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I need to convolve an exponential decay (defined as the exponential $Ae^{-\lambda t}$ from $0$ to $+\infty$) with a Gaussian of known standard deviation $\sigma$, in other words I need to compute the following integral:

$$ g(\tau) = \int_{0}^\infty \exp(-\lambda t) \exp\left(-\frac{(t-\tau)^2}{2\sigma^2} \right)\mathrm d t $$

which is almost the same integral as in this question, but with a $0$ as the lower limit.

The answer to the above question does not seem to apply here, or at least not for the whole range of the convolution: naively I would expect an exponential increase up to $\tau=0$, then a Gaussian-like peak and finally an exponential decay for large $\tau$.

Anyway know how to get the full mathematical expression?

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You're right that the integral should be from 0 to $\infty$, and this is precisely what is done in my answer (see Jul 20, 2014 answer to https://math.stackexchange.com/a/872561/165312 by TimeVariant.)

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  • $\begingroup$ Be aware that while your decay equation is on defined for $t>=0$, your convolved version will necessarily be defined for all $\tau$ including negative values. In certain physics problems, this causes a causality issue. $\endgroup$ – Jerry Guern Oct 18 '15 at 1:41

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