I should prove the Frullani integral equality
$$
\int_{0}^{\infty} (1-e^{zx}) \frac{\beta}{x} e^{-\gamma x}dx = \beta \log \left(1- \frac{z}{\gamma}\right)
$$
for $z \in \mathbb{C}$ with non-positive real part.
I should first consider $z \leq 0$ and use
$$
\frac{e^{-\gamma x} - e^{-(\gamma - z)x}}{x}=\int_{\gamma}^{\gamma - z}e^{-y x}dy$$
and then change the order of integration. These steps are clear (see also Frullani integral for $f(x)=e^{-\lambda x }$).
But then I should use analytic extension to show that the formula is valid for $z \in \mathbb{C}$ with non-positive real part. I need help for this step.
Thank you in advance!
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