convolution computation involving $e^{-x^2}$

Question

In working a problem involving convolution, I have arrived at the following integral, but do not know how to compute it: $$2\int_0^{\infty}e^{-a(x-y)^2-by^2}dy$$ I thought that this integrand did not have an antiderivative. Based on the way the problem is formulated, however, there must be an actual solution. Thank you for your help!

Answer 1

Expand the argument of the exponential:

$$\begin{align}\int_0^{\infty} dy \: e^{-a (x-y)^2} e^{-b y^2} &= e^{-a x^2} \int_0^{\infty} dy \: e^{-(a+b)y^2 + 2 a x y}\end{align}$$

Now complete the square in the exponential:

$$\begin{align}(a+b) y^2 - 2 a x y &= (a+b) \left[y^2 - \frac{2 a x}{a+b} y + \left( \frac{a x}{a+b} \right )^2 \right ] - \frac{a^2 x^2}{a+b} \\ &=(a+b) \left(y - \frac{a x}{a+b} \right)^2- \frac{a^2 x^2}{a+b} \end{align}$$

so that the integral becomes

$$\begin{align}e^{-(a b/(a+b)) x^2} \int_0^{\infty} dy \: e^{-(a+b) \left(y - \frac{a x}{a+b} \right)^2} &= e^{-(a b/(a+b)) x^2} \int_{-\frac{a x}{a+b}}^{\infty} du \: e^{-(a+b) u^2}\\ &= \frac{e^{-(a b/(a+b)) x^2}}{\sqrt{a+b}} \int_{-\frac{a x}{\sqrt{a+b}}}^{\infty} dv \: e^{-v^2}\\ &= \frac{e^{-(a b/(a+b)) x^2}}{\sqrt{a+b}} \frac{\sqrt{\pi}}{2} \left[ 1+ \text{erf}\left (\frac{a x}{\sqrt{a+b}} \right ) \right ]\end{align}$$

where erf is the standard error function.

Answer 2

You have: $$ -a(x - y)^2 - b y^2 = - (a + b) \left(y - \frac{a x}{a + b} \right)^2 + \frac{a b x^2}{(a + b)^2} $$ The substitution $u = \sqrt{a + b} \left(y - \frac{a x}{a + b} \right)$ reduces this to an integral involving the error function.