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Integral of Jacobi Theta function

Let \vartheta_4(z;q) denote one of the Jacobi Theta functions. Prove that

    \[\int_{0}^{1}\vartheta_4\left ( 0;q \right ) \, \mathrm{d}q = \frac{\pi}{\sinh \pi}\]

Solution

We have successively,

    \begin{align*} \int_{0}^{1} \vartheta_4\left ( 0;q \right )\, \mathrm{d}q &= \int_{0}^{1} \sum_{n=-\infty}^{\infty} (-1)^n q^{n^2} \, \mathrm{d}q \\ &= \sum_{n=-\infty}^{\infty} (-1)^n \int_{0}^{1} q^{n^2} \, \mathrm{d}q\\ &=\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+1} \\ &= \frac{\pi}{\sinh \pi} \end{align*}

The sum is evaluated as follows. Consider the function

    \[f(z) = \frac{\pi \csc \pi z}{z^2+1}\]

and integrate it around a square \Gamma_N with vertices \left ( N+\frac{1}{2} \right )\left ( \pm 1\pm i \right ). The function f has poles at every integer z=n with residue \frac{(-1)^n}{n^2+1} as well as at z=\pm i with residues -\frac{\pi}{2 \sinh \pi}. We also note that as N \rightarrow +\infty the contour integral of f tends to 0. Thus,

    \begin{align*} \frac{1}{2\pi i}\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z &= \sum_{n=-N}^{N} \mathfrak{Res}_{z=n} \frac{\pi \csc \pi z}{z^2+1} + \mathfrak{Res}\left ( f ; i \right ) + \mathfrak{Res}\left ( f;-i \right ) \\ &= \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{2\sinh \pi} -\frac{\pi}{2 \sinh \pi} \\ &= \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{\sinh \pi} \end{align*}

Hence,

    \begin{align*} 0 &=\lim_{N \rightarrow +\infty} \oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z \\ &= \lim_{N \rightarrow +\infty} \left ( \sum_{n=-N}^{N} \frac{(-1)^n}{n^2+1} - \frac{\pi}{\sinh \pi} \right )\\ &= \sum_{n=-\infty}^{\infty} \frac{(-1)^n}{n^2+1} - \frac{\pi}{ \sinh \pi} \end{align*}

and the exercise is complete.

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2 Comments

  1. I’m giving the 3rd one a try. We have that

        \begin{align*} \int_{0}^{1} \vartheta_2\left ( 0;q \right ) \ln \frac{1}{q} \, \mathrm{d}q &= \int_{0}^{1} \ln \frac{1}{q} \sum_{n=-\infty}^{\infty} q^{(n+1/2)^2} \, \mathrm{d}q \\   &=-\sum_{n=-\infty}^{\infty} \int_{0}^{1} q^{(n+1/2)^2} \ln q \, \mathrm{d}q \\   &= 16\sum_{n=-\infty}^{\infty} \frac{1}{\left ( 4n^2+4n+5 \right )^2} \\   &=\frac{\pi}{2} \left ( \tanh \pi - \frac{\pi}{\cos^2 \pi} \right )  \end{align*}

    The series is standard. Initiate a contour integration by invoking the kernel \displaystyle f(z) = \frac{\pi \cot \pi z}{(4z^2+4z+5)^2} and the result follows.

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