\left(-k, \frac{1}{2}(1+a\pi+x)\right) \right) $$

where the variables $k, a, x$ are general complex numbers and the Pochhammer symbol $(-k)_p$ is given in equation (5.2.5) in [7]. The derivations follow the method used by us in [8]. This method involves using a form of the generalized Cauchy's integral formula given by

$$ (1.3) \qquad \frac{y^k}{\Gamma(k+1)} = \frac{1}{2\pi i} \int_C \frac{e^{wy}}{w^{k+1}} dw, $$

where $y, w \in \mathbb{C}$ and $C$ is in general an open contour in the complex plane where the bilinear concomitant [8] has the same value at the end points of the contour. This method involves using a form of equation (1.3) then multiplies both sides by a function, then takes the definite integral of both sides. This yields a definite integral in terms of a contour integral. Then we multiply both sides of equation (1.3) by another function and take the infinite sum of both sides such that the contour integral of both equations are the same.

## 2. Left-hand side contour integral

We use the method in [8]. The cut and contour are in the first quadrant of the complex w-plane with $Re(w) > 0$. The cut approaches the origin from the interior of the first quadrant and goes to infinity vertically and the contour goes round the origin with zero radius and is on opposite sides of the cut. Using a generalization of Cauchy's integral formula (1.3) we first replace $y \to \log(a)$, $k \to k - 2p - 1$ then multiply both sides by $\frac{(-1)^p E_{2p+1}(x)}{(2p+1)!}$ and take the infinite sum over $p \in [0, \infty)$ to get

$$ (2.1) \quad \sum_{p=0}^{\infty} \frac{(-1)^p E_{2p+1}(x) \log^{k-2p-1}(a)}{(2p+1)! (k-2p-1)!} = \frac{1}{2\pi i} \int_C \sum_{p=0}^{\infty} \frac{(-1)^p a^w E_{2p+1}(x) w^{2p-k}}{(2p+1)!} dw $$

where $Re(w) > 0$. We apply Fubini's theorem for multiple integrals and sums, see page 178 in [10] as the summand is of bounded measure over the space $\mathbb{C} \times [0, \infty)$.
**2.1. The Hurwitz-Lerch zeta Function.** The Hurwitz-Lerch zeta function (25.14) in [7] has a series representation given by

$$ (2.2) \qquad \Phi(z, s, v) = \sum_{n=0}^{\infty} (v+n)^{-s} z^n $$

where $|z|<1, v \neq 0, -1, ..$ and is continued analytically by its integral representation given by

$$ (2.3) \qquad \Phi(z, s, v) = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1} e^{-vt}}{1 - ze^{-t}} dt = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1} e^{-(v-1)t}}{e^t - z} dt $$

where $Re(v) > 0$, and either $|z| \le 1, z \neq 1, Re(s) > 0$, or $z = 1, Re(s) > 1$. Where $\zeta(s, a) = \Phi(1, s, a)$ for $Re(s) > 1, a \neq 0, -1, -2, ...$, from equation (25.14.3) in [7].

**3. The Hurwitz zeta Function**

The Hurwitz zeta function (25.11)(i) in [7] is defined by the infinite sum

$$ (3.1) \qquad \zeta(s, a) = \sum_{n=0}^{\infty} \frac{1}{(n + a)^s}, $$

where $\zeta(s, a)$ has a meromorphic continuation in the s-plane, its only singularity in $\mathbb{C}$ being a simple pole at $s = 1$ with residue 1. As a function of $a$, with $s(\neq 1)$ fixed, $\zeta(s, a)$ is analytic in the half-plane $Re(a) > 0$.

The Hurwitz zeta function is continued analytically with a definite integral representation (25.11.25) in [7] given by

$$ (3.2) \qquad \zeta(s, a) = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{x^{s-1} e^{-ax}}{1 - e^{-x}} dx, $$

where $Re(s) > 1, Re(a) > 0$.

