Laplace transform and Cauchy integral formula
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 $(\mathcal L_Tf(z)-\mathcal Lf(z) )(1+\frac{z^2}{R^2})e^{Tz}$