Two sided Laplace transform of convolution integral

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 $${\operatorname {Re} } \, s > -1$$.

• you are right I messed with the sign Dec 12 '19 at 21:11