I have to prove that for $0<\alpha<1$ and $\beta>0$,


\begin{equation}
 \int\_{0}^{\infty} x^{-\alpha-1}\left(e^{-\beta x}-1\right)dx=\beta^\alpha\Gamma(-\alpha),
\end{equation}


and I have to show that the same equality is valid when $-\beta$ is replaced by any
complex number $w \neq 0$ with $Re(w)\leq 0$.


In the first case


$$\int\_{0}^{\infty} x^{-\alpha-1}\left(e^{-\beta x}-1\right)dx=\left[-\frac {1}{\alpha x^\alpha} (e^{-\beta x}-1) \right]\_0^\infty-\frac {\beta}{\alpha} \int\_{0}^{\infty} x^{-\alpha}e^{-\beta x}dx=0-\frac {\beta}{\alpha} \int\_{0}^{\infty} x^{(1-\alpha)-1}e^{-\beta x}dx=-\frac {\beta}{\alpha} \frac{\Gamma(1-\alpha)}{\beta^{1-\alpha}}=\beta^\alpha\Gamma(-\alpha),$$


where we use that
\begin{equation}
\int\_{0}^{\infty} \frac {\beta^\lambda}{\Gamma(\lambda)}x^{\lambda-1}e^{ -\beta x}dx=1,
\end{equation}
for $\lambda>0$ and $\beta>0$.


Now, if $w=c+id$ with $c\leq0$ then


$$\int\_{0}^{\infty} x^{-\alpha-1}\left(e^{w x}-1\right)dx=\left[-\frac {1}{\alpha x^\alpha} (e^{w x}-1) \right]\_0^\infty-\frac {w}{\alpha} \int\_{0}^{\infty} x^{-\alpha}e^{w x}dx=0-\frac {w}{\alpha} \int\_{0}^{\infty} x^{(1-\alpha)-1}e^{-(-c-id)x}dx$$
Now, my question is:


Is true that
\begin{equation}
\int\_{0}^{\infty} \frac {(c+id)^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-(-c-id) x}dx=1,
\end{equation}
for $\alpha>0$ and $-c>0$?
Because if this is true then I can prove also the second request.


Thanks



Integral calculus with Gamma function
You fix $\alpha$ and denote your integral to the left by $I(\beta )$. Then $I$ is convergent and analytic on the semi-plane $H=\{\beta\in{\mathbb C}\mid\Re (\beta )>0\}$. The right hand side too is analytic on $H$. Since the two analytic functions are the same on $(0,\infty )$, they coincide on the whole $H$.


