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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: $$(f*g)(t)=\int_{-\infty}^\infty f(\tau)\,g(t-\tau)\,\mathrm{d}\tau=\int_{-\infty}^0 f(\tau)\,g(t-\tau)\,\mathrm{d}\tau+\int_{0}^\infty f(\tau)\,g(t-\tau)\,\mathrm{d}\tau=0+\int_{0}^\infty f(\tau)\,g(t-\tau)\,\mathrm{d}\tau$$

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  • $\begingroup$ What do you mean by "even $f(t)$ is zero, the integral is not"? $\endgroup$ – MSobak Aug 27 '18 at 15:58
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    $\begingroup$ Maybe this will help. If $y<0$, then $f(y)=0$, so also $f(y)g(t-y)=0$, so $\int_{-\infty}^0 f(y)g(t-y)\text{d}y = 0$ $\endgroup$ – Jakobian Aug 27 '18 at 15:58
  • $\begingroup$ @Sobi at t<0, g(t) still overlap with the positive part of f(t), note that the integration is not with respect to t. $\endgroup$ – 7E10FC9A Aug 27 '18 at 16:11
  • $\begingroup$ @7E10FC9A But you're integrating with respect to $\tau$, and you have $f(\tau)$ in your integral. So when $\tau < 0,$ then $f(\tau) = 0$ and the integral is $0$. $\endgroup$ – MSobak Aug 27 '18 at 16:12
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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 $\tau < 0$.

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  • $\begingroup$ Yes, you are right. Thx. $\endgroup$ – 7E10FC9A Aug 27 '18 at 16:35

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