Curious integrals for Jacobi Theta Functions $\int_0^1 \vartheta_n(0,q)dq$

Question

There are various identities for the Jacobi Theta Functions $\vartheta_n(z,q)$ on the MathWorld page and on the Wikipedia page. But I found no integral identities for these functions.

Meanwhile, there are beautiful identities for the simple case of $z=0$:

$$\int_0^1 \vartheta_2(0,q)dq=\pi \tanh \pi$$

$$\int_0^1 \vartheta_3(0,q)dq=\frac{\pi}{ \tanh \pi}$$

$$\int_0^1 \vartheta_4(0,q)dq=\frac{\pi}{ \sinh \pi}$$

I found these identities using the series approach and Mathematica for summation.

Surprisingly enough Mathematica can't take the integrals themselves, and numerically for $\vartheta_2(0,q),\vartheta_3(0,q)$ they are extremely hard to compute because of the sharp increase around $q=1$.

Using the same method, it's possible to find some other interesting integrals, for example:

$$\int_0^1 \vartheta_2(0,q) \ln \frac{1}{q} dq=\frac{\pi}{2} \left( \tanh \pi-\frac{\pi}{\cosh^2 \pi} \right)$$

$$\int_0^1 \vartheta_3(0,q) \ln \frac{1}{q} dq=\frac{\pi}{2} \left( \frac{1}{ \tanh \pi}+\frac{\pi}{\sinh^2 \pi} \right)$$

$$\int_0^1 \vartheta_4(0,q) \ln \frac{1}{q} dq=\frac{\pi}{2 \sinh \pi} \left( \frac{\pi}{ \tanh \pi}+1 \right)$$

Where can I find out more about the integral identities for the Jacobi Theta Functions? Are there some identities for the general case of $z \neq 0$?

And more, is there some intuition behind the relationship between theta functions and hyperbolic functions? (I can undertand $\pi$, since they are related to elliptic functions, but where do $\tanh, \sinh, \cosh$ come from?)

Answer 1

The connection to hyperbolic functions is due the following Fourier series, valid when $|z|<\pi$:

$$\frac{\pi}{x} \frac{\cosh z x}{\sinh \pi x} = \sum_{n \in \mathbb{Z}}\frac{(-1)^n \cos( z n)}{n^2+x^2} .\tag{1}$$

It is obtained quite straightforwardly. For instance, see @Machinato's related calculation in this answer, or the calculation on this page.

Now, using the series expansion of $\vartheta_4(z,q)$ and integrating term by term we obtain: $$\int_0^1 q^{s-1} \vartheta_4(z,q)\,dq= \int_0^1 q^{s-1} \sum_{n \in \mathbb{Z}} (-1)^n q^{n^2} \cos(2 n z) \, dq \\= \sum_{n \in \mathbb{Z}} \frac{(-1)^n \cos(2 n z)}{n^2+s} = \frac{\pi}{\sqrt{s}} \frac{\cosh 2 z \sqrt{s}}{\sinh \pi \sqrt{s}}.\tag{2}$$

Multiplication of the integrand by powers of $\ln q$ is done by successive differentiation with respect to $s$, and to obtain the similar cases for $\vartheta_2$ and $\vartheta_3$ we only need to keep in mind that $$\vartheta_3(z,q) = \vartheta_4(z-\pi/2, q) \tag{3}$$ and $$\vartheta_2(z,q) = \vartheta_3(z,q)-\vartheta_4(z/2,q^{1/4}) \tag{4}.$$