**4. Contour integral representations for the Hurwitz-zeta function**

**4.1. Derivation of the generalized contour integral.** We use the method in [8]. Using equation (1.3) we first form a new equation by replacing $y$ by $\log(a) + ix + y$ and multiplying both sides by $e^{imx}$. Next form a second equation by replacing $x$ by $-x$ and subtract the second equation and the first to get

$$ (4.1) \qquad \frac{e^{-imx} ((\log(a) - ix + y)^k - e^{2imx} (\log(a) + ix + y)^k)}{\Gamma(k+1)} \\ = -\frac{1}{2\pi i} \int_C 2ia^w w^{-k-1} e^{wy} \sin(x(m+w)) dw $$

Next replace $y$ by $ib(2y+1)$ and multiply both sides by $(-1)^y e^{ibm(2y+1)}$ then take the infinite sum over $y \in [0, \infty)$ and simplify in terms of the Hurwitz-Lerch zeta function to get

$$ (4.2) \qquad \frac{2^k (ib)^k e^{im(b-x)}}{\Gamma(k+1)} \left( \Phi \left( -e^{2ibm}, -k, \frac{b-x-i \log(a)}{2b} \right) - e^{2imx} \Phi \left( -e^{2ibm}, -k, \frac{b+x-i \log(a)}{2b} \right) \right) $$
$$
\begin{align*}
&= -\sum_{y=0}^{\infty} \frac{1}{2\pi i} \int_C 2i(-1)^y a^w w^{-k-1} e^{ib(2y+1)(m+w)} \sin(x(m+w)) dw \\
&= -\frac{1}{2\pi i} \int_C \sum_{y=0}^{\infty} 2i(-1)^y a^w w^{-k-1} e^{ib(2y+1)(m+w)} \sin(x(m+w)) dw \\
&= -\frac{1}{2\pi i} \int_C ia^w w^{-k-1} \sec(b(m+w)) \sin(x(m+w)) dw
\end{align*}
$$

from equations (1.232.2) and (1.411.1) in [11] where Im(w) > 0 in order for the
sum to converge. We apply Fubini's theorem for integrals and sums, see page 178
in [10] as the summand is of bounded measure over the space C × [0, ∞).

**4.2. Derivation of the right-hand side contour integral.** Here we use equation (4.2) and set $b = \frac{1}{2}$, $m = 0$, $x = x - \frac{1}{2}$ and simplify in terms of the Hurwitz zeta function to get

$$
\begin{equation}
\begin{split}
& (4.3) \quad \frac{i^{k+1}}{\Gamma(k+1)} \left( 2^k \zeta \left( -k, \frac{1}{2}(-x-i\log(a)+1) \right) \right. \\
& \qquad \left. -2^k \zeta \left( -k, \frac{1}{2}(-x-i\log(a)+2) \right) - 2^k \zeta \left( -k, \frac{1}{2}(x-i\log(a)) \right) \right. \\
& \qquad \left. +2^k \zeta \left( -k, \frac{1}{2}(x-i\log(a)+1) \right) \right) \\
& = \frac{1}{2\pi i} \int_C a^w w^{-k-1} \sec\left(\frac{w}{2}\right) \sin\left(w\left(x-\frac{1}{2}\right)\right) dw
\end{split}
\end{equation}
$$

5. The infinite sum of the Euler polynomial in terms of Hurwitz zeta function

In this section we will derive the finite sum of the Euler polynomial in terms
of the Hurwitz zeta function.

THEOREM 5.1. For all $k, a, m \in \mathbb{C}$ then,

$$
\begin{equation}
\begin{split}
& (5.1) \quad \sum_{p=0}^{\infty} \frac{(-1)^p \left(\frac{i}{\pi a}\right)^{2p} (1-k)_{2p} E_{1+2p}(x)}{\Gamma(2+2p)} \\
& = -\frac{a\pi}{k} \left(\frac{2}{a\pi}\right)^k \left( \zeta\left(-k, \frac{1}{2}(1+a\pi-x)\right) - \zeta\left(-k, \frac{1}{2}(2+a\pi-x)\right) \right. \\
& \qquad \left. -\zeta\left(-k, \frac{1}{2}(a\pi+x)\right) + \zeta\left(-k, \frac{1}{2}(1+a\pi+x)\right) \right)
\end{split}
\end{equation}
$$

PROOF. Applying equation (1.1), the right-hand side of relation (2.1) is equal to the right-hand side of equation (4.2); hence, the left-hand sides of the same are identical too. Simplifying with the Gamma function and Pochhammer symbol yields the desired conclusion. $\square$

