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.
 A: Integrating by parts we get $-\frac 1 {2a^{2}} e^{-a^{2}x^{2}} \sin(px)|_0^{\infty}+ \frac p {2a^{2}}\int_0^{\infty}  e^{-a^{2}x^{2}} cos(px)dx$. The first term is $0$ and the second term is the real part of a Gaussian characteristic function up to a constant factor. 
The answer is $\frac {p\sqrt{\pi}} {4a^{3}} e^{-p^{2}/2a^{2}}$
A: 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
$$-{I}'\left( p \right)=-\frac{d}{dp}\left( \frac{\sqrt{\pi }}{a}\exp \left( -\frac{{{p}^{2}}}{4{{a}^{2}}} \right) \right)$$
A: $$
\begin{align}
\int_0^\infty x\sin{px}\exp{(-a^2x^2)}\ dx
& =\int_0^\infty x\sum_{n\geq0}\dfrac{(-1)^n(px)^{2n+1}}{(2n+1)!}\exp{(-a^2x^2)}\ dx \\
& =\sum_{n\geq0}\dfrac{(-1)^np^{2n+1}}{(2n+1)!}\int_0^\infty x^{2n+2}\exp{(-a^2x^2)}\ dx \\
& =\sum_{n\geq0}\dfrac{(-1)^np^{2n+1}}{(2n+1)!}\dfrac{1}{2a^{2n+3}}\int_0^\infty u^{n+\frac12}e^{-u}\ du ~~~;~~~{a^2x^2=u}\\
& =\sum_{n\geq0}\dfrac{(-1)^np^{2n+1}(2n+2)}{\Gamma(2n+3)}\dfrac{1}{2a^{2n+3}}\Gamma(n+\frac32)\\
& =\sum_{n\geq0}\dfrac{(-1)^np^{2n+1}(2n+2)}{2^{2n+2}\sqrt{\pi}^{-1}\Gamma(n+2)\Gamma(n+\frac32)}\dfrac{1}{2a^{2n+3}}\Gamma(n+\frac32)\\
& =\sqrt{\pi}\sum_{n\geq0}\dfrac{(-1)^np^{2n+1}}{2^{2n+2}}\dfrac{1}{2a^{2n+3}}\\
& =\sqrt{\pi}\dfrac{p}{4a^3}\sum_{n\geq0}\left(\dfrac{-p^2}{4a^2}\right)^n\dfrac{1}{n!}\\
& =\sqrt{\pi}\dfrac{p}{4a^3}\exp\left(\dfrac{-p^2}{4a^2}\right)
\end{align}
$$
