Using the integral of this question, we can conclude easily that for $x >0$ and $1>r>0,$ we have $$x^{r-1} = \frac{\sin(\pi r)}{\pi} \int_0^\infty \frac{s^{r-1}}{s+x}ds,$$ I wonder whether we can extend the above formula on the subset $$\Gamma := \{z\in \mathbb{C}; Re(z) >0\},$$ or maybe in another subset contained in $\Gamma.$
The "simple" way to go is: let $z\in \Gamma,$ then we have $|z|>0$ and $$|z|^{r-1} = \frac{\sin(\pi r)}{\pi} \int_0^\infty \frac{s^{r-1}}{s+|z|}ds.$$ on the other hand,
$$z^{r-1} := |z|^{r-1}e^{i(r-1)Arg(z)} = \frac{\sin(\pi r)}{\pi} \int_0^\infty \frac{s^{r-1}e^{i(r-1)Arg(z)}}{s+|z|}ds.$$ I don't see why the term inside the integral must be $\frac{s^{r-1}}{s+z}.$
Maybe this is not the best way to go ...
Thank you for any hint.
Let $U=\{z\in \mathbb{C}; z\not \in ]-\infty,0]\}$. For $z\in U$, $z=|z|\exp(i\theta)$, $-\pi<\theta<+\pi$, we put $\log z=\log |z|+i\theta$, this function is analytic in $U$ (this is the "principal determination of the logarithm"). The function $z\to F(z)= \frac{\sin(\pi r)}{\pi}\int_0^{+\infty}\frac{s^{r-1}}{z+s}ds$ is also analytic in $U$. Then as $\exp((r-1)\log z)-F(z)$ is analytic in the connected open subset $U$ of $\mathbb{C}$, and zero on $]0,+\infty[$, it is $0$ on $U$, and your formula is valid on $U$.