I want to calculate the convolution $F * G$ of two Gaussian functions without resorting to Fouritertransforms:
$F(t) := \exp(-at^2), G(t) := \exp(-bt^2) \qquad a,b>0$
But intuitively I expected the convolution to result again in a non constant function. Can anyone find my mistake / confirm that this calculation is correct?
Let $\Omega = \mathbb R$, then
$\begin{align*} (F*G)(x) &= \int_\Omega F(t)G(x-t)dt &\\ & = \int_\Omega e^{-at^2-b(x-t)^2} dt \qquad\qquad \quad \text{substitute }u = t+\frac{1}{2} x \implies "du=dt" \\ &=\int_\Omega e^{-a(u-\frac{1}{2}x)^2-b(\frac{1}{2}x-u)^2}dt \qquad \text{substitute }v = u-\frac{1}{2} x \implies "du=dv"\\ &=\int_\Omega e^{-(a+b)v^2} dv \qquad\qquad \qquad\,\, \text{substitute } w = \sqrt{a+b}v \implies"dw = \sqrt{a+b}dv" \\ &=\frac{1}{\sqrt{a+b}}\int_\Omega e^{-w^2}dw \\ &=\frac{\sqrt{\pi}}{\sqrt{a+b}} \end{align*}$
1 Answer
You seem to have lost the constant term modified by the completion of the square: $$ \begin{align} \int_{-\infty}^\infty e^{-a(x-t)^2}e^{-bt^2}\,\mathrm{d}t &=\int_{-\infty}^\infty e^{-ax^2+2axt-at^2-bt^2}\,\mathrm{d}t\\ &=e^{-ax^2+\frac{a^2x^2}{a+b}}\int_{-\infty}^\infty e^{-\frac{a^2x^2}{a+b}+2axt-(a+b)t^2}\,\mathrm{d}t\\ &=e^{\frac{-abx^2}{a+b}}\int_{-\infty}^\infty e^{-(a+b)\left(t-\frac{ax}{a+b}\right)^2}\,\mathrm{d}t\\ &=\sqrt{\tfrac\pi{a+b}}\,e^{\frac{-abx^2}{a+b}} \end{align} $$
