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BLAISE PASCAL

STEFAN CZEKALSKI

The new properties of the theta functions

Volume 20, n° 2 (2013), p. 391-398.

<http://ambp.cedram.org/item?id=AMBP_2013__20_2_391_0>

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# The new properties of the theta functions

STEFAN CZEKALSKI

## Abstract

It is shown, that the function

$$H(x) = \sum_{k=-\infty}^{\infty} e^{-k^2 x}$$

satisfies the relation

$$H(x) = \sum_{n=0}^{\infty} \frac{(2\pi)^{2n}}{(2n)!} H^{(n)}(x).$$

The theta function is defined by the following equation:

$$\Theta_0 = \sum_{k=1}^{\infty} e^{-k^2 \pi t} \quad t > 0. \qquad (1.1)$$

For computational purposes, it is convenient to introduce the function

$$\Theta(x) = \Theta_0\left(\frac{x}{\pi}\right) = \sum_{k=1}^{\infty} e^{-k^2 x}, \quad x > 0.$$

**Theorem 1.1.** *The following formula holds*

$$\sum_{k=1}^{\infty} e^{-k^2 x} = \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} \frac{e^{-x-y^2}}{1 - e^{2(-x+i\sqrt{xy})}} dy. \qquad (1.2)$$
*Proof.* For $x > 0$ we have $|e^{2(-x+i\sqrt{xy})}| < 1$, therefore

$$ 
\begin{align*}
\frac{1}{\sqrt{\pi} \int_{-\infty}^{\infty} \frac{e^{-x-y^2}}{1-e^{2(-x+i\sqrt{xy})}} dy} &= \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} e^{-x-y^2} \sum_{k=0}^{\infty} e^{2(-x+i\sqrt{xy})k} dy \\
&= \frac{1}{\sqrt{\pi}} \sum_{k=0}^{\infty} e^{-(2k+1)x} \int_{-\infty}^{\infty} e^{-y^2+2iy\sqrt{xk}} dy \\
&= \frac{1}{\sqrt{\pi}} \sum_{k=0}^{\infty} e^{-(2k+1)x} \left( \int_{-\infty}^{\infty} e^{-y^2} \cos 2y \sqrt{xk} dy \right. \\
& \qquad \left. + i \int_{-\infty}^{\infty} e^{-y^2} \sin 2y \sqrt{xk} dy \right)
\end{align*}
$$

here the first integral is equal $\sqrt{\pi}e^{-k^2x}$; second integral - zero; therefore we have

$$ = \frac{1}{\sqrt{\pi}} \sum_{k=0}^{\infty} e^{-(k^2+2k+1)x} \sqrt{\pi} = \sum_{k=1}^{\infty} e^{-k^2 x}. $$

□

By inserting $x = 1$ into (1.2), we obtain

$$ \sum_{k=1}^{\infty} e^{-k^2} = \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} \frac{e^{-1-y^2}}{1-e^{2(-1+iy)}} dy. \quad (1.3) $$

In the same way we prove the relation

$$ \sum_{k=1}^{\infty} (-1)^{k-1} e^{-k^2 x} = \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} \frac{e^{-x-y^2}}{1+e^{2(-x+iy\sqrt{x})}} dy. \quad (1.4) $$

When $x = 1$ we have

$$ \sum_{k=1}^{\infty} (-1)^{k-1} e^{-k^2} = \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} \frac{e^{-1-y^2}}{1+e^{2(-1+iy)}} dy. \quad (1.5) $$
THE NEW PROPERTIES OF THE THETA FUNCTIONS

By virtue of (1.2) we easy find the formula

$$
\sum_{k=1}^{\infty} ke^{-k^2 x} = \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} \frac{e^{-x-y^2}[2 - e^{2(-x+iy\sqrt{x})}]}{[1 - e^{2(-x+iy\sqrt{x})}]^2} dy. \quad (1.6)
$$

By taking $x = 1$, we obtain

$$
\sum_{k=1}^{\infty} ke^{-k^2} = \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} \frac{e^{-1-y^2}[2 - e^{2(-1+iy)}]}{[1 - e^{2(-1+iy)}]^2} dy. \quad (1.7)
$$

