Would the inverse Laplace transform of $\frac{e^{-s}}{s(s+a)}$ be $u(t-1)e^{-at}$? I'm stuck on a problem and I'm not sure if I'm approaching it correctly. I'f I'm trying to find the Inverse Laplace Transform of:
$$f(s) = \frac{e^{-s}}{s(s+a)}$$
I would think that the answer would be:
$$F(t) = u(t-1)e^{-at}$$
because $\frac{e^{-s}}{s}=u(t-1)$ and $\frac{1}{s+a}=e^{-at}$, but I feel like I'm wrong. Is my answer correct or am I doing something wrong?
 A: The inverse Laplace transform of a product of functions is not equal to the product of the inverse Laplace transforms of the functions in general. That is, given $\mathcal L[f(t)]=F(s),\mathcal L[g(t)]=G(s)$, we have $\mathcal L^{-1}[F(s)\cdot G(s)]\ne f(t)\cdot g(t)$. What you require is the convolution theorem, which states that the inverse Laplace transform of a product is equal to the convolution of the inverse Laplace transforms. That is,$$\mathcal L^{-1}[F(s)\cdot G(s)]=f(t)*g(t)=\int_0^tf(\tau)g(t-\tau)d\tau$$Thus$$\mathcal L^{-1}\left[\frac{e^{-s}}{s(s+a)}\right]=u(t-1)*e^{-at}\\=\int_0^tu(\tau-1)e^{-a(t-\tau)}d\tau\\=\begin{cases}0,&t<1\\\displaystyle\int_1^te^{a(\tau-t)}d\tau,&t>1&\end{cases}\\=\frac{u(t-1)[1-e^{a(1-t)}]}a$$Note that you could have alternatively found the answer by noting that$$\mathcal L[f(t-a)\cdot u(t-a)]=e^{-as}F(s)$$Thus, we first find the inverse Laplace transform of $\frac1{s(s+a)}=\frac1a\left(\frac1s-\frac1{s+a}\right)$, which is $f(t)=\frac1a(1-e^{-at})$. Since we also had $e^{-s}$ in the numerator, the given function must be the Laplace transform of $f(t-1)\cdot u(t-1)$, which is the same as obtained above.
