TITLE: Evaluating $\int_{\mathbb{R}}\frac{\exp(-x^2)}{1+x^2}\,\mathrm{d}x$
QUESTION [13 upvotes]: I would like to evaluate in a closed form the integral 
$$\int_{\mathbb{R}}\frac{\exp(-x^2)}{1+x^2}\,\mathrm{d}x$$
I tried various methods : 

integration by parts
some changes of variables ($y=x^2$, $x=\tan$)
residue calculus (but the factor $\exp(-x^2)$ forbid to send the contour to infinity)
developping $\exp(-x^2)$ or $\frac{1}{1+x^2} $ in power series (but in both cases one cannot exchange sum and integral)

Does anyone knows if this integral is known, or how to evaluate it ?
At least I would like to find an exact expression.
The reason for me to believe that a closed form exists is that this integral arose in a problem of probability where I expect - if I haven't made any mistake previously - a very simple expression.

REPLY [7 votes]: Using Parseval's theorem, we get
$$
\int_{-\infty}^\infty \frac{e^{-x^2}}{1+x^2}dx=
\frac{1}{2\pi}\int_{-\infty}^\infty \sqrt{\pi}e^{-\xi^2/4}\pi e^{-|\xi|}\,d\xi=
\sqrt{\pi}\int_0^\infty e^{-\xi^2/4-\xi}\,d\xi=
$$
$$
e\sqrt{\pi}\int_0^\infty e^{-(\xi^2+4\xi+4)/4}\,d\xi=\left[u=\frac{\xi+2}{2}\right]=
2e\sqrt{\pi}\int_1^\infty e^{-u^2}du=
e\pi(1-\operatorname{erf}(1)).
$$