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The question is about Laplace Transform and the inverse transform formula.

Can the inverse transform formula be proved using Cauchy's integral formula?

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Context please? –  Jonas Teuwen May 5 '11 at 17:48
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Cauchy's integral formula only applies to holomorphic functions, whereas the Laplace transform works on much more general functions, so I don't see how this would work without some nontrivial extra steps. –  Qiaochu Yuan May 5 '11 at 18:31
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The thing is, Laplace transform is an analytic function (at least in some domain), even if the previous function wasn't. I can apply Cauchy's integral formula to actually compute the inverse transform integral, but only for specific cases (and not for a general case). –  Maxim May 6 '11 at 0:03

1 Answer 1

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}

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