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 $(\mathcal L_Tf(z)-\mathcal Lf(z) )(1+\frac{z^2}{R^2})e^{Tz}$