Frullani integral for $f(x)=e^{x}$ in a complex context 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!
 A: Consider the integral
$$f(\gamma,z)=\int_{0}^{\infty}  \frac{(1-e^{zx})}{x} e^{-\gamma x}\,dx\tag{1}$$
For $x\to0$ the integrand remains finite, for $x\to\infty$ convergence requires that 
$$\Re(\gamma)\gt0\tag{2a}$$ and $$\Re(\gamma - z)\gt0\tag{2b}$$.
Now the derivative with respect to $\gamma$ is
$$\frac{\partial f(\gamma,z)}{\partial\gamma}=-\int_{0}^{\infty}  (1-e^{zx}) e^{-\gamma x}\,dx\tag{3}$$
Notice that the $x$ in the denominator has vanished, and the resulting integral is elementary
$$\frac{\partial f(\gamma,z)}{\partial\gamma}=-\int_{0}^{\infty}  (1-e^{zx}) e^{-\gamma x}\,dx=-\frac{1}{\gamma }+\frac{1}{\gamma -z}\tag{4}$$
Now integrating with respect to $\gamma $ gives
$$ f(\gamma,z)= -\log(\gamma) + \log(\gamma - z) + C= \log\left(\frac{\gamma-z}{\gamma}\right)=\log\left(1-\frac{z}{\gamma}\right) \tag{5}$$
The constant $C$ was found to vanish considering the limit $\gamma \to \infty$.
Hence we have proven (apart from the trivial factor $\beta$) the relation of the OP under the conditions $(2)$.
$(5)$ then gives the analytic continuation to the complex $z$-plane.
A: Using the power series expansion
$$\frac{(1-e^{zx})}{x} = - \sum_{n=1}^\infty \frac{1}{n!} z^n x^{n-1}\tag{1}$$ 
the integral
$$f(\gamma,z)=\int_{0}^{\infty}  \frac{(1-e^{zx})}{x} e^{-\gamma x}\,dx\tag{2}$$
becomes after interchanging sum and integral 
$$f(\gamma,z)=- \sum_{n=1}^\infty \frac{1}{n!} z^n \int_{0}^{\infty} x^{n-1}  e^{-\gamma x}\,dx\tag{3}$$
Now 
$$\int_{0}^{\infty} x^{n-1}  e^{-\gamma x}\,dx = \frac{1}{\gamma^n} \Gamma(n)= \frac{1}{\gamma^n} (n-1)!\tag{4} $$
so that
$$f(\gamma,z)=- \sum_{n=1}^\infty \frac{1}{n!}(n-1)! \left(\frac{z}{\gamma}\right)^n = - \sum_{n=1}^\infty \frac{1}{n} \left(\frac{z}{\gamma}\right)^n= \log\left(1-\frac{z}{\gamma}\right)\tag{5}$$
QED.
