# Convolution of Gaussian with exponential decay?

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 Math exchange question above 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? SOLVED

REAL QUESTION: And by the way - am I doing this right? I have a lot of data for particle momenta, creation and end vertices: from these I'm calculating the rest lifetime for each particle event and plotting it in a histogram. The tail does look like an exponential decay but close to $t =0$ there's a Gaussian-like behaviour. If we assume that the detector resolution will smear the data with a Gaussian, the final plot of lifetimes should look like the convolution of an exponential decay with a Gaussian, right?