Convolute exponential with a gaussian 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.
 A: 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.
A: Note that the algebraic identity
$$\lambda t+\frac{(t-\tau)^2}{2\sigma^2}=\tau\lambda-\frac12\sigma^2\lambda^2+\frac{(t-\tau+\sigma^2\lambda)^2}{2\sigma^2}
$$
and the change of variable $s=t-\tau+\sigma^2\lambda$ yield
$$
\int_{-\infty}^{\infty}\exp\left(-\lambda t\right)\,\exp\left(-\frac{(t-\tau)^2}{2\sigma^2}\right)\mathrm dt=\exp\left(-\tau\lambda+\frac12\sigma^2\lambda^2\right)\cdot\int_{-\infty}^{\infty}\exp\left(-\frac{s^2}{2\sigma^2}\right)\mathrm ds,
$$
that is,
$$
g(\tau)=\sqrt{2\pi\sigma^2}\cdot\exp\left(-\tau\lambda+\frac12\sigma^2\lambda^2\right).
$$
This assumes that the function $g$ is defined as
$$
g(\tau) = \int_{-\infty}^\infty \exp(-\lambda t) \exp\left(-\frac{(t-\tau)^2}{2\sigma^2} \right)\mathrm d t,
$$
since the current formula in the question makes no sense (subscript $-$ in the integral, presumably instead of $-\infty$, $\mathrm d\tau$ to integrate a function of $t$, presumably instead of $\mathrm dt$).