Let us consider the following formula

$$
\sum_{k=0}^{\infty} e^{-k^2 x} e^{-2axk} = \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} \frac{e^{-y^2}}{1 - e^{-2ax + 2iy\sqrt{x}}} dy \quad a > 0. \quad (1.8)
$$

Let $h = e^{-x}$ and $z = e^{-ax}$ be, then

$$
\sum_{k=0}^{\infty} h^{k^2} z^{2k} = \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} \frac{e^{-y^2}}{1 - z^2 e^{2iy\sqrt{-\ln h}}} dy. \quad (1.9)
$$

If we introduce $h = e^{-x}$ and $z = e^{-ax - \frac{1}{2}x}$ in (1.8) then we obtain

$$
\sum_{k=0}^{\infty} h^{(k-\frac{1}{2})^2} z^{2k-1} = \frac{1}{\sqrt{\pi}} \int_{-\infty}^{\infty} \frac{e^{-y^2} z h^{-\frac{1}{4}}}{1 - z^2 h^{-1} e^{2iy \sqrt{-\ln h}}} dy. \quad (1.10)
$$

We define the function $H(x) = 2\Theta(x) + 1$.

**Theorem 1.2.** $H(x)$ can be expanded in a series

$$
H(x) = e^{-x} \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} H^{(n)}(x). \quad (1.11)
$$

*Proof.* Let us consider the functions

$$
\Theta(x) = \sum_{k=1}^{\infty} e^{-k^2 x} \quad \text{and} \quad v(x) = \sum_{k=1}^{\infty} ke^{-k^2 x}.
$$

We evaluate the sums

$$
\sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} \Theta^{(n)}(x) = \sum_{k=1}^{\infty} e^{-k^2 x} \left[ 1 + \frac{k^2(2x)^2}{2!} + \frac{k^4(2x)^4}{4!} + \dots \right]
$$
$$ \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!} v^{(n)}(x) = \sum_{k=1}^{\infty} e^{-k^2 x} \left[ \frac{k2x}{1!} + \frac{k^3(2x)^3}{3!} + \frac{k^5(2x)^5}{5!} + \dots \right]. $$

The above two series are absolutely convergent, therefore we can change the order of the terms. Addition and subtraction of both sides of equations yields

$$ \begin{aligned} & \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} \Theta^{(n)}(x) + \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!} v^{(n)}(x) \\ & = \sum_{k=1}^{\infty} e^{-k^2 x} e^{2kx} = e^x \Theta(x) + e^x \end{aligned} $$

$$ \begin{aligned} & \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} \Theta^{(n)}(x) + \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!} v^{(n)}(x) \\ & = \sum_{k=1}^{\infty} e^{-k^2 x} e^{-2kx} = e^x \Theta(x) - 1. \end{aligned} $$

We add and subtract the equations once again. Thus we get

$$ \begin{aligned} & \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} \Theta^{(n)}(x) = e^x \Theta(x) + \frac{1}{2}e^x - \frac{1}{2} \\ & \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!} v^{(n)}(x) = \frac{1}{2}e^x + \frac{1}{2}. \end{aligned} $$

Introduction $H(x) = 2\Theta(x) + 1$ and $U(x) = 2v(x)$ yields

$$ \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} H^{(n)}(x) = e^x H(x) \quad (1.12) $$

$$ \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!} U^{(n)}(x) = e^x + 1. \quad (1.13) $$

□
THE NEW PROPERTIES OF THE THETA FUNCTIONS

The following relations for $H(x)$ and $U(x)$ hold

$$ \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n+1}}{(2n+1)!} H^{(n+1)}(x) = e^x H(x) \quad (1.14) $$

$$ \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} U^{(n)}(x) = e^x U(x) + e^x. \quad (1.15) $$

Proof of these relations is identical as (1.12) and (1.13).

We give now a generalization of (1.12). Let us consider the following extension of $\Theta(x)$

$$ \Theta_1(x) = \sum_{k=1}^{\infty} e^{-k^2 x} e^{-ka} \quad x > 0, a \in \mathbb{R}. $$

Then the more general version of (1.12) is equal

$$ \operatorname{cha} H_1(x) - \operatorname{sha} = e^{-x} \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n}}{(2n)!} H_1^{(n)}(x) \quad (1.16) $$

where $H_1(x) = 2\Theta_1(x) + 1$.