EXAMPLE 5.2. Infinite sum in terms of the logarithm of quotient gamma func-
tions.
$$
(5.2) \qquad \sum_{p=0}^{\infty} \frac{(-1)^{3p} \pi^{-2p-1} (ia)^{-2p} E_{2p+1}(x)}{a(2p+1)} = \log \left( \frac{\Gamma\left(\frac{1}{2}(\pi a - x + 1)\right) \Gamma\left(\frac{1}{2}(\pi a + x + 1)\right)}{\Gamma\left(\frac{1}{2}(\pi a - x + 2)\right) \Gamma\left(\frac{1}{2}(\pi a + x)\right)} \right)
$$

PROOF. Use equation (5.1) and apply l'Hopital's rule to the right-hand side as $k \to 0$ and simplify using equation (25.11.18) in [7]. Similar form is given in equation (5.17.5) in [7]. $\square$

EXAMPLE 5.3. Infinite product of the exponential of Euler’s polynomial in
terms of quotient gamma functions.

$$
(5.3) \quad \prod_{p=0}^{\infty} \exp \left( \frac{(-1)^p \left(\frac{i}{a\pi}\right)^{2p} E_{2p+1}(x)}{(a\pi)(2p+1)} \right) = \frac{\Gamma\left(\frac{1}{2}(1+a\pi-x)\right) \Gamma\left(\frac{1}{2}(1+a\pi+x)\right)}{\Gamma\left(\frac{1}{2}(2+a\pi-x)\right) \Gamma\left(\frac{1}{2}(a\pi+x)\right)}
$$

PROOF. Use equation (5.2) and take the exponential function of both sides and simplify. The right-hand side can be also be represented using the Knar's formula in [12]. $\square$

EXAMPLE 5.4. Asymptotic approximation where $Re(x) < 0 < Re(a)$. Similar forms and evaluations are given in the works of section (5.6) in [7], Erdélyi et al. [13], Laforgia et al. [14], Namias, [15] and Chen at al. [16].

$$
(5.4) \quad \prod_{p=0}^{\infty} \exp \left( \frac{(-1)^p \left(\frac{i}{a\pi}\right)^{2p} E_{1+2p}(2x)}{a\pi(1+2p)} \right) < \frac{x^{\frac{1}{2}-x}}{\sqrt{-x}}
$$

PROOF. Use equation (5.3) and apply equation (4) in [17] and simplify.

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

EXAMPLE 5.5. A limiting case involving an infinite sum in terms of the digamma function.

$$
(5.5) \quad \sum_{p=0}^{\infty} \frac{(-1)^p \left(\frac{i}{a\pi}\right)^{2p} E_{1+2p}(x)(2)_{2p}}{\Gamma(2+2p)} \\
= -\frac{1}{2} a^2 \pi^2 \left( \begin{aligned}[t]
& \psi^{(0)}\left(\frac{1}{2}(1+a\pi-x)\right) - \psi^{(0)}\left(\frac{1}{2}(2+a\pi-x)\right) \\
& - \psi^{(0)}\left(\frac{1}{2}(a\pi+x)\right) + \psi^{(0)}\left(\frac{1}{2}(1+a\pi+x)\right)
\end{aligned} \right)
$$

PROOF. Use equation (5.1) and take the limit of both sides as $k \to -1$ and simplify using (25.11.12) in [7]. $\square$

EXAMPLE 5.6. A limiting case of an infinite sum in terms of the Stieltjes constant $\Gamma_n$.

