Show that $\int_{0}^{\infty}\left(x\over e^{ax}-x^2\right)^2dx=\sum_{n=1}^{\infty}{n(2n)!\over [a(n+1)]^{2n+1}}$ $$\int_{0}^{\infty}\left(x\over e^{ax}-x^2\right)^2dx=\sum_{n=1}^{\infty}{n(2n)!\over [a(n+1)]^{2n+1}}\tag1$$
Any hints?
 A: Hint. Assume $a>0$. By starting with the expansion
$$
\left(x\over e^{ax}-x^2\right)^2=\left(\frac{xe^{-ax}}{1-x^2e^{-ax}}\right)^2 = \sum_{n=0}^\infty n x^{2n}e^{-a(n+1)x}
$$ one is allowed to integrate termwise obtaining
$$
\int_{0}^{\infty}\left(x\over e^{ax}-x^2\right)^2dx=\sum_{n=0}^\infty n \int_{0}^{\infty}x^{2n}e^{-a(n+1)x}\:dx=\sum_{n=1}^{\infty}{n(2n)!\over [a(n+1)]^{2n+1}}
$$ where we have used the standard integral representation of the Euler gamma function.
A: HINT: $$ \frac{1}{(e^{ax}-x^2)^2}= \frac{1}{x^4}\cdot \frac{1}{(\frac{e^{ax}}{x^2}-1)^2} $$
Now set $y:=\frac{e^{ax}}{x^2}$ and notice that by the geometric series you have
$$\frac{1}{(\frac{e^{ax}}{x^2}-1)^2}=\frac1{(y-1)^2}=-\frac{d}{dy}\frac1{(y-1)}=\frac{d}{dy} \sum_{k=0}^\infty y^k=\sum_{k=0}^\infty (k+1)y^{k}.$$
Then evaluating the integrals of the various summands should be easy.
PS: The idea comes from Jack d'Aurizio's answer to the following question Evaluate the integral $\int_{0}^{+\infty}\left(\frac{x^2}{e^x-1}\right)^2dx$
Another possibility might be integration under the integral sign (i.e. seeing the integral as a function of $a$ and differentiating with respect to $a$). But I haven't been able to run the full argument.
