I'm trying to calculate the convolution between a Gaussian and a discontinuous function: $$(f*g)(t)=\int_{-\infty}^\infty f(\tau)\,g(t-\tau)\,\mathrm{d}\tau$$
Where
$$f(t\geq0)=\mathrm{exp}(-kt),\,f(t<0)=0$$
$$g(t)=\frac{1}{\sigma\sqrt(2\pi)}\mathrm{exp}(-\frac{1}{2}(\frac{t- \mu}{\sigma})^2)$$
I've managed to calculate the positive part by limiting the integral domain:
$$\int_0^\infty f(\tau)\,g(t-\tau)\,\mathrm{d}\tau =\frac{1}{2}\mathrm{exp}(-kt)\,\mathrm{exp}(\frac{k}{2}(k\sigma^2+2\mu))\,\left(\mathrm{erf}\left(\frac{t-\,k\,\sigma ^2-\,\mu}{\sqrt{2}\,\sigma }\right)+1\right)$$
Is there any way to calculate the negative part ?
Update:
$$(f*g)(t)=\int_{-\infty}^\infty f(\tau)\,g(t-\tau)\,\mathrm{d}\tau=\int_{-\infty}^0 f(\tau)\,g(t-\tau)\,\mathrm{d}\tau+\int_{0}^\infty f(\tau)\,g(t-\tau)\,\mathrm{d}\tau=0+\int_{0}^\infty f(\tau)\,g(t-\tau)\,\mathrm{d}\tau$$
You have $$ (f*g)(t) = \int_{-\infty}^\infty f(\tau)g(t-\tau) \, d\tau = \int_{0}^\infty f(\tau)g(t-\tau) \, d\tau + \int_{-\infty}^0 f(\tau)g(t-\tau) \, d\tau. $$ You already computed the first integral, and the second integral is $0$, since $f(\tau) = 0$ for all $\tau < 0$.