I am looking at an old exam about Laplace transforms which starts with the following exercise:
Let $a \in \mathbb{R}$ and $\mathcal{L}$ denote the Laplace transform.
Compute the inverse Laplace transform of $$\frac{1}{s(s-a)^2}$$ using
the convolution theorem.
Firstly, I note that $\mathcal{L}[1](s) = \frac{1}{s}$ and $\mathcal{L}[t] = \frac{1}{s^2}$. Using the contraction property, we have that
$$\mathcal{L}[e^{at} t](s) = \mathcal{L}[t](s-a).$$
Therefore,
$$\frac{1}{s(s-a)^2} = \mathcal{L}[1](s) \cdot \mathcal{L}[e^{at}t](s) = \mathcal{L}[1 * e^{at} t](s)$$
and thus
$$\mathcal{L}^{-1}\left(\frac{1}{s(s-a)^2}\right) = 1 * e^{at} t.$$
However,
$$1*e^{at} t = \int_{-\infty}^\infty e^{a \tau} \tau \, d\tau,$$
which firstly will not depend on $t$ anymore and secondly doesn't converge.
Wolframalpha claims the inverse Laplace transform is given by
$$\frac{at e^{at} – e^{at} + 1}{a^2}.$$
What is wrong with my calculation?
Best Answer
The Laplace transform is defined only for functions $f: \mathbb{R}_{\geq 0} \to \mathbb{C}$. Therefore, if I write an expression such as $$\mathcal{L}[e^{at} t](s),$$ implicitly, $t \mapsto e^{at} t$ is only defined on $\mathbb{R}_{\geq 0}$ (or defined on $\mathbb{R}$ but $0$ for $t < 0$). On that domain, the convolution reduces to $$\int_0^t f(\tau) g(t-\tau) \, d\tau,$$ which for this exercise yields the correct solution. Thanks to daulomb's comments which pointed me to the right direction.