# Inverse Laplace transform question

#### dwsmith

##### Well-known member
With a Laplace transform, we can remember common set ups; for example,
$\mathcal{L}\{e^{-at}\} = \frac{1}{s + a}.$
When it comes to the inverse Laplace transform, I can only find the tables to remember in a book. However, if we go back to the Laplace transform, we can always do
$\int_0^{\infty}f(t)e^{-st}dt$
to determine the transform with out a table. Is there an inverse analog? What if I can't remember the inverse Laplace of $$\mathcal{L}^{-1}\big\{\frac{s}{s^2 + a^2}\big\}$$? Can I work it like I could with Laplace transform?

#### ThePerfectHacker

##### Well-known member
With a Laplace transform, we can remember common set ups; for example,
$\mathcal{L}\{e^{-at}\} = \frac{1}{s + a}.$
When it comes to the inverse Laplace transform, I can only find the tables to remember in a book. However, if we go back to the Laplace transform, we can always do
$\int_0^{\infty}f(t)e^{-st}dt$
to determine the transform with out a table. Is there an inverse analog? What if I can't remember the inverse Laplace of $$\mathcal{L}^{-1}\big\{\frac{s}{s^2 + a^2}\big\}$$? Can I work it like I could with Laplace transform?
Yes, there is an integral formula for the inverse Laplace transform known as Mellin Transform.
It is not exactly pleasant to compute,
$$L^{-1}[ F(s) ] = \frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} F(s) e^{ts} ~ ds$$

Note, there are conditions on the function $F(s)$ but let us not worry about that. You also need to understand how the integral on RHS is interpreted. And in order to evaluate the integral on RHS you typically use residue calculus from complex analysis. This is why it is not exactly pleasant to compute inverse transforms using an integral.