Integral $\int\limits_0^\infty\frac{e^{-Ak^{2}}}{k}\sin(kr)dk$

Question

I have the following integral

$$ f(r)=\int_{0}^{\infty}\frac{\exp(-Ak^{2})}{k}\,\sin(kr)\,\mathrm{d}k $$

with $A>0$ and $r>0$. I know from Wolfram that the result should be

$$ f(r)=\frac{\pi}{2}\text{erf} \left (\frac{r}{2\sqrt{A}} \right) $$

The problem is I have no clue how to obtain this result. I would be very curious how to do this by myself. I know that this is a sine transform, and I should be able to use a similar way like shown here Fourier transform of the error function, erf (x). But already when introducing the sgn function I get lost. Any hints are appreciated.

Answer 1

Hint. By setting $$ I(r):=\int_{0}^{\infty}\frac{e^{-Ak^{2}}}{k}\sin(kr)dk,\qquad A>0,\,r>0, $$one is allowed to differentiate under the integral sign with respect to $r$ to get $$ I'(r)=\int_{0}^{\infty}e^{-Ak^{2}}\cos(kr)dk,\qquad A>0,\,r>0, $$ that is $$ I'(r)=\frac12\sqrt{\frac{\pi}A}e^{\large-\frac{r^2}{4A}},\qquad A>0,\,r>0, $$ giving $$ I(r)=\frac{\pi}{2}\text{erf} \left (\frac{r}{2\sqrt{A}} \right),\qquad A>0,\,r>0. $$