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