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?
 A: 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.
A: Here is a slightly longer method (compared to @Robert Z's slick series solution approach) that uses Feynman's trick of differentiating under the integral sign.
Let 
$$I(a) = \int_0^\infty \frac{\sin (ax)}{x} e^{-x^2} \, dx, \quad a > 0.$$
Note that $I(0) = 0$ and we are required to find $I(1)$.
Differentiating $I(a)$ with respect to the parameter $a$ gives
$$I'(a) = \int_0^\infty \cos (ax) e^{-x^2} \, dx. \tag1$$
On integrating (1) by parts leads to
$$I'(a) = \frac{2}{a} \int_0^\infty x e^{-x^2} \sin (ax) \, dx.$$
Also, differentiating (1) again with respect to the parameter $a$ yields
$$I''(a) = -\int_0^\infty x e^{-x^2} \sin (ax) \, dx = -\frac{a}{2} I'(a).$$
If we set $u(a) = I'(a)$ the above second-order differential equation can be reduced to the following first-order differential equation
$$u'(a) = -\frac{a}{2} u(a).$$
Solving yields
$$u(a) = I'(a) = K e^{-a^2/4}, \tag1$$
where $K$ is a constant to be determined. To find this constant setting $a = 0$ in $I'(a)$ leads to
$$I'(0) = \int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2} \cdot \frac{2}{\sqrt{\pi}} \int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2} \cdot \text{erf} (\infty) = \frac{\sqrt{\pi}}{2}.$$ 
So on setting $a= 0$ in (1) we find $K = \sqrt{\pi}/2$. Thus
$$I'(a) = \frac{\sqrt{\pi}}{2} e^{-a^2/4}.$$
Now as $I(0) = 0$, we have
$$I(1) = \int_0^1 I'(a) \, da = \frac{\sqrt{\pi}}{2} \int_0^1 e^{-a^2/4} \, da.$$
Enforcing a substitution of $a \mapsto 2a$ leads to
$$I(1) = \sqrt{\pi} \int_0^{1/2} e^{-a^2} \, da = \frac{\pi}{2} \cdot \frac{2}{\sqrt{\pi}} \int_0^{1/2} e^{-a^2} \, da.$$
And since from the integral representation for the error function which is given by
$$\text{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \, dt,$$
one has
$$\int_0^\infty \frac{\sin x}{x} e^{-x^2} \, dx = \frac{\pi}{2} \text{erf} \left (\frac{1}{2} \right ),$$
as expected.
A: Following Sangchul Lee's hint.
\begin{align}
I:=\int^\infty_0 \frac{\sin(x)}{x}e^{-x^2}\,dx=\frac{1}{2}\int^\infty_{-\infty}\frac{\sin(x)}{x}e^{-x^2}\,dx=\frac{1}{4}\int^\infty_{-\infty}\int^1_{-1}e^{ixt}e^{-x^2}\,dt\,dx
\end{align}
Since $|e^{ixt}e^{-x^2}|=e^{-x^2}$ the double integral is clearly absolutely convergent so by Tonelli-Fubini we can interchange the integration order to obtain:
\begin{align}
I=\frac{1}{4}\int^{1}_{-1}\int^{\infty}_{-\infty}e^{ixt}e^{-x^2}\,dx\,dt=\frac{1}{4}\int^{1}_{-1}\int^{\infty}_{-\infty}e^{-\frac{t^2}{4}-(x-it/2)^2}\,dx\,dt
\end{align}
where we have completed the square on the last integral. Now by an easy contour integration on a rectangle we get
\begin{align}
\int^\infty_{-\infty} e^{ixt}e^{-x^2}\,dx=\sqrt[]{\pi}e^{-t^2/4}
\end{align}
So with substitutation and parity (P) we can conclude:
\begin{align}
I=\frac{\sqrt[]{\pi}}{4}\int^1_{-1}e^{-t^2/4}\,dt\stackrel{\color{red}{u=t/2}}{=}\frac{\sqrt[]{\pi}}{2}\int^{1/2}_{-1/2}e^{-u^2}\,du\stackrel{\color{red}{P}}{=}\frac{\pi}{2}\cdot\frac{2}{\sqrt[]{\pi}}\int^{1/2}_0e^{-u^2}\,du=\frac{\pi}{2}\operatorname{erf}\left(\frac 12\right)
\end{align}
