I have data measuring an exponential decay that is convoluted by a gaussian response function.
I have the measured shape of the gaussian, and want an analytical expression for the exponential post-convolution that I can use to compare to the data.
I need to calculate the following, but am having trouble.
$g(\tau) = \int_-^\infty \exp(-\lambda t) \exp(-\frac{(t-\tau)^2}{2\sigma^2} ) d \tau$
Where $\sigma$ is known.
$g(\tau) = \int_-^\infty \exp(-\lambda t -\frac{t^2}{2\sigma^2} +\frac{t \tau}{\sigma^2}) \exp(-\frac{\tau^2}{2\sigma^2} ) d \tau$
The last term looks like the Error function, but Im not sure is it.
Best Answer
The answers are all helpful so far, but the original question was on the right track: there is indeed an error function that results. The reason is that the original integral, which Henry correctly pointed out is over t, not $\tau$, is from 0 to $\infty$, and not from $-\infty$ to $\infty$. This changes everything.
In what follows, I took the liberty of using an exponential and a Gaussian both of which are individually normalized to 1 when integrated over their full range. For instance, the original exponential must be multiplied by a $\lambda$. Similarly, the Gaussian is to be divided by $\sigma\sqrt{2\pi}$.
When you work it out, you see that the correct answer is, in this notation, $$ {\lambda\over 2}e^{\sigma^2\lambda^2\over 2}e^{-\lambda\tau} \left(1 - \hbox{erf}\left(\overline \tau\over\sigma\sqrt{2}\right)\right) $$ where $\overline \tau\equiv -(\tau - \sigma^2/\lambda)$.
It is easy to see that this gives the right answer, for instance, by taking the limit in which $\sigma\to 0$. In this limit the "erf" function is either -1 or +1. [Wikipedia gives a good definition of erf() with a nice plot.] And then you recover the exponential, as you should.