Laplace inverse transform formula and Cauchy's integral formula The question is about Laplace Transform and the inverse transform formula.
Can the inverse  transform formula be proved using Cauchy's integral formula?
 A: I wish I could draw figures here (can I with "tikz"?) or import. 
Let me start by describing the figure. Start at the Brownwich contour
from $\gamma - \mathrm{i}R $ to $\gamma + \mathrm{i} R$ (we want to
let $R \to \infty$). Then on the right hand side of the complex plane make a semicircle down back to $\gamma - \mathrm{i} R$. We know that all the singularities for the inverse Laplace transform are at the left of the contour. Here we apply the Cauchy theorem as follows:
\begin{equation}
- 2 \pi \mathrm{i} F(s) = \int_{C_-} \frac{F(z)}{z -s} d z
= \int_{\gamma - \mathrm{i} R}^{\gamma + \mathrm{i} R}  
\frac{F(z)}{z - s} dz + \int_{C_R} \frac{F(z)}{z - s}  dz
\end{equation}
The minus "-" in front is because were are going clockwise. The whole contour is $C_{-}$ and and $C_R$ is the semicircular arc. 
We show that the last integral goes to zero as the radius of the
circle goes to infinity. We first parameterize the contour which
is centered at $z_0=(\gamma , 0)$, as
\begin{equation}
C_R = \{  z : z - z_0 = R \mathrm{e}^{\mathrm{i} \theta} \;
, \; \pi/2 > \theta > - \pi/2  \}.
\end{equation}
where $dz = \mathrm{i} \, R \, \mathrm{e}^{i} \,  d \theta$.
In the denominator we have
\begin{equation}
  z - s = z - z_0 + z_0 -s = R \mathrm{e}^{\mathrm{i} \theta }
  + z_0 -s
\end{equation}
and dividing numerator and denominator by $R$ we find
\begin{equation}
 \int_{C_R} \frac{F(z)}{z - s}  dz
 = \int_{-\pi/2}^{\pi/2} \frac{\mathrm{i}
   \mathrm{e}^{\mathrm{i} \theta} F(z_0 + R \mathrm{e}^{\mathrm{i} \theta})}
 {\mathrm{e}^{\mathrm{i} \theta} + (z_0 -s)/R} d \theta.
\end{equation}
We see that as $R \to \infty$ the absolute value of the numerator goes to zero.
That is,
\begin{equation}
\lim_{R \to \infty} | \mathrm{i} \mathrm{e}^{\mathrm{i} \theta} F(z_0 + R \mathrm{e}^{i \theta}) | = \lim_{R \to \infty} | F(z_0 + R \mathrm{e}^{i \theta})| =  0.
\end{equation}
and the denominator goes to $\mathrm{e^{\theta }}$ which is always
a number with size $1$. Hence the integral over the arc of circle $C_R$
goes to zero.  We have so far that
\begin{equation}
  F(s) = \frac{1}{2 \pi \mathrm{i}} \int_{\gamma - \mathrm{i} \infty}^
  {\gamma + \mathrm{i} \infty} \frac{F(z)}{s - z} d z.
\end{equation}
and then we have the following chain of equations
\begin{equation}
f(t) = \mathcal{L}^{-1} F(s) = \mathcal{L}^{-1} 
  \frac{1}{2 \pi \mathrm{i}} \int_{\gamma - \infty}^{\gamma+\infty}
  \frac{F(z)}{s-z} dz
  = \frac{1}{2 \pi \mathrm{i}} \int_{\sigma-\mathrm{i} \infty}
  ^{\sigma + \mathrm{i} \infty} F(z) 
  \left [ \mathcal{L}^{-1} \left ( \frac{1}{s - z} \right ) \right ] dz 
\end{equation}
We know that:
\begin{equation}
\mathcal{L}^{-1} \left ( \frac{1}{s -z } \right ) = \mathcal{e}^{zt}.
\end{equation}
The reason the Laplace inverse can go inside the integral (Fubinis' rule)
is because being Analytic the integral converges uniformly. Please fill details and or correct me if I am wrong.
So:
\begin{equation}
  f(t)  = \frac{1}{2 \pi \mathrm{i}} \int_{\gamma-\mathrm{i} \infty}
  ^{\gamma + \mathrm{i} \infty} F(z) \mathrm{e}^{z t} dz.
\end{equation}
