How can I solve this? $$ \int_{-\infty}^{\infty} \frac{x^2e^{-\alpha x^2}}{x^2+b^2} dx $$
 A: 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.
A: 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:
$$\boxed{\int_{-\infty}^{+\infty}\frac{x^2}{x^2+b^2}e^{-a x^2}\,dx = \color{red}{\sqrt{\frac{\pi}{a}}-\pi b e^{ab^2}\text{Erfc}\left(b\sqrt{a}\right)}.}$$
