I have to prove that
$\beta \log(1-\frac{z}{\gamma})=\int_0^\infty (1-e^zx) \frac{\beta}{x} e^{-\gamma x} dx$, where $z\in \mathbb{C}$.
On they book they suggest to note that in $z= 0$ then
$\frac{e^{-(\gamma-z)x}-e^{-\gamma x}}{x}=\int^\gamma_{\gamma-z} e^{-yx} dy$. How can I conclude ?
Interchange the order of integration: $$ \begin{align} \int_0^{\infty}\int^\gamma_{\gamma-z} e^{-yx} \, dy \, dx &= \int^\gamma_{\gamma-z} \int_0^{\infty} e^{-yx} \, dx \, dy \\ &= \int_{\gamma-z}^{\gamma} \frac{dy}{y} \\ &= \log{\gamma}-\log{(\gamma-z)} \\ &= -\log{\left( 1-\frac{z}{\gamma} \right)}. \end{align} $$