Integrating $\int^\infty_ 0\frac{x^2}{e^{x^2} - 1}\,\mathrm dx$ How can I integrate
$$\int^{\infty}_ 0 \frac{x^2}{{e^{x^2}} - 1}\,\mathrm{d}x?$$
I have tried using the Gaussian Integral Formula and Integration by parts, but have made no progress so far.
 A: As was mentioned in the comments, I'll have fun generalizing the result to
$$I(s)=\int_0^{+\infty} \frac{x^s}{e^{x^2}-1}\,dx \qquad s\in\mathbb{C}, \text{Re }(s)> 1$$
We'll expand the denominator into a geometric series before using the Gamma function to evaluate the final integral:
\begin{align}
I(s)&=\int_0^{+\infty} \frac{x^s}{e^{x^2}-1}\,dx \\ 
&=\int_0^{+\infty} \frac{x^s e^{-x^2}}{1-e^{-x^2}}\,dx \\
&=\int_0^{+\infty} x^s e^{-x^2} \sum_{n=0}^{+\infty} e^{-nx^2},dx \\
&=\sum_{n=0}^{+\infty}\int_0^{+\infty} x^s e^{-x^2} e^{-nx^2},dx \\
&=\sum_{n=0}^{+\infty}\int_0^{+\infty} x^s e^{-x^2(n+1)},dx \\
&=\sum_{n=1}^{+\infty}\int_0^{+\infty} x^s e^{-nx^2}dx \qquad u=x^2\\ 
&=\frac{1}{2}\sum_{n=1}^{+\infty}\int_0^{+\infty} u^{s/2} e^{-nu} \frac{du}{\sqrt{u}} \\
&=\frac{1}{2}\sum_{n=1}^{+\infty}\int_0^{+\infty} u^{(s-1)/2} e^{-nu}\,du \\
&=\frac{1}{2}\Gamma\left(\frac{s+1}{2}\right)\sum_{n=1}^{+\infty} \frac{1}{n^{(s+1)/2}}\\ \\
I(s)&=\frac{1}{2}\Gamma\left(\frac{s+1}{2}\right) \zeta\left(\frac{s+1}{2}\right)
\end{align}
Thus, for $s\in\mathbb{C}, \text{Re }(s)>1$, 
\begin{align}
\boxed{\int_0^{+\infty} \frac{x^s}{e^{x^2}-1}\,dx=\frac{1}{2}\Gamma\left(\frac{s+1}{2}\right) \zeta\left(\frac{s+1}{2}\right)}
\end{align}
A: Using that 
$$\frac1{e^{x^2}-1}=\sum_{n=1}^\infty e^{-nx^2}$$
it follows that
$$\int_0^\infty \frac{x^2}{e^{x^2}-1}\,dx
=\sum_{n=1}^\infty\underbrace{\int_0^\infty e^{-nx^2}x^2\,dx}_{=\sqrt{\pi}/(4n^{3/2})}
=\frac{\sqrt\pi}4\sum_{n\geq1}\frac1{n^{3/2}}
=\frac{\sqrt\pi}4\zeta\left(\frac32\right)$$
Edit: the calculation of 
$$\int_0^\infty e^{-nx^2}x^2\,dx$$
can be done using the classical result $\int_0^\infty e^{-n x^2}\,dx=\sqrt{\pi}/(2\sqrt{n})$ with $n>0$. Indeed, 
$$\int_0^\infty e^{-nx^2}x^2\,dx
=-\int_0^\infty \frac{d}{dn}e^{-nx^2}\,dx
=-\frac{d}{dn}\frac{\sqrt{\pi}}{2\sqrt{n}}
=\frac{\sqrt\pi}{4 n^{3/2}}.$$
