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.

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