Frullani integral for $f(x)=e^{x}$ in a complex context

Question

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!

Answer 1

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.

Answer 2

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.