I'm trying to find the two sided Laplace transform of
$$ \int_{-\infty}^te^{-(t-\tau)-\tau^2} d\tau = \int_{-\infty}^te^{-(t-\tau)}e^{-\tau^2} d\tau $$
which seems to be some kind of convolution integral. I figured that I could apply the theorem
$$\mathcal{L}\{f \ast g \, (t)\} = \mathcal{L}\{f(t)\} \, \cdot \mathcal{L}\{g(t)\}$$
if I could just change the upper limit to $\infty$ (because of two sided laplace transform). The way I tried to do this was by subtracting the part of the integrand $>t$ using a step function, which gives me
$$ \int_{-\infty}^{\infty}e^{-(t-\tau)}e^{-\tau^2}\big(1-H(\tau-t)\big) \, d\tau. $$
This is were I am stuck as I can't see how to correctly apply the theorem here. It seems like it would work if I had $H(t-\tau)$ instead of $H(\tau-t)$.
Maybe someone can point out what I'm missing?
The key realization is that
$$ 1 - H(\tau-t) = 1 - H(-(t-\tau)) = H(t-\tau). $$
The integral thus becomes
$$ \int_{-\infty}^{t} e^{-(t-\tau)-\tau^2} d\tau = \int_{-\infty}^{\infty} H(t-\tau)e^{-(t-\tau)} e^{-\tau^2} \, d\tau. $$
So if $f(t) = H(t)e^{-t}$ and $g(t) = e^{-t^2}$, then the laplace transform of the integral above is
$$ \mathcal{L}\{f\ast g \, (t)\} = \mathcal{L}\{H(t)e^{-t}\} \, \cdot \, \mathcal{L}\{e^{-t^2}\} = \frac{1}{s+1} \sqrt{\pi}e^{s^2/4} $$
for ${\operatorname {Re} } \, s > -1$.