How to perform the integral $\int_{-\infty}^\infty\frac{e^{-x^2}\sin(x) }{x }$? Does anybody know how  to perform the integral
$$
\int_{-\infty}^\infty\frac{e^{-x^2}\sin(x) }{x }
$$
Thanks.
 A: Write
$$ \frac{\sin x}{x} = \int_{-1}^1 \frac12 e^{i x y}\,dy, $$
so that the integral is
$$ \frac12\int_{-\infty}^\infty dx \int_{-1}^1\,dy \,e^{-x^2+i x y} = \frac12\int_{-1}^{1}\sqrt\pi e^{-y^2/4}\,dy = \frac\pi2 \mathop{\text{erf}}(x/2)\Big|_{-1}^{1} = \pi \mathop{\text{erf}}(1/2). $$
A: Consider :
$$f(a)=\int_{-\infty}^\infty \frac{e^{-x^2}\sin(ax)}x\,dx$$
then 
\begin{align}
f'(a)&=2\int_0^\infty e^{-x^2}\cos(ax)\,dx\\
&=\int_{-\infty}^\infty e^{-x^2}e^{iax}\,dx\\
&=\int_{-\infty}^\infty e^{-(x-ia/2)^2-a^2/4}\,dx\\
&=e^{-a^2/4}\int_{-\infty}^\infty e^{-(x-ia/2)^2}\,dx\\
&=e^{-a^2/4}\int_{-\infty}^\infty e^{-y^2}\,dy\quad(*)\\
&=\sqrt{\pi}\;e^{-a^2/4}
\end{align}
($(*)$ a justification for this rewriting is provided in the comments)
Integrating again we get :
\begin{align}
f(a)&=\sqrt{\pi}\int_0^a\,e^{-a^2/4}\,da\\
f(a)&=\pi\;\operatorname{erf}\left(\frac a2\right)
\end{align}
(from the definition of the error function)
and the answer will be  :
$$f(1)=\pi\;\operatorname{erf}\left(\frac 12\right)$$
