Finding a generalization for $\int_{0}^{\infty}e^{- 3\pi x^{2} }\frac{\sinh(\pi x)}{\sinh(3\pi x)}dx$

Question

$\;\;\;\;$I was reading the introduction of Paul J. Nain's book "Dr. Euler's fabulous formula" where he talks about the sense of beauty in mathematics and quotes the G.N.Watson as saying that a particular formula gave him "... a thrill which is indistinguishable from the thrill which I feel when I enter the Sagrestia Nuova of the Capelle Medicee and see before me the austere beauty of the four statues representing Day, Night, Evening, and Dawn which Michelangelo has set over the tombs of Guiliano de'Medici and Lorenzo de'Medici". The formula is \begin{align*} \int_{0}^{\infty}e^{-3\pi x^{2} }\frac{\sinh(\pi x)}{\sinh(3\pi x)}dx=\frac{1}{e^{2\pi/3}\sqrt{3}}\sum_{n=0}^{\infty}\frac{e^{-2n(2n+1)\pi}}{(1+e^{-\pi})^{2}\cdots(1+e^{-(2n+1)\pi})^{2}} \end{align*} I have a question relating this formula: Is it possible to get a generalization of the following form \begin{align*} \int_{0}^{\infty}e^{- m\pi x^{2} }\frac{\sinh(\pi x)}{\sinh(m\pi x)}dx \end{align*} where $3$ is replaced by $m$?

Answer 1

I refer to this question in the following.

This looks like the Plancherel-Parseval Theorem may apply. That is, given functions $f(x)$, $g(x)$ and their respective Fourier transforms $\hat{f}(k)$, $\hat{g}(k)$, we have

$$\int_{-\infty}^{\infty} dx \, f(x) g(x)^* = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, \hat{f}(k) \hat{g}(k)^* $$

When $f(x) = e^{-m \pi x^2}$, we may show that $\hat{f}(k) = m^{-1/2} e^{-k^2/(4 m \pi)}$ with little trouble. However, we need the help of that question for the next FT:

When

$$g(x) = \frac{\sinh{\pi x}}{\sinh{m \pi x}}$$

then

$$\hat{g}(k) = \frac{1}{m} \frac{\sin{(\pi/m)}}{\sin{(\pi/m)} + \cosh{(k/m)}}$$

(I credit @Sasha with the accepted result.) Therefore we have

$$\int_{-\infty}^{\infty} dx \, e^{-m \pi x^2} \frac{\sinh{\pi x}}{\sinh{m \pi x}} = \frac{\sin{(\pi/m)}}{2 \pi m^{3/2}} \int_{-\infty}^{\infty} dk \, \frac{e^{-k^2/(4 m \pi)}}{\sin{(\pi/m)} + \cosh{(k/m)}}$$

And I will leave it at this for now; I suspect something can be done with this.

Answer 2

The integral in the question is connected to one of the several mock theta functions defined by Ramanujan.

Let $q$ be a real number with $|q| < 1$ and define the mock theta function $\omega(q)$ by equation $$\omega(q) = \sum_{n = 0}^{\infty}\frac{q^{2n(n + 1)}}{(1 - q)^{2}(1 - q^{3})^{2}\dots(1 - q^{2n + 1})^{2}}\tag{1}$$ so that $$\omega(-q) = \sum_{n = 0}^{\infty}\frac{q^{2n(n + 1)}}{(1 + q)^{2}(1 + q^{3})^{2}\dots(1 + q^{2n + 1})^{2}}\tag{2}$$ If $q = e^{-\pi}$ then the series in the question is $\omega(-q)$.

In his paper G. N. Watson proved the following transformation formula for $\omega(-q)$. Let $\alpha, \beta$ be positive real numbers with $\alpha\beta = \pi^{2}$ and let $q = e^{-\alpha}, q_{1} = e^{-\beta}$ then $$q^{2/3}\omega(-q) + \sqrt{\frac{\pi}{\alpha}}q_{1}^{2/3}\omega(-q_{1}) = 2\sqrt{\frac{3\alpha}{\pi}}I(\alpha)\tag{3}$$ where $$I(\alpha) = \int_{0}^{\infty}e^{-3\alpha x^{2}}\frac{\sinh \alpha x}{\sinh 3\alpha x}\,dx\tag{4}$$ The integral $I(\alpha)$ is the correct a fruitful generalization of the integral $I$ given in the question. In fact $I = I(\pi)$ and putting $\alpha = \beta = \pi$ in equation $(3)$ we get $$I = I(\pi) = \frac{q^{2/3}}{\sqrt{3}}\omega(-q)$$ where $q = e^{-\pi}$ and this is the result given in question.