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?


1 Answer 1


Looks like the Laplace transform of a Gaussian function, which is well known, e.g. done here, and here. You will have to expand the square to make use of the identities, of course.

  • $\begingroup$ Thanks, so that's solved. What about the physical interpretation question? $\endgroup$
    – SuperCiocia
    Dec 9, 2014 at 21:00
  • $\begingroup$ I don't really have a good feel for what you are measuring, but probably because I haven't had time to give it enough thought. Are you sure you don't need a Poissonnian distribution? $\endgroup$ Dec 9, 2014 at 22:02

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