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

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.

• Because you prove that you can. The proof is long and somewhat delicate, perhaps using the Riemann-Lebesgue lemma, etc, etc, and so you shouldn't expect anything more than a reference. Commented Feb 18, 2016 at 16:50
• Perhaps the question could be improved by focusing more narrowly on some aspect of the proof that you don't understand. $\qquad$ Commented Feb 18, 2016 at 16:59
• @MichaelHardy I have never seen a proof of this actually. Commented Feb 18, 2016 at 19:39
• @user119264 : But you are asking about an aspect of a particular kind of proof of this particular proposition. I haven't read a proof of this either. You're asking how to justify a particular thing, and not having read a proof myself, I'd want to know with some specificity what it is that is to be justified, before thinking about how to justify it. However, it seems perfectly plausible to me that contour integration would be used, for two reasons: (1) I've seen integrals of functions of a real variable done that way, and (2) the statement of this theorem involves $\ldots\qquad$ Commented Feb 18, 2016 at 20:33
• $\ldots\,{}$an integral along a path in the complex plane. $\qquad$ Commented Feb 18, 2016 at 20:33

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.

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.

• I found this answer helpful. But, I've realised that I did not state the question as clearly as possible. What I really meant to say is why it possible that this recovers the original function, that is, the function before the Laplace transform was applied? Commented Feb 19, 2016 at 17:00
• @user119264: Aha. OK, I will post an addendum to this post to answer your question. In the meantime, in this answer lies a preview of what I will be posting: math.stackexchange.com/questions/1277296/… Commented Feb 19, 2016 at 17:19
• How am I just now seeing this. Great answer as always Ron! Commented Mar 4, 2016 at 23:21