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?


1 Answer 1


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

  • $\begingroup$ you are right I messed with the sign $\endgroup$
    – reuns
    Commented Dec 12, 2019 at 21:11

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