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
$\sum_{-1}^{\infty}\frac{(-i)^{n}+(i)^{n}}{2(1+n)}\int_{0}^{\infty}x^{n}e^{-bx^{2}}dx=\sum_{-1}^{\infty}\frac{(-i)^{n}+(i)^{n}}{2(1+n)} \frac{\Gamma(\frac{n+1}{2})}{2b^{\frac{n+1}{2}}}.$

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.
