How to evaluate the following integral involving a gaussian? I want to evaluate the following integral:
$$\int\limits_0 ^\infty {x \sin{px} \exp{(-a^2x^2})} dx$$
Now I am unsure how to proceed. I know that this is an even function so I can extend the limit terms to $-\infty, \infty $ and then divide by 2. I have tried to evaluate this on Wolfram Alpha, but it only shows the answer while I am interested in the procedure.

Start with:
$$I\left( p \right)=\int_{0}^{\infty }{\cos \left( px \right)\exp (-{{a}^{2}}{{x}^{2}})dx}$$
We can use differentiation under the integral sign:
$${I}'\left( p \right)=-\int_{0}^{\infty }{x\sin \left( px \right)\exp (-{{a}^{2}}{{x}^{2}})dx}$$
Integration by parts using $u=\sin \left( px \right)\quad and\quad dv=-x\exp \left( -{{a}^{2}}{{x}^{2}} \right)dx$
$${I}'\left( p \right)=\left. \sin \left( px \right)\frac{\exp \left( -{{a}^{2}}{{x}^{2}} \right)}{2{{a}^{2}}} \right|_{0}^{\infty }-\frac{p}{2{{a}^{2}}}\int_{0}^{\infty }{\cos \left( px \right)\exp \left( -{{a}^{2}}{{x}^{2}} \right)dx}$$
The first term on the right vanishes, and  we have the first-order differential equation:
$$\frac{{I}'\left( p \right)}{I\left( p \right)}=-\frac{p}{2{{a}^{2}}}\Rightarrow \ln \left( I\left( p \right) \right)=-\frac{{{p}^{2}}}{4{{a}^{2}}}+C$$
Using  $$I\left( 0 \right)=\frac{\sqrt{\pi }}{a}$$
We can find $C=\ln \left( \frac{\sqrt{\pi }}{a} \right)$
hence
$$\ln \left( I\left( p \right) \right)=-\frac{{{p}^{2}}}{4{{a}^{2}}}+\ln \left( \frac{\sqrt{\pi }}{a} \right)$$
So
$$I\left( p \right)=\frac{\sqrt{\pi }}{a}\exp \left( -\frac{{{p}^{2}}}{4{{a}^{2}}} \right)$$
Finally the integral in question equals
$$-{I}'\left( p \right)=-\frac{d}{dp}\left( \frac{\sqrt{\pi }}{a}\exp \left( -\frac{{{p}^{2}}}{4{{a}^{2}}} \right) \right)$$
