## Monday Math 118

The Laplace Transform
Part 19: Inverse Laplace Transform

Recall from here the Fourier transform
$g(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-\imath\omega{t}}\,dt$
and its inverse
$f(t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}g(\omega)e^{\imath\omega{t}}\,d\omega$.
Combining these, we see
$f(\tau)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty}f(t)e^{-\imath\omega{t}}\,dt\right]e^{\imath\omega{\tau}}\,d\omega$.
Now, letting $f(t)=0$ for t<0, the above becomes
$f(\tau)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\int_{0}^{\infty}f(t)e^{-\imath\omega{t}}\,dt\right]e^{\imath\omega\tau}\,d\omega$.

Now, suppose this f(t) has a Laplace transform F(s) with region of convergence $\Re(s)>\sigma$. Thus, for a real number c>σ, the line $\Re(s)=c$ lies entirely in the region of convergence. So let us make in the change of variables $s=c+\imath\omega$ in the above integral; then $ds=\imath\,d\omega$, so $d\omega=\frac{ds}{\imath}$, and we get, with some rearranging, that
$\begin{array}{rcl}f(\tau)&=&\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\int_{0}^{\infty}f(t)e^{-\imath\omega{t}}\,dt\right]e^{\imath\omega\tau}\,d\omega\\&=&\frac{1}{2\pi}\int_{c-\imath\infty}^{c+\imath\infty}\left[\int_{0}^{\infty}f(t)e^{(c-s)t}\,dt\right]e^{(s-c)\tau}\,\frac{ds}{\imath}\\f(\tau)&=&\frac{1}{2\pi\imath}\int_{c-\imath\infty}^{c+\imath\infty}\left[\int_{0}^{\infty}f(t)e^{-st}\,dt\right]e^{s\tau}\,ds\end{array}$,
and thus, the inverse of the Laplace transform is given by
$f(t)=\mathcal{L}^{-1}[F(s)]{t}=\frac{1}{2\pi\imath}\int_{c-\imath\infty}^{c+\imath\infty}F(s)e^{st}\,ds$,
where the contour of integration is over the line $\Re(s)=c$. This is known as the Bromwich integral. However, actually evaluating this integral is generally very difficult.