I am struggling with the derivation of the Lévy-measure of a Gamma-process $X_t$ with law $p_t(x)= \frac{\lambda^{ct}}{\Gamma(ct)}x^{ct-1}e^{-\lambda x}1_{\lbrace x>0 \rbrace }$. The paper I am reading says it is shown "easily" via the characteristic function and using the Levy-Khintchine formula:
$\Phi(s)=E[e^{isX_1}] = exp( ias - \frac{1}{2}\sigma^2s^2 + \int_{\mathbb{R}}(1-e^{isx} - is\mathbb{1}_{|x|<1})d\nu(x))$
I computed via the pdf of the Gamma-distribution $\Phi(s) = E[e^{isX_t}]= (1-\frac{is}{\lambda})^{-ct}$
But how is the Lévy-measure then derived to be $\nu(x) = \frac{ce^{-\lambda s}}{x}1_{\lbrace 1 > 0\rbrace}$?
I know this answer comes quite late. A nice explanation is given in these great course notes.
You can show this result by using the Frullani integral, which under some conditions tells you that $$ \int_0^\infty \frac{f(ax)-f(bx)}{x}dx = (f(0)-f(\infty))\log\left(\frac{b}{a}\right), \quad b > a > 0. $$ Here, $f(\infty)$ of course represents the limit $\underset{x \rightarrow \infty}\lim f(x)$.
Applying this result on the function $e^{-x}$ for the values $a=\lambda, b=\lambda-z$ where $z<0$, we obtain \begin{align*} \int_0^\infty \frac{e^{-\lambda x}-e^{-(\lambda - z)x}}{x}dx &= \log\left(\frac{\lambda - z}{\lambda}\right), \end{align*} which can be rewritten as $$ \int_0^\infty (1-e^{z x})c\frac{e^{-\lambda x}}{x}dx = - \int_{-\infty}^\infty (e^{z x}-1)c\frac{e^{-\lambda x}}{x}1_{x>0}dx = -c \log\left(1-\frac{\lambda}{z}\right).$$ Using analytic continuation, you can show that this result holds for any $z$ such that $\text{Re}(z) \leq 0$ from which the result quite easily follows!