Evaluation of $\int_{0}^\infty \frac{\sin(x)}{x}e^{- x²} dx$ I have tried to evaluate the integral
$$\int_{0}^\infty \frac{\sin(x)}{x}e^{- x^2} dx.$$
I used integration by parts but I did not succeed. Wolfram Alpha says that is convergent and it is equal to: $\frac \pi 2 \text{erf}({\frac 12})$ . 
Is there any simple way for evaluate it?

We have that
$$\begin{align}\int_{0}^\infty \frac{\sin(x)}{x}e^{-x^2} dx
&=
\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}\int_{0}^\infty x^{2k}e^{-x^2} dx\\
&=
\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}\int_{0}^\infty t^{k-1/2}e^{-t} dt\\
&=\frac{1}{2}\sum_{k=0}^{\infty}\frac{(-1)^k\Gamma(k+1/2)}{(2k+1)!}
\\
&=\sqrt{\pi}\sum_{k=0}^{\infty}\frac{(-1)^k\cdot(1/2)^{2k+1}}{(2k+1)k!}=
\frac{\pi}{2}\text{erf}(1/2).\end{align}$$
For the last step see the Taylor series of erf.
