Inverse Laplace transform

At a basic level, the inverse Laplace transform doesn’t need to be computed by hand. We can consult a table of transforms and work backwards. By definition it’s given by:

$$f(t)={\mathcal{L}}^{-1}\{F(s)\}=\frac{1}{2\pi j}{\int }_{\sigma -j\infty }^{\sigma +j\infty }F(s)e^{st}ds$$

As with the forward transform, the inverse transform is linear:

$${\mathcal{L}}^{-1}\{\alpha F(s)+\beta G(s)\}=\alpha {\mathcal{L}}^{-1}\{F(s)\}+\beta {\mathcal{L}}^{-1}\{G(s)\}$$

In MATLAB, we can use ilaplace(exp).¹

Proof

By theorem, suppose $f(t)$ is continuous and $|f(t)|\le M{e}^{\alpha t}$ for some $M,\alpha$, i.e., is of exponential type $\alpha$. Then we can define $F(s)$ for $s\in \mathbb{C}$, more precisely:

$$F(z)={\int }_0^{\infty }{e}^{-zt}f(t)dt$$

is well defined and analytic for $\text{Re}(z)>\alpha$.

Then we say:

$$|{\int }_0^{\infty }{e}^{-zt}f(t)dt|\le M{\int }_0^{\infty }\left|e^{-zt}\right||f(t)|dt\le M{\int }_0^{\infty }{e}^{-\text{Re}(z)t}dt$$

This integral converges if $\text{Re}(z)>\alpha$. Moreover, one can show it is differentiable there:

$$F^{\prime }(z)={\int }_0^{\infty }(-t)e^{-zt}f(t)dt$$

We should think about $s$ as a complex plane, which we’ll sometimes call $z$. In particular, it lives in the shaded region and is analytic there: By theorem, suppose $f(t)$ (as in the previous theorem) is exponential and of exponential type $\alpha$, then we can recover $f(t)$ by integrating on a vertical line $\text{Re}(z)=\sigma$ from $\text{Im}(z)=\pm \infty$. Then:

$$f(t)=\frac{1}{2\pi i}{\int }_{\sigma -i\infty }^{\sigma +i\infty }{e}^{zt}F(z)dz$$

where $\sigma >\alpha$.

The formula is obtained as followed. Let ${\gamma }_R$ be the closed contour of the vertical line and a semicircle towards the right. Apply Cauchy’s integral formula to $F(z)$, which we’ve seen is analytic inside ${\gamma }_R$ if $\sigma >\alpha$. Then:

$$F(s)=-\frac{1}{2\pi i}{\int }_{{\gamma }_R}\frac{F(z)}{s-z}dz$$

From the intuitive understanding of the integral formula as a sum of many functions that look like:

$$\frac{1}{z-z_0}$$

Something converges. Lost around minute 25 of this lecture by Prof Nachman.

Footnotes

1. ”MATLAB has a powerful inverse Laplace transform engine.” - Prof Najm ↩