Hi I am trying to compute the following integral that I have encountered in one of my calculations.
$$\int_\mathbb{R}{y e^{-\frac{y^2}{2}} e^{\frac{-(x-y)^2}{4t}} dy} $$
The main problem I am having with this is that the exponential term complicates everything and I would like a way to "take out" the dependence on $t$ and $x$ out of the integral but I am not really not sure on how to go about this.
First complete the square:
$$ \frac{y^2}{2}+\frac{(x-y)^2}{4t}=\frac{2ty^2+x^2-2xy+y^2}{4t} $$
$$ =\frac{(2t+1)y^2-(2x)y+x^2}{4t} $$
$$ = \frac{2t+1}{4t}\left(y^2-\frac{2x}{2t+1}y+\frac{x^2}{2t+1}\right) $$
$$ =\frac{2t+1}{4t}\left[\left(y-\frac{x}{2t+1}\right)^2-\frac{x^2}{(2t+1)^2}+\frac{x^2}{2t+1}\right] $$
$$ = \frac{2t+1}{4t}\left[\left(y-\frac{x}{2t+1}\right)^2+\frac{2tx^2}{(2t+1)^2}\right] $$
$$ \frac{2t+1}{4t}\left(y-\frac{x}{2t+1}\right)^2+\frac{x^2}{4t+2} $$
I'll use the substitutions $u=y-x/(2t+1)$ and $A^2=(2t+1)/(4t)$ so this is
$$ I=\int_{\mathbb{R}} \left(u+\frac{x}{2t+1}\right)\exp\left(-(Au)^2-\frac{x^2}{4t+2}\right)\,\mathrm{d}y $$
which becomes
$$ e^{x^2/(4t+2)} I=\int_{\mathbb{R}} ue^{-(Au)^2}\,\mathrm{d}u+\frac{x}{2t+1}\int_{\mathbb{R}} e^{-(Au)^2}\,\mathrm{d}u = \frac{x}{2t+1}\frac{\sqrt{\pi}}{A}$$
which becomes
$$ I=xe^{-x^2/(4t+2)} \sqrt{\frac{4\pi t}{(2t+1)^3}} $$