Integrating $e^{a/x^2-x^2}/(1-e^{b/x^2})$

Question

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.

Answer 1

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.