Evaluate $\int_{0}^{\infty} \frac{x \exp\left\{-\beta^2 x^2\right\}}{\sinh \left(\frac{\pi x}{2}\right)} \mathrm{d} x$ 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.
 A: If we name the OP integral $J$ then let me introduce another integral:
$$I= \frac{\pi}{2} \int_{-\infty}^\infty \frac{x e^{- \beta^2 x^2}}{\sinh (\frac{\pi}{2} x)} dx=\pi J$$
We can use the following result:
$$\int_0^\infty \frac{\cos (x t)}{\cosh^2 t}dt=\frac{\frac{\pi}{2} x}{\sinh (\frac{\pi}{2} x)}$$
Which leads to:
$$I=\int_0^\infty \int_{-\infty}^\infty \frac{\cos (x t) e^{- \beta^2 x^2}}{\cosh^2 t} dx dt=2 \sqrt{\pi} \int_0^\infty \frac{e^{- t^2}}{\cosh^2 (2 \beta t)} dt$$
Where we used a well known result:
$$\int_{-\infty}^\infty \cos (a x) e^{- x^2} dx=\sqrt{\pi} e^{-a^2/4}$$
Using integration by parts for the last integral, we get a different form:
$$I= \frac{2 \sqrt{\pi}}{\beta} \int_0^\infty t e^{-t^2} \tanh(2 \beta t) dt$$
Using geometric series:
$$\tanh(2 \beta t)=\left(2 \sum_{n=0}^\infty (-1)^n e^{-4  \beta n t}\right)-1$$
Substituting and taking the integrals (I omit this part), we obtain the following series:

$$I=\frac{2 \sqrt{\pi}}{\beta} \sum_{n=0}^\infty (-1)^n \left(1-2 \sqrt{\pi} \beta n ~e^{4 \beta^2 n^2} \operatorname{erfc} (2 \beta n) \right)-\frac{\sqrt{\pi}}{\beta}  $$

Where $\operatorname{erfc}=1-\operatorname{erf}$ is the complementary error function.
The series shows good convergence, even though it looks awkward.

Note that the OP already considered a similar form of the integral in another question: Series Expansion of $\int_{\mathbb{R}} \tanh\left(\beta x\right) x \exp\left(-\frac{x^2}{2}\right) \mathrm{d} x$
