Computing $\int_0^\infty \frac{\sin(u)}{u}e^{-u^2 b} \, du$ I want to compute
$\int_0^\infty u^{-1}(1-e^{\frac{-u^2 t}{2}})\sin(u(|x|-r))\,du$ and so ,as shown below, I want to compute 
$$\int_0^\infty \frac{\sin(u)}{u}e^{-u^2 b} \, du$$
Attempt
We split the first integral into two:


*

*$\displaystyle \int_0^\infty u^{-1} \sin(u(|x|-r))\, du=\frac{\pi}{ 2}$

*$\displaystyle \int_0^\infty u^{-1} e^{\frac{-u^2 t}{2}}\sin(u(|x|-r))\,du = \int_0^\infty \frac{\sin(z)}{z} e^{-z^2 b} \, dz$
where $b=\dfrac{t}{2(|x|-r)^2}$ is a positive real constant.
any suggestions?
How to show that the value is $\frac{\pi}{2}erf(\frac{1}{2\sqrt{b}})$?
Given that I can swap integral and sum we have
$\sum_{-1}^{\infty}\frac{(-i)^{n}+(i)^{n}}{2(1+n)}\int_{0}^{\infty}x^{n}e^{-bx^{2}}dx=\sum_{-1}^{\infty}\frac{(-i)^{n}+(i)^{n}}{2(1+n)} \frac{\Gamma(\frac{n+1}{2})}{2b^{\frac{n+1}{2}}}.$
 A: Are you wanting just an answer or the full solution.  Assuming $b$ is positive and Real, Mathematica says:
$$
\int_0^\infty \frac{\sin{u}}{u}e^{-u^2 b}du\quad= \quad\frac{1}{2}\pi \,Erf(\frac{1}{2\sqrt{b}})
$$
Where $Erf$ is the error function.
A: Hint. Here is an approach.
Let $b$ be a real number such that $b>0$ and set 
$$
f(b):=\int_0^\infty \frac{\sin{u}}{u}e^{-u^2 b}du. \tag1
$$ We are allowed to write that
$$
\begin{align}
f'(b) &=-\int_0^\infty u\sin{u}\: e^{-u^2 b}du \tag2 \\\\
&=\frac{1}{2b}\left[\sin{u}\:e^{-u^2 b}\right]_{0}^{\infty} -\frac{1}{2b}\int_0^\infty \cos{u}\: e^{-u^2 b}du\\\\
&=0-\frac{1}{4b}\Re \int_{-\infty}^\infty  e^{-u^2 b-iu}du\\\\
&=-\frac{e^{-\frac{1}{4b}}}{4b}\Re \int_{-\infty}^\infty  e^{-\left(\sqrt{b}\:u+\dfrac{i}{\sqrt{b}}\right)^2}du\\\\
&=-\frac{e^{-\frac{1}{4b}}}{4b} \int_{-\infty}^\infty  e^{-U^2}\frac{dU}{\sqrt{b}}\\\\
&=-\frac{e^{-\dfrac{1}{4b}}}{4b}\frac{\sqrt{\pi}}{\sqrt{b}} \tag3\\\\
\end{align}
$$ and, since from $(1)$ we have $\displaystyle \lim\limits_{b\to +\infty}f'(b)=0 $, then from $(3)$ we obtain
$$
\begin{align}
f(b) &=-\frac{\sqrt{\pi}}{4}\int_b^\infty \frac{e^{\large -\frac{1}{4\:t}}}{t\sqrt{t}}dt \tag4 \\\\
&=\sqrt{\pi}\int_0^{\large \frac{1}{2\sqrt{b}}} e^{-x^2}dx \qquad \left(x:=\frac{1}{2\sqrt{t}},\quad dx:=-\frac{1}{4\:t\sqrt{t}}dt\right)\\\\
&=\frac{\pi}{2}{\rm{erf}}\left(\frac{1}{2\sqrt{b}}\right)
\end{align}
$$ as desired.
A: Let $I(a)=\displaystyle\int_0^\infty\cos(au)~e^{-bu^2}~du$, and evaluate it using Euler's formula, completing the square, and the value of the Gaussian integral. Then express your integral in terms of $\displaystyle\int I(a)~da$ and let $a=1$. The final result will be a simple error function in b.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}{\sin\pars{u} \over u}\,\expo{-u^{2}b}\,\dd u}
=\half\int_{-\infty}^{\infty}\expo{-u^{2}b}\,{\sin\pars{u} \over u}\,\dd u
=\half\int_{-\infty}^{\infty}\expo{-u^{2}b}\
\overbrace{\half\int_{-1}^{1}\expo{\ic k u}\,\dd k}
^{\ds{=\ \dsc{\sin\pars{u} \over u}}}\
\,\dd u
\\[5mm]&={1 \over 4}\int_{-1}^{1}
\int_{-\infty}^{\infty}\exp\pars{-b\bracks{u^{2} - {\ic k \over b}u}}
\,\dd u\,\dd k
\\[5mm]&={1 \over 4}\int_{-1}^{1}
\int_{-\infty}^{\infty}
\exp\pars{-b\braces{\bracks{u - \ic\,{k \over 2b}}^{2} + {k^{2} \over 4b^{2}}}}
\,\dd u\,\dd k
\\[5mm]&={1 \over 4}\int_{-1}^{1}\exp\pars{-\,{k^{2} \over 4b}}
\int_{-\infty - \ic k/\pars{2b}}^{\infty - \ic k/\pars{2b}}\exp\pars{-bu^{2}}
\,\dd u\,\dd k
\\[5mm]&=\half\,\pars{2\root{b}}\int_{0}^{1/\pars{\root{2}b}}\exp\pars{-k^{2}}
{1 \over \root{b}}\ \overbrace{\int_{-\infty}^{\infty}\exp\pars{-u^{2}}\,\dd u}
^{\dsc{\root{\pi}}}\ \,\dd k
\\[5mm]&=\dsc{\root{\pi}}{\root{\pi} \over 2}\bracks{{2 \over \root{\pi}}
\int_{0}^{1/\pars{\root{2}b}}\exp\pars{-k^{2}}\,\dd k}
=\color{#66f}{\large{\pi \over 2}\,\,{\rm erf}\pars{1 \over \root{2b}}}
\end{align}
