Integrating $e^{a/x^2-x^2}/(1-e^{b/x^2})$ I want to solve the following two integrals analytically
\begin{aligned}
I_1 = & \int\limits_0^{\infty}\frac{e^{a/x^2}}{1-e^{b/x^2}}e^{-x^2}dx \\
I_2 = & \int\limits_0^{\infty}\frac{e^{a/x^2}}{1-e^{b/x^2}}e^{-x^2}(x^2-\frac{3}{2})dx
\end{aligned}
where
\begin{aligned}
a,b \in & \mathbb{C} \\
Re[a] < & 0 \\
Re[b] < & 0.
\end{aligned}
My Approach
I know the following integrals,
\begin{aligned}
\int\limits_0^{\infty} e^{a/x^2}e^{-x^2}dx & = \frac{\sqrt{\pi}}{2}e^{-2\sqrt{-a}}\\
\int\limits_0^{\infty} e^{a/x^2}e^{-x^2}(x^2-\frac{3}{2})dx & = \frac{\sqrt{\pi}}{2}e^{-2\sqrt{-a}}(-1+\sqrt{-a}),
\end{aligned}
but from here on I don't know how to progress.
 A: Set
$$
I(\alpha,\beta,\tau): =  \int_0^{\infty}\frac{e^{-\alpha/x^2}}{1-e^{-\beta/x^2}}e^{-\tau \:x^2}\mathrm{d}x, \quad \Re(\alpha)>0,\Re(\beta)>0,\Re(\tau)>0.
$$
We obtain
$$
\begin{align}
I(\alpha,\beta,\tau): & =  \int_0^{\infty}\frac{e^{-\alpha/x^2}}{1-e^{-\beta/x^2}}e^{-\tau x^2}\mathrm{d}x \\
& = \int_0^{\infty}e^{-\alpha/x^2}\sum_{n=0}^{\infty}e^{-\beta n/x^2}e^{-\tau \:x^2}\mathrm{d}x \\
& = \sum_{n=0}^{\infty} \int_0^{\infty}e^{-\tau \:x^2-(\alpha+\beta n)/x^2}\mathrm{d}x\\
&= \sum_{n=0}^{\infty}e^{-2\sqrt{\tau(\alpha+\beta n)}} \int_0^{\infty}e^{-\tau \left( x- \tfrac{\sqrt{\alpha+\beta n}}{\sqrt{\tau} \:x}\right)^2}\mathrm{d}x  \\
& =\frac{\sqrt{\pi}}{2\sqrt{\tau}} \sum_{n=0}^{\infty}e^{-2\sqrt{\tau(\alpha+\beta n)}}
\end{align}
$$
and I would be surprised that the latter infinite sum could be further reduced. 
You readily get $I_1$ and $I_2$ by
$$
\begin{align}
I_1 & = I(-a,-b,1) \\
I_2 & = -\frac{3}{2}I(-a,-b,1)-\partial_{\tau}I(-a,-b,\tau)|_{\tau=1}
\end{align}
$$
Here we have used the general fact that

$$
\int_{-\infty}^{+\infty}f\left(x-\frac{s}{x}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x, \quad s>0. \quad (*)
$$

Proof (of (*)). The function $$(-\infty,0) \cup (0,+\infty) \ni x \longmapsto u = x-\frac{s}{x} $$ has two local inverses, one for $(-\infty,0)$ and the second one for $(0,+\infty)$, both mapping the corresponding half-line to $(-\infty,+\infty)$. Then we may write 
$$
\begin{align}
\int_{-\infty}^{+\infty}f\left(x-\frac{s}{x}\right)\mathrm{d}x & =  \int_{-\infty}^{0}f\left(x-\frac{s}{x}\right)\mathrm{d}x+\int_{0}^{+\infty}f\left(x-\frac{s}{x}\right)\mathrm{d}x \\
&  \qquad \qquad \qquad \qquad u=x-\frac{s}{x}\\
& x_-=\frac{u-\sqrt{u^2+4s^2}}{2} \qquad x_+=\frac{u+\sqrt{u^2+4s^2}}{2}\\
& = \int_{-\infty}^{+\infty}f(u)\left(\frac{1}{2}-\frac{u}{2\sqrt{\cdots}} \right) \!\mathrm{d}u+\int_{-\infty}^{+\infty}f(u)\left(\frac{1}{2}+\frac{u}{2\sqrt{\cdots}} \right)\!\mathrm{d}u  \\
& =\int_{-\infty}^{+\infty} f(u)\: \mathrm{d}u.
\end{align}
$$ Applying $(*)$ to  $f(x)=e^{-\tau x^2}$ gives the last line in the above computation of $I(\alpha,\beta,\tau)$ . 
The transformation $(*)$ is a result due to A. Cauchy, then generalised by G. Boole.
