Problem: find the inverse Laplace transform of

$$F(s) = \frac{s}{s-1}$$

Solution: Well, this one should be easy and one wonders why we are even bothering. Just split $F$ up as follows:

$$F(s) = \frac1{s-1} + 1$$

The ILT of $1/(s-1)$ is simply $e^t$, and the ILT of $1$ is $\delta(t)$. Done.

Or are we? Why indeed is the ILT of $1$ equal to $\delta(t)$? Well, what is traditionally stated is that

$$\int_0^{\infty} dt \, \delta(t) e^{-s t} = e^{-s (0)} = 1$$

This equation makes me nervous. I think it should make you nervous. We are sifting the exponential as instructed by the delta function at an integration endpoint. How valid is that really? Isn’t there a bit of indeterminacy built into the sifting based on more subtle considerations? For example, is the $0$ in the lower limit really a $0^+$, representing a one-sided limit? Without additional information about the nature of the integration used in the Laplace transform, I think that the above result is simply hand-waving and hindsight reasoning.

Fortunately, we do not need to worry about any of this. There is a way to derive this result directly from the definitions of the inverse Laplace transform and the delta function.

The clue is that $s/(s-1)$ doesn’t vanish as $|s| \to \infty$. That alone should put the problem solver on some alert regarding the existence of nicely-behaved analytic functions in the inverse transform.

However, let’s do some analysis to see where we can derive a delta function without knowing that we need to do a partial fraction decomposition to properly obtain the inverse LT. Consider the integral

$$\oint_C dz \frac{z}{z-1} e^{z t} $$

where $t \gt 0$ 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

$$\int_{c-i \sqrt{R^2-c^2}}^{c+i \sqrt{R^2-c^2}} ds \, \frac{s}{s-1} e^{s t} + i R \int_{\frac{\pi}{2} – \arcsin{\frac{c}{R}}}^{\frac{3\pi}{2} + \arcsin{\frac{c}{R}}} d\theta \, e^{i \theta} \frac{R e^{i \theta}}{R e^{i \theta}-1} e^{R t e^{i \theta}}$$

The delta function will emerge from this second integral.

The contour integral is also equal to $i 2 \pi$ times the residue of the integrand at the pole $z=1$, or $e^t$. Thus,

$$\begin{align}\int_{c-i \sqrt{R^2-c^2}}^{c+i \sqrt{R^2-c^2}} ds \, \frac{s}{s-1} e^{s t} &= i 2 \pi e^t – i R \int_{\frac{\pi}{2} – \arcsin{\frac{c}{R}}}^{\frac{3\pi}{2} + \arcsin{\frac{c}{R}}} d\theta \, e^{i \theta} \frac{R e^{i \theta}}{R e^{i \theta}-1} e^{R t e^{i \theta}}\\ &= i 2 \pi e^t – i R \int_{\frac{\pi}{2} – \arcsin{\frac{c}{R}}}^{\frac{3\pi}{2} + \arcsin{\frac{c}{R}}} d\theta \, e^{i \theta} \frac{e^{R t e^{i \theta}}}{1-\frac1{R e^{i \theta}}}\end{align}$$

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

$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \frac{s}{s-1} e^{s t} &= e^t + \lim_{R \to \infty}\frac{R}{i 2 \pi} \int_0^{\pi} d\theta \, e^{i \theta} \, e^{i R t e^{i \theta}}\\ &= e^t +\lim_{R \to \infty}\frac{R}{i 2 \pi} (-i) \int_0^{\pi} d(e^{i \theta}) \, e^{i R t e^{i \theta}}\\ &= e^t +\lim_{R \to \infty}\frac{R}{i 2 \pi} (-i) \frac1{i R t} \left [e^{i R t e^{i \theta}} \right ]_0^{\pi} \\ &= e^t+\lim_{R \to \infty}\frac{R}{i 2 \pi} (-i) \frac1{i R t} (-i 2 \sin{R t}) \\ &= e^t +\lim_{R \to \infty} \frac{\sin{R t}}{\pi t} \end{align} $$

The latter term on the RHS is equal to $\delta(t)$.

Although this seemed like a chore, it does illustrate how the delta function emerges from the inverse Laplace transform.

Dear Ron,

Could you look at this question?

http://math.stackexchange.com/questions/1491027/inverse-laplace-transform-of-a-hypergeometric-function