The proof of (1.16) is patterned after this of the Theorem 1.2.

By virtue of the relation

$$ \sum_{k=-\infty}^{\infty} e^{-k^2 x} e^{-ka} = \sum_{k=1}^{\infty} e^{-k^2 x} e^{-ka} + \sum_{k=1}^{\infty} e^{-k^2 x} e^{ka} + 1 $$

it is easy to verify that, $\Psi(x) = \sum_{k=-\infty}^{\infty} e^{-k^2 x} e^{-ka}$ satisfies (1.16) with the member *sha* removed.

**Theorem 1.3.** The formula (1.12) can be written in the form

$$ H(x) = \sum_{n=0}^{\infty} \frac{(2\pi)^{2n}}{(2n)!} H^{(n)}(x). \quad (1.17) $$

To prove theorem 0.3 we need the lemma

**Lemma 1.4.**

$$ [x^{-\frac{1}{2}} H(x)]^{(n)} = (-1)^n \frac{(2n)!}{2^{2n}} \sum_{k=0}^{n} \frac{2^{2k}}{(n-k)!} \frac{x^{-n-k-\frac{1}{2}}}{(2k)!} \frac{d^k}{d(\frac{1}{x})^k} H(\frac{1}{x}). \quad (1.18) $$
*Proof.* Proof of Lemma 1.4 (by induction) The case $n = 1$ is obvious. We suppose, that (1.18) is true for any $n$ and differentiate it.

$$ 
\begin{aligned}
& [x^{-\frac{1}{2}} H(x)]^{(n+1)} \\
&= (-1)^{n+1} \frac{(2n)!}{2^{2n}} \sum_{k=0}^{n} \frac{2^{2k}}{(n-k)!} \frac{(n+k+\frac{1}{2})x^{-n-k-\frac{3}{2}}}{(2k)!} \frac{d^k}{d(\frac{1}{x})^k} H(\frac{1}{x}) \\
&\quad + (-1)^{n+1} \frac{(2n)!}{2^{2n}} \sum_{k=0}^{n} \frac{2^{2k}}{(n-k)!} \frac{x^{-n-k-\frac{5}{2}}}{(2k)!} \frac{d^{k+1}}{d(\frac{1}{x})^{k+1}} H(\frac{1}{x}) \\
&= (-1)^{n+1} \frac{(2n)!}{2^{2n}} \left[ \frac{2n+1}{2n!} x^{-n-\frac{3}{2}} H(\frac{1}{x}) \right. \\
&\quad \left. + \sum_{k=1}^{n} \frac{2^{2k}}{(n-k)!} \frac{(n+k+\frac{1}{2})x^{-n-k-\frac{3}{2}}}{(2k)!} \frac{d^k}{d(\frac{1}{x})^k} H(\frac{1}{x}) \right] \\
&\quad + (-1)^{n+1} \frac{(2n)!}{2^{2n}} \sum_{k=0}^{n} \frac{2^{2k-2}}{(n+1-k)!} \frac{x^{-n-k-\frac{3}{2}}}{(2k-2)!} \frac{d^k}{d(\frac{1}{x})^k} H(\frac{1}{x}) \\
&\quad + (-1)^{n+1} x^{-2n-\frac{5}{2}} \frac{d^{n+1}}{d(\frac{1}{x})^{n+1}} H(\frac{1}{x})
\end{aligned}
$$

and consequently

$$ 
\begin{aligned}
[x^{-\frac{1}{2}} H(x)]^{(n+1)} = & (-1)^{n+1} \frac{(2n+2)!}{2^{2n+2}} \frac{x^{-n-\frac{3}{2}}}{(n+1)!} H(\frac{1}{x}) \\
& + \sum_{k=1}^{n} \frac{2^{2k}}{(n+1-k)!} \frac{x^{-n-k-\frac{3}{2}}}{(2k)!} \frac{d^k}{d(\frac{1}{x})^k} H(\frac{1}{x}) \\
& + (-1)^{n+1} x^{-2n-\frac{5}{2}} \frac{d^{n+1}}{d(\frac{1}{x})^{n+1}} H(\frac{1}{x})
\end{aligned}
$$

