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

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.
