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?
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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.
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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}
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Double integral $$ \begin{aligned} \int_0^{\infty} \frac{\sin x}{x} e^{-x^2} d x= & \int_0^{\infty} e^{-x^2}\left(\int_0^1 \cos(x y) d y\right) d x \\ = & \int_0^1 \int_0^{\infty} e^{-x^2} \cos (x y) d x d y \\ = & \int_0^1 \frac{\sqrt{\pi}}{2} e^{-\frac{y^2}{4}} d y \quad \textrm{(See footnote for details)} \\ = & \frac{\sqrt{\pi}}{2} \int_0^1 e^{-\frac{y^2}{4}} d y \\ = & \frac{\pi}{2} \operatorname{erf}\left(\frac{1}{2}\right) \end{aligned} $$ Footnote: $$ \begin{aligned}\int_0^{\infty} e^{-x^2} \cos (x y) d x = & \Re\int_0^{\infty} e^{-x^2} e^{x y i} d x \\ = & \Re\int_0^{\infty} e^{-\left(x-\frac{y}{2} i\right)^2-\frac{y^2}{4}} d x \\ = & e^{-\frac{y^2}{4}} \Re \int_0^{\infty} e^{-\left(x-\frac{y}{2} i\right)^2} d x \\ = & e^{-\frac{y^2}{4}} \Re\left[\frac{\sqrt{\pi}}{2}\left(1+i \operatorname{erfi}\left(\frac{y}{2}\right)\right)\right] \\ = & \frac{\sqrt{\pi}}{2} e^{-\frac{y^2}{4}} \end{aligned} $$
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Feynman’s Trick
Let’s consider the parameterized integral $$ I(a)=\int_0^{\infty} \frac{\sin (a x) e^{-x^2}}{x} d x $$ Then differentiating $I(a)$ w.r.t. $a$ yields $$ \begin{aligned} I^{\prime}(a) & =\int_0^{\infty} \cos (a x) e^{-x^2} d x \\ & =\Re \int_0^{\infty} e^{a x i} e^{-x^2} d x \\ & =\Re\int_0^{\infty} e^{-\left(x-\frac{a}{2} i\right)^2-\frac{a^2}{4}} d x \\ & =e^{-\frac{a^2}{4}}\Re \int_0^{\infty} e^{-\left(x-\frac{a}{2} i\right)^2} d x \\ & =e^{-\frac{a^2}{4}} \Re\left[\frac{\sqrt{\pi}}{2}\left(1+i \operatorname{erfi}\left(\frac{a}{2}\right)\right)\right] \\ & =\frac{\sqrt{\pi}}{2} e^{-\frac{a^2}{4}} \end{aligned} $$ Integrating $I’(a)$ from $a=0$ to $1$ yields $$ \begin{aligned} \int_0^{\infty} \frac{\sin (x) e^{-x^2}}{x} d x =I(1)-I(0) & =\int_0^1 \frac{\sqrt{\pi}}{2} e^{-\frac{a^2}{4}} d a =\frac{\pi}{2} \operatorname{erf}\left(\frac{1}{2}\right) \end{aligned} $$
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.