I am trying to evaluate the integral $$ \int_{0}^{\infty} \frac{x \exp\left\{-\beta^2 x^2\right\}}{\sinh \left(\frac{\pi x}{2}\right)} \mathrm{d} x, $$ where $\beta \in \mathbb{R}$ is some constant.

After many unfruitful attempts and checking up several integral handbooks, I am still not sure how to handle it. Does there exist an analytic solution?

Thanks in advance!

**Edit:**

Inspired by the discussions in this question, I tried to make the following manipulation (assume we can interchange the order of the infinite sum and the definite integral): $$ \begin{aligned} \int_{0}^{\infty} \frac{x \exp(-\beta^2 x^2)}{\sinh(\pi x/2)}\mathrm{d}x =& \frac{16}{\pi^3} \int_{0}^{\infty}\sum_{n=0}^{\infty} x \exp(-\beta^2x^2) \exp(-(2n+1)x) \mathrm{d} x \\ =&\frac{16}{\pi^3} \sum_{n=0}^{\infty} \exp\left(\frac{(2n+1)^2}{4\beta^2} \right) \int_{0}^{\infty} x \exp\left\{ -\frac{1}{2\frac{1}{2\beta^2}} \left(x + \frac{2n+1}{2\beta^2} \right)^2 \right\} \mathrm{d}x\\ =& \frac{16}{\pi^3}\sum_{n=0}^{\infty} \exp\left(\frac{(2n+1)^2}{4\beta^2}\right) \left\{ \frac{\sqrt{2\pi}}{2\beta^2}\exp\left(-(2n+1)^2\beta^2\right) \right. \\ &\left.- \frac{\sqrt{\pi}(2n+1)}{2\beta^3} \left(1-\Phi\left(\frac{\sqrt{2}}{2}(2n+1)\beta\right)\right) \right\} \end{aligned}, $$ where $\Phi(\cdot)$ is the standard normal cdf. This still seems not very promising: I have little idea how to deal with this infinite series.