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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

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2 Answers 2

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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 $H$.

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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]}]

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

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