$$\int_{-\infty}^\infty \!\!\text{d} r\dfrac{1}{r}e^{\frac{-(r-r_0)^2}{\delta^2}}\sin(k r)$$
Where $\delta>0$, $r_0\in \mathbb{R}$.
Can anyone help me with this? it seems to me there has to be a closed expression. I can find it for the case $r_0=0$, but not for any real $r_0$.
I'm a bit desperate. I'd thank very much any help ;).
Hint: A trick is to note that $$2\frac{\sin(kr)}{r}=\int_{-k}^k\mathrm e^{\mathrm isr}\mathrm ds=2\Re\int_0^k\mathrm e^{\mathrm isr}\mathrm ds,$$ hence the integral to be computed is $$\Re\int_0^kJ(s)\mathrm ds,\qquad J(s)=\int_\mathbb R\mathrm e^{-(r-r_0)^2/\delta^2}\mathrm e^{\mathrm isr}\mathrm dr.$$ To compute $J(s)$, recall that, for every complex number $z$, $$\int_\mathbb R\frac1{\sqrt{2\pi}}\mathrm e^{-r^2/2}\mathrm e^{\mathrm izr}\mathrm dr=\mathrm e^{-z^2/2}.$$ If you try to follow these indications and still meet some difficulties, just say so.