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

Question

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.

Answer 1

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, $$\begin{align} \int\limits_{\Bbb R} dx e^{-x^2-(x-\xi)^2} &= e^{-\frac{\xi^2}{2}}\int\limits_{\Bbb R} dx e^{-\left(\sqrt{2}x-\frac{\xi}{\sqrt{2}} \right)^2} \\ &= \dfrac{e^{-\frac{\xi^2}{2}}}{\sqrt{2}} \int\limits_{\Bbb R} dx e^{-x^2}\\ &= \sqrt{\dfrac{\pi}{2}} e^{-\frac{\xi^2}{2}}. \end{align}$$

Answer 2

@Will M. provides help.

This is from Mathematica: $$ \int \exp \left(-x^2-(x-\xi )^2\right) \, dx = \frac{1}{2} \sqrt{\frac{\pi }{2}} e^{-\frac{\xi ^2}{2}} \text{erf}\left(\frac{2 x-\xi }{\sqrt{2}}\right) $$ Therefore $$ \int_{-\infty}^{\infty} \exp \left(-x^2-(x-\xi )^2\right) \, dx = \sqrt{\frac{\pi }{2}} e^{-\frac{\xi ^2}{2}} $$

As you noted, the shape is Gaussian. For this plot $\xi=-2$.