improper integral calculation

Question

I do not know how to solve this integral and then expand the result in powers of $t$

$$(1-t)^{-1/2}\int_{-\infty}^{\infty}\frac{\exp\left(\frac{2x^2t}{1+t}\right)}{1+x^2}dx=?,~~~~|t|<1.$$

any idea?

Answer 1

I will assume that $t$ is such that the integral converges. Now let's consider

$$\int_{-\infty}^{\infty} dx \frac{e^{-a x^2}}{1+x^2}$$

where $a \gt 0$. We may evaluate this integral using Parseval's theorem, which states that, for functions $f$ and $g$, each having Fourier transforms $F$ and $G$, respectively, then

$$\int_{-\infty}^{\infty} dx \, f(x) \, g^*(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) \, G^*(k)$$

Now, if $f(x) = e^{-a x^2}$ and $g(x)=1/(1+x^2)$, then

$$F(k) = \sqrt{\frac{\pi}{a}} e^{-k^2/(4 a)}$$

$$G(k) = \pi \, e^{-|k|}$$

Then

$$\begin{align}\int_{-\infty}^{\infty} dx \frac{e^{-a x^2}}{1+x^2} &= \frac12 \sqrt{\frac{\pi}{a}} \int_{-\infty}^{\infty} dk \, e^{-k^2/(4 a)} e^{-|k|}\\ &= \frac12 \sqrt{\frac{\pi}{a}} 2 \int_0^{\infty} dk \, e^{-k^2/(4 a)} e^{-k} \\ &= \sqrt{\frac{\pi}{a}} e^a \int_0^{\infty} dk \, e^{-(k+2 a)^2/(4 a)} \\ &= \sqrt{\frac{\pi}{a}} \sqrt{\pi a} e^a \text{erfc}{\sqrt{a}}\\ &= \pi e^a \text{erfc}{\sqrt{a}} \end{align}$$

Now, $a=-2 t/(1+t)$ for $t \in (-1,0)$. To plug in, use the fact that $\text{erfc}{(y)} = 1-\text{erf}{(y)}$, where

$$\text{erf}{\sqrt{a}} = 2 \sqrt{\frac{a}{\pi}} \sum_{k=0}^{\infty} (-1)^k \frac{a^k}{k! (2 k+1)}$$