Q:

About the integral $\int_\mathbb{R} e^{-x^2 -(x-\xi)^2}\,dx$

everyone.
While solving a PDE, I used Poisson's formula for the diffusion equation, which eventually gave me an integral:
$$\frac{1}{4\sqrt{\pi t}}\int\limits_{-\infty}^{\infty} e^{-\xi^2}e^{\frac{-(x-\xi)^2}{4t}}d\xi$$
where $t$ and $x$ are parameters.
(This is essentially a convolution of a gaussian with another gaussian, shifted by $x$ and rescaled by $4t$.)
I kind of struggle to take this integral.
Any help would be appreciated.

A:

Complete square, translate and use the fact that $\int e^{-x^2} dx = \sqrt{\pi}.$
First, notice that $x^2+(x-\xi)^2= \left(\sqrt{2}x-\frac{\xi}{\sqrt{2}} \right)^2+\frac{\xi^2}{2}$ ("complete the square"). Then, $e^{-x^2-(x-\xi)^2}=e^{-\frac{\xi^2}{2}}e^{-\left(\sqrt{2}x-\frac{\xi}{\sqrt{2}} \right)^2}.$ Integrate now,
$$\begin{align}
\int\limits_{\Bbb R} dx e^{-x^2-(x-\xi)^2} &= e^{-\frac{\xi^2}{2}}\int\limits_{\Bbb R} dx e^{-\left(\sqrt{2}x-\frac{\xi}{\sqrt{2}} \right)^2} \\
&= \dfrac{e^{-\frac{\xi^2}{2}}}{\sqrt{2}} \int\limits_{\Bbb R} dx e^{-x^2}\\
&= \sqrt{\dfrac{\pi}{2}} e^{-\frac{\xi^2}{2}}.
\end{align}$$

