Integral calculus with Gamma function [closed]

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

$$$$\int_{0}^{\infty} x^{-\alpha-1}\left(e^{-\beta x}-1\right)dx=\beta^\alpha\Gamma(-\alpha),$$$$

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

Thanks

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 $$H$$.

That is true.

Using Mathematica I got

$$\int_{0}^{\infty} x^{\alpha-1}e^{-(-c-id) x}dx=\frac{\Gamma(\alpha)}{(c+id)^\alpha}.$$

The code I used:

Integrate[x^(a - 1) * Exp[-(-c - I*d)*x], {x, 0, Infinity},
Assumptions -> {a > 0, c < 0, Element[d, Reals]}]


• Thanks. But how can I prove this? Commented Mar 26, 2023 at 16:50
• Constantin-Nicolae Beli gave a simple correct answer.
– Medo
Commented Mar 26, 2023 at 17:01