therefore

$$ [x^{-\frac{1}{2}} H(x)]^{(n+1)} = (-1)^{n+1} x^{-n-\frac{3}{2}} \frac{(2n+2)!}{2^{2n+2}} \sum_{k=0}^{n+1} \frac{2^{2k}}{(n+1-k)!} \frac{x^{-k}}{(2k)!} \frac{d^k}{d(\frac{1}{x})^k} H(\frac{1}{x}). \quad \square $$
THE NEW PROPERTIES OF THE THETA FUNCTIONS

*Proof.* Proof of the Theorem 1.3 The known functional relation

$$H(\pi^2 x) = \pi^{-\frac{1}{2}} x^{-\frac{1}{2}} H(\frac{1}{x})$$

we apply to (1.12) and obtain

$$H(\pi^2 x) = \pi^{-\frac{1}{2}} e^{-\pi^2 x} \sum_{n=0}^{\infty} (-1)^n \frac{(2x)^{2n} (\pi^2)^n}{(2n)!} [x^{-\frac{1}{2}} H(\frac{1}{x})]^{(n)}$$

We use now (1.18) and have

$$H(\pi^2 x) = \pi^{-\frac{1}{2}} x^{-\frac{1}{2}} e^{-\pi^2 x} \sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{(2\pi)^{2k}}{(2k)!} \frac{(\pi^2 x)^{n-k}}{(n-k)!} \frac{d^k}{d(\frac{1}{x})^k} H(\frac{1}{x})$$

and consequently

$$H(\frac{1}{x}) = e^{-\pi^2 x} \left( \sum_{n=0}^{\infty} \frac{(\pi^2 x)^n}{n!} H(\frac{1}{x}) + \sum_{n=1}^{\infty} \frac{(\pi^2 x)^{n-1}}{(n-1)!} \frac{(2\pi)^2}{2!} \frac{d}{d(\frac{1}{x})} H(\frac{1}{x}) \right. \\ \left. + \sum_{n=2}^{\infty} \frac{(\pi^2 x)^{n-1}}{(n-2)!} \frac{(2\pi)^4}{4!} \frac{d^2}{d(\frac{1}{x})^2} H(\frac{1}{x}) + \dots \right)$$

therefore

$$H(\frac{1}{x}) = \sum_{n=0}^{\infty} \frac{(2\pi)^{2n}}{(2n)!} \frac{d^n}{d(\frac{1}{x})^n} H(\frac{1}{x}). \quad \square$$

If we replace $\frac{1}{x}$ by $x$ we obtain (1.17).

The following formula for $\Theta_1(x)$ holds

$$\sum_{k=1}^{\infty} e^{-k^2 x} e^{-ka} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} x^n \left( \frac{1}{e^a - 1} \right)^{(2n)} \quad a > 0. \qquad (1.19)$$

*Proof.* We transform the right-hand side of (1.19) and have

$$\sum_{n=0}^{\infty} \frac{(-1)^n}{n!} x^n \left( \sum_{k=1}^{\infty} e^{-ka} \right)^{(2n)} = \sum_{k=1}^{\infty} e^{-ka} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} (k^2 x)^n \\ = \sum_{k=1}^{\infty} e^{-k^2 x} e^{-ka}.$$
S. CZEKALSKI

Using the formula (1.19) with $a = 2x$, we find following expression for the
theta function [1, 2]

$$
\Theta(x) = e^{-x} \left[ 1 + \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \left(\frac{x}{4}\right)^n \left(\frac{1}{e^{2x}-1}\right)^{(2n)} \right]. \quad (1.20)
$$

□

References

[1] R. BELLMAN – *A Brief Introduction to Theta Functions*, Hall, Rinehart and Winston, New York, 1961.

[2] A. KRAZER – *Lehrbuch der Theta - Funktionen*, Chelsea, New York, 1971.

STEFAN CZEKALSKI
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