# Laplace transform and Cauchy integral formula

cauchy-integral-formulacomplex-analysislaplace transform

Let $$f : \Bbb R_+ \to \Bbb C$$ be a bounded and continuous function such that its Laplace transform is holomorphic on all $$z \in \Bbb C$$ such that $$Re(z)>0$$ and has a meromorphic extension on $$\Bbb H_{-\delta}:=\{z \in \Bbb C \lvert Re(z)>-\delta\}$$ for some positive $$\delta$$. Let $$\mathcal L_Tf(z):=\int_0^T f(t)e^{-tz}dt$$. Let $$R>0$$, show that on $$\Omega_R = \Bbb H_{-\frac{\delta}2} \cap D(0,R)$$ $$\mathcal L_Tf(0)-\mathcal Lf(0) = \frac{1}{2\pi i} \oint_{\partial\Omega_R}(\mathcal L_Tf(z)-\mathcal Lf(z))\bigg(1+\frac{z^2}{R^2}\bigg)e^{Tz} \frac{dz}{z}$$

I thought of using Cauchy integral formula for $$\mathcal L_Tf(z)-\mathcal Lf(z)$$ but this would just give me that $$\mathcal L_Tf(0)-\mathcal Lf(0) = \frac{1}{2\pi i} \oint_{\partial\Omega_R}(\mathcal L_Tf(z)-\mathcal Lf(z)) \frac{dz}{z}$$. How does the $$\bigg(1+\frac{z^2}{R^2}\bigg)e^{Tz}$$ part appear ?

You probably meant $$\mathcal Lf(z)$$ has an holomorphic extension to (an open containing) $$\Re(z)\ge 0$$.
(as it fails with $$f(x)=1$$ such that $$\mathcal L_Tf(z)=1/z$$ has a pole at $$z=0$$)
Then given $$R$$ for $$\delta$$ small enough this is just the Cauchy integral formula (or residue theorem) for $$(\mathcal L_Tf(z)-\mathcal Lf(z) )(1+\frac{z^2}{R^2})e^{Tz}$$