$$
(5.6) \quad \sum_{p=0}^{\infty} \frac{(-1)^p \left(\frac{i}{a\pi}\right)^{2p} E_{1+2p}(x)(2)_{2p}(-1+\gamma+\psi^{(0)}(2+2p))}{\Gamma(2+2p)} \\
= -\frac{1}{2}a^2\pi^2 \left(-\psi^{(0)}\left(\frac{1}{2}(1+a\pi-x)\right) - \log(2)\psi^{(0)}\left(\frac{1}{2}(1+a\pi-x)\right) \right. \\
\left. - \log\left(\frac{1}{a}\right)\psi^{(0)}\left(\frac{1}{2}(1+a\pi-x)\right) + \log(\pi)\psi^{(0)}\left(\frac{1}{2}(1+a\pi-x)\right) \right)
$$
$$ 
\begin{aligned}
&+\psi^{(0)}\left(\frac{1}{2}(2+a\pi-x)\right) + \log(2)\psi^{(0)}\left(\frac{1}{2}(2+a\pi-x)\right) + \log\left(\frac{1}{a}\right)\psi^{(0)}\left(\frac{1}{2}(2+a\pi-x)\right) \\
&-\log(\pi)\psi^{(0)}\left(\frac{1}{2}(2+a\pi-x)\right) + \psi^{(0)}\left(\frac{1}{2}(a\pi+x)\right) + \log(2)\psi^{(0)}\left(\frac{1}{2}(a\pi+x)\right) \\
&+\log\left(\frac{1}{a}\right)\psi^{(0)}\left(\frac{1}{2}(a\pi+x)\right) - \log(\pi)\psi^{(0)}\left(\frac{1}{2}(a\pi+x)\right) - \psi^{(0)}\left(\frac{1}{2}(1+a\pi+x)\right) \\
&-\log(2)\psi^{(0)}\left(\frac{1}{2}(1+a\pi+x)\right) - \log\left(\frac{1}{a}\right)\psi^{(0)}\left(\frac{1}{2}(1+a\pi+x)\right) \\
&+\log(\pi)\psi^{(0)}\left(\frac{1}{2}(1+a\pi+x)\right) + \gamma_1\left(\frac{1}{2}(1+a\pi-x)\right) \\
&-\gamma_1\left(\frac{1}{2}(2+a\pi-x)\right) - \gamma_1\left(\frac{1}{2}(a\pi+x)\right) + \gamma_1\left(\frac{1}{2}(1+a\pi+x)\right)
\end{aligned}
$$

PROOF. Use equation (5.1) and take the first partial derivative with respect to $k$ and take the limit as $k \to -1$ and simplify using using equations (25.6.12) and (25.11.12) in [7]. $\square$

EXAMPLE 5.7. Infinite sum identity in terms of the Euler polynomial.

$$ (5.7) \qquad \sum_{p=0}^{\infty} \frac{(-1)^p \left(\frac{i}{a\pi}\right)^{2p} (E_{1+2p}(-x) + E_{1+2p}(1+x))(1-k)_{2p}}{\Gamma(2+2p)} = 0 $$

PROOF. Use equation (5.1) and form a second and third equation by replacing $x \to -x$ and $x \to x+1$ respectively. Next add these two new equations and simplify. Similar form is given in equation (20.5.4) in [18]. $\square$

EXAMPLE 5.8. Infinite sum in terms of Catalan's constant.

$$ (5.8) \qquad \sum_{p=0}^{\infty} \frac{(1+p)E_{1+2p}\left(\frac{9}{2}\right)}{2^{4p}} \approx 8 \left( 16C - \zeta\left(2, \frac{17}{4}\right) + \zeta\left(2, \frac{19}{4}\right) \right) $$

PROOF. Use equation (5.1) and set $k = -2, a = \frac{4}{\pi}, x = \frac{9}{2}$ and simplify using equation (2.2.1.2.7) in [19]. $\square$

EXAMPLE 5.9. A special case of the infinite sum of a generating function.

$$ (5.9) \qquad \sum_{p=0}^{\infty} \frac{E_{2p+1}\left(-\frac{1}{2}\right) (1-k)_{2p}}{\Gamma(2p+2)} = -\frac{2^{-k} (3^k - 1)}{k} $$

PROOF. Use equation (5.1) and set $x = -a/2, a = 1/\pi$ and simplify. $\square$

## 6. Conclusion

This paper uses a contour integral approach to deriving and evaluating the infinite sum involving the Euler polynomial. We use special functions, mathematical constants, and contour integration to derive the desired result. In section (5), we obtain a closed form solution for the infinite sum of Euler's polynomial by employing the Hurwitz zeta function. Additionally, we present intriguing derivations involving the infinite product of the exponential function, which, to the best of our knowledge, have not been explored in existing literature. Moreover, we establish the relationship between the ratio of the Gamma function and the infinite product
of the exponential function. To validate our findings, we perform numerical verifi-
cation using Mathematica by Wolfram for a range of parameter values, including
real, imaginary, and complex values, within the integrals.

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(Robert Reynolds) DEPARTMENT OF MATHEMATICS AND STATISTICS, YORK UNIVERSITY, TORONTO,
ON, CANADA, M3J1P3

Email address, Corresponding author: milver73@gmail.com