Using complex analysis to find the Inverse Laplace transform 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?
 A: 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".
