Integrals of Gradshteyn and Ryzhik: 6.161 – Mellin Transforms of Theta Functions with Respect to the Lattice Parameter

We use definitions of the theta functions, shown below, from GR. Note that there is no standard notation for the theta functions.

\(z = \) argument, \(\tau = \) lattice parameter (\(\mathfrak{I}(\tau) \gt 0\)), and
\(q = \mathrm{e}^{i\pi \tau}\) (\(|q| \lt 1\))

\begin{align}
\tag{1a}
\label{eq:theta1e}
\theta_{1}(z|\tau) &= \theta_{1}(z,q) = 2 \sum_{n=0}^{\infty} (-1)^{n} q^{(n+1/2)^{2}} \sin[(2n+1)z] \\
\tag{1b}
\label{eq:theta1t}
&= -i \sum_{n=-\infty}^{\infty} (-1)^{n} q^{(n+1/2)^{2}} \mathrm{e}^{i(2n+1)z}
\end{align}

\begin{align}
\tag{2a}
\label{eq:theta2e}
\theta_{2}(z|\tau) &= \theta_{2}(z,q) = 2 \sum_{n=0}^{\infty} q^{(n+1/2)^{2}} \cos[(2n+1)z] \\
\tag{2b}
\label{eq:theta2t}
&= \sum_{n=-\infty}^{\infty} q^{(n+1/2)^{2}} \mathrm{e}^{i(2n+1)z}
\end{align}

\begin{align}
\tag{3a}
\label{eq:theta3e}
\theta_{3}(z|\tau) &= \theta_{3}(z,q) = 1 + 2 \sum_{n=1}^{\infty} q^{n^{2}} \cos(2nz) \\
\tag{3b}
\label{eq:theta3t}
&= \sum_{n=-\infty}^{\infty} q^{n^{2}} \mathrm{e}^{i2nz}
\end{align}

\begin{align}
\tag{4a}
\label{eq:theta4e}
\theta_{4}(z|\tau) &= \theta_{4}(z,q) = 1 + 2 \sum_{n=1}^{\infty} q^{n^{2}} \cos(2nz) \\
\tag{4b}
\label{eq:theta4t}
&= \sum_{n=-\infty}^{\infty} q^{n^{2}} \mathrm{e}^{i2nz}
\end{align}

6.161.1

\begin{equation}
\int_{0}^{\infty} x^{s-1} \theta_{2}(0|ix^2) dx
= 2\int_{0}^{\infty} x^{s-1} \sum_{n=0}^{\infty} \mathrm{e}^{-\pi x^{2}(n+1/2)^2} dx
\end{equation}

\begin{align}
\int_{0}^{\infty} x^{s-1} \mathrm{e}^{-\pi x^{2}(n+1/2)^2} dx
&= \frac{1}{2(\pi (n+1/2)^2)^{s/2}} \int_{0}^{\infty} y^{-1+s/2} \mathrm{e}^{-y} dy \\
&= \frac{1}{2(\pi (n+1/2)^2)^{s/2}} \Gamma(s/2)
\end{align}

We used the substitution \(y = \pi x^{2}(n+1/2)^2\).

Now we have
\begin{align}
\int_{0}^{\infty} x^{s-1} \theta_{2}(0|ix^2) dx &=
\frac{1}{\pi ^{s/2}} \Gamma(s/2) \sum_{n=0}^{\infty} \frac{1}{(n+1/2)^s} \\
&= \frac{2^s}{\pi ^{s/2}} \Gamma(s/2) \sum_{n=0}^{\infty} \frac{1}{(2n+1)^s} \\
&= \frac{2^s}{\pi ^{s/2}} \Gamma(s/2) (1-2^{-s}) \zeta(s)
\end{align}

6.161.2

\begin{align}
\int_{0}^{\infty} x^{s-1} [\theta_{3}(0|ix^2) \,- 1] dx
&= \int_{0}^{\infty} x^{s-1} \left(\left[1 + 2\sum_{n=1}^{\infty} \mathrm{e}^{-\pi x^{2} n^2} \right] -1 \right) dx \\
&= 2\sum_{n=1}^{\infty} \int_{0}^{\infty} x^{s-1} \mathrm{e}^{-\pi x^{2} n^2} dx \\
&= 2\sum_{n=1}^{\infty} \frac{\Gamma(s/2)}{2\pi^{s/2} n^{s}} \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} \sum_{n=1}^{\infty} \frac{1}{n^{s}} \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} \zeta(s)
\end{align}

We used the substitution \(y = \pi x^{2} n^2\) and the same work as above to evaluate the integral.

6.161.3

\begin{align}
\int_{0}^{\infty} x^{s-1} [1 – \theta_{4}(0|ix^2)] dx
&= \int_{0}^{\infty} x^{s-1} \left(1 – \left[1 + 2\sum_{n=1}^{\infty} (-1)^{n} \mathrm{e}^{-\pi x^{2} n^2} \right] \right) dx \\
&= -2\sum_{n=1}^{\infty} (-1)^{n} \int_{0}^{\infty} x^{s-1} \mathrm{e}^{-\pi x^{2} n^2} dx \\
&= -2\sum_{n=1}^{\infty} (-1)^{n} \frac{\Gamma(s/2)}{2\pi^{s/2} n^{s}} \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{s}} \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} (1 – 2^{1-s}) \zeta(s)
\end{align}

We used the substitution \(y = \pi x^{2} n^2\) and the same work as above to evaluate the integral.

6.161.4

\begin{align}
\int_{0}^{\infty} x^{s-1} [\theta_{4}(0|ix^2) + \theta_{2}(0|ix^2) \,- \theta_{3}(0|ix^2)] dx
&= \int_{0}^{\infty} x^{s-1} ([\theta_{4}(0|ix^2) \,-1] + \theta_{2}(0|ix^2) \,- [\theta_{3}(0|ix^2) \,-1]) dx \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} (2^{1-s} – 1) \zeta(s)
+ \frac{2^s}{\pi ^{s/2}} \Gamma(s/2) (1-2^{-s}) \zeta(s)
\,- \frac{\Gamma(s/2)}{\pi^{s/2}} \zeta(s) \\
&= -\frac{\Gamma(s/2)}{\pi^{s/2}} \zeta(s) [1 – 2^{1-s} – 2^{-s} + 1 + 1] \\
&= -\frac{\Gamma(s/2)}{\pi^{s/2}} \zeta(s) (2^{-s} – 1)(2^{1-s} – 1)
\end{align}
Here we used the previous 3 results.

All Riemann zeta function expressions can be found here.

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