How can I solve this?

Question

$$ \int_{-\infty}^{\infty} \frac{x^2e^{-\alpha x^2}}{x^2+b^2} dx $$

Answer 1

A 'sort-of' elementary way to proceed [without complex analysis] is as follows. Write $$ \frac{1}{x^2+b^2}=\int_0^\infty ds\ e^{-s(x^2+b^2)} $$ and exchange the order of integrals to get $$ I= \int_0^\infty ds\ e^{-s b^2}\int_{-\infty}^\infty dx\ x^2 e^{-(\alpha+s)x^2} $$ $$ =\partial_\alpha\int_0^\infty ds\ e^{-s b^2}\int_{-\infty}^\infty dx\ e^{-(\alpha+s)x^2} $$ $$ =\partial_\alpha\int_0^\infty ds\ e^{-s b^2}\sqrt{\frac{\pi}{\alpha+s}} $$ and changing variables $\alpha+s=u$ $$ =\sqrt{\pi}\partial_\alpha \left[e^{\alpha b^2}\int_\alpha^\infty du\frac{e^{-b^2 u}}{\sqrt{u}}\right]\ . $$ The integrand admits a quite simple antiderivative $$ \int du \frac{e^{-b^2 u}}{\sqrt{u}}=\frac{\sqrt{\pi }\ \text{erf}\left(b \sqrt{u}\right)}{b}+C\ , $$ in terms of the error function. The finishing touch should be easy.

Answer 2

Assuming $a,b\in\mathbb{R}^+$ the given integral is simply $$\begin{eqnarray*} 2\int_{0}^{+\infty}\frac{x^2}{x^2+b^2}e^{-a x^2}\,dx &\stackrel{x\mapsto bz}{=}&2b\int_{0}^{+\infty}\frac{z^2}{z^2+1}e^{-a b^2 z^2}\,dz\\&=&\sqrt{\frac{\pi}{a}}-b\int_{-\infty}^{+\infty}\frac{\exp\left(-a b^2 z^2\right)}{z^2+1}\,dz\end{eqnarray*}$$ and we may tackle the last integral through the Fourier transform: $$ \int_{-\infty}^{+\infty}\frac{\exp\left(-a b^2 z^2\right)}{z^2+1}\,dz =\sqrt{\frac{\pi}{ab^2}} \int_{0}^{+\infty}e^{-s}\exp\left(-\frac{s^2}{4ab^2}\right)\,ds$$ getting a value of the (complementary) error function by completing the square: $$\boxed{\int_{-\infty}^{+\infty}\frac{x^2}{x^2+b^2}e^{-a x^2}\,dx = \color{red}{\sqrt{\frac{\pi}{a}}-\pi b e^{ab^2}\text{Erfc}\left(b\sqrt{a}\right)}.}$$