Why is it justifiable to use contour integration to find the inverse Laplace transform?

Question

I asked this on Quora, but I want to see what the answers here will be. I've always wondered why it is possible to represent the inverse Laplace transform as a contour integral.

Answer 1

I think you are a little misinformed. The ILT is an integral over a vertical line in the complex plane. It is not a closed contour integral by definition. Rather, we can use a closed contour integral in the complex plane, a piece of which is the vertical line along which the ILT is defined, to evaluate the ILT. Let me illustrate a bit. Let's say that we want to evaluate

$$f(t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, F(s) e^{s t} $$

What we do to make our lives easier, rather than parametrize the line and attempt a direct evaluation (which is messy), is to consider a closed contour integral in the complex plane:

$$\oint_C dz \, F(z) e^{z t} $$

where $C$ is what we call a Bromwich contour. Let's assume that $F$ is meromorphic and single-valued in the complex plane. Then when $t \gt 0$, $C$ consists of (a) the line segment $z=c+i y$ where $y \in\left [ -\sqrt{R^2-c^2}, \sqrt{R^2-c^2} \right ]$ and (b) the arc $z=R e^{i \theta}$, where $\theta \in \left [\frac{\pi}{2} - \arcsin{\frac{c}{R}},\frac{3 \pi}{2} + \arcsin{\frac{c}{R}} \right ]$. We consider the limit as $R \to \infty$. In this limit, is $F$ behaves appropriately, e.g., $\lim_{R \to \infty} |F(R)| = 0$, then the integral over the circular arc, i.e., part (b), vanishes. In this case, the contour integral is equal to the integral over the line in part (a). By the residue theorem, the contour integral is also equal to $i 2 \pi$ times the sum of the residues of the poles to the left of the line.

Thus, it is possible to say that the ILT is equal to $1/(i 2 \pi)$ times the contour integral because the integral over the arc vanishes.

ADDENDUM

The OP wants to know how we recover the original function from the inverse LT. This will be somewhat nontrivial but I hope it is a sufficient explanation. Let's begin by inserting the expression for the LT in the ILT integral:

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \left [\int_0^{\infty} dt' \, f(t') \, e^{-s t'} \right ] \, e^{s t} $$

We assume that $f$ is well-behaved enough that we can reverse the order of integration and get

$$ \int_0^{\infty} dt' \, f(t') \, \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{s (t-t')} $$

So we need to consider the ILT of $e^{s (t-t')}$. To do this, consider the integral

$$\oint_C dz \, e^{z (t-t')} $$

where $t-t' \ge 0$ and $c$ consists of the line $[c-i \sqrt{R^2-c^2},c+ i \sqrt{R^2-c^2}]$ and the circular arc $R e^{i \theta}$, $\theta \in \left [ \frac{\pi}{2} - \arcsin{\frac{c}{R}},\frac{3\pi}{2} + \arcsin{\frac{c}{R}} \right ]$, as illustrated in the Figure below.

The contour integral is thus equal to

$$\int_{c-i \sqrt{R^2-c^2}}^{c+i \sqrt{R^2-c^2}} ds \, e^{s (t-t')} + i R \int_{\frac{\pi}{2} - \arcsin{\frac{c}{R}}}^{\frac{3\pi}{2} + \arcsin{\frac{c}{R}}} d\theta \, e^{i \theta} \, e^{R (t-t') e^{i \theta}}$$

The delta function will emerge from this second integral.

The contour integral is equal to zero by Cauchy's theorem because there are no poles of the integrand inside the contour (or at all). Thus,

$$\begin{align}\int_{c-i \sqrt{R^2-c^2}}^{c+i \sqrt{R^2-c^2}} ds \, e^{s (t-t')} &= - i R \int_{\frac{\pi}{2} - \arcsin{\frac{c}{R}}}^{\frac{3\pi}{2} + \arcsin{\frac{c}{R}}} d\theta \, e^{i \theta} \, e^{R (t-t') e^{i \theta}}\end{align}$$

Now take the limit as $R \to \infty$ and shift the integral limits by $\pi/2$. We may then write down the ILT:

$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{s t} &= \lim_{R \to \infty}\frac{R}{i 2 \pi} \int_0^{\pi} d\theta \, e^{i \theta} \, e^{i R (t-t') e^{i \theta}}\\ &= \lim_{R \to \infty}\frac{R}{i 2 \pi} (-i) \int_0^{\pi} d(e^{i \theta}) \, e^{i R (t-t') e^{i \theta}}\\ &= \lim_{R \to \infty}\frac{R}{i 2 \pi} (-i) \frac1{i R (t-t')} \left [e^{i R (t-t') e^{i \theta}} \right ]_0^{\pi} \\ &= \lim_{R \to \infty}\frac{R}{i 2 \pi} (-i) \frac1{i R (t-t')} (-i 2 \sin{R (t-t')}) \\ &= \lim_{R \to \infty} \frac{\sin{R (t-t')}}{\pi (t-t')} \end{align} $$

The term on the RHS is equal to $\delta(t-t')$. Thus, the ILT becomes

$$\int_0^{\infty} dt' \, f(t') \delta(t-t') = f(t)$$

as expected.