The question is about Laplace Transform and the inverse transform formula.
Can the inverse transform formula be proved using 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?
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}