Using complex analysis to find the Inverse Laplace transform

Question

I have been reviewing for my comprehensive graduation exam where I have been solving the Inverse Laplace transform via complex analysis. Consider $$ H(s) = \frac{s^2 - s + 1}{(s + 1)^2} $$ Then we have two poles at $s = -1$. $$ \frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{s^2 - s + 1}{(s + 1)^2}e^{st}ds = \sum\text{Res}(F(s)e^{st}) $$ I get $$ \lim_{s\to -1}[2s - 1 + t(s^2 - s + 1)]e^{st} = -3e^{-t} + 3te^{-t} $$ but if I solve the problem by partial fractions and tables, I also pick up Dirac delta; that is, the solution should be $$ \delta(t) -3e^{-t} + 3te^{-t} $$ How did I lose a Dirac delta function?

Answer 1

This is not so much a proof as a demonstration. We'll take the Bromwich integral and stick a constant 1 in it. We have the Inverse Laplace transform as

$$ F(t) = \frac{1}{2 \pi i} \int_{\gamma-i \infty}^{\gamma+i \infty} e^{st} f(s) ds $$

where $\gamma$ is a real number that must be chosen so that all the poles of $f(s)$ lie to the left of the vertical contour it defines.

Set $f(s)=1$, and we can take $\gamma=0$ for convenience since our function has no poles. I'll introduce a limit in the integration limits, $p$. You'll then have

$$ F(t) = \frac{1}{2 \pi i} \lim_{p \rightarrow \infty} \int_{-i p}^{+i p} e^{st} ds \\ = \frac{1}{2 \pi i} \lim_{p \rightarrow \infty} \left. \frac{e^{st}}{t} \right|_{-ip}^{ip} \\ = \frac{1}{2 \pi i} \lim_{p \rightarrow \infty} \left( \frac{e^{ipt}-e^{-ipt}}{t} \right) \\ = \frac{1}{2 \pi i t} \lim_{p \rightarrow \infty} 2 i \sin(pt) \\ = \lim_{p \rightarrow \infty} \frac{\sin pt}{\pi t} $$

Now one of the "definitions" of the Dirac-delta function as a "limit" is $$ \lim_{e \rightarrow 0} \frac{1}{\pi x} \sin \left( \frac{x}{e} \right) $$

Which is precisely our result, if you set $e=1/p$. So if you assume the integral makes sense, the inverse Laplace transform of 1 should be the Dirac delta "function".