Contour integral of $\int_0^\infty \frac{1}{x^p(2+x)}$ with $p \in \mathbb{R}$

Question

I am trying to evalute this integral: $$I = \int_0^\infty \frac{1}{x^p(2+x)} \quad p \in \mathbb{R}$$

For the integral to be convergent, it must be: $0<p<1$, right?

I use a pacman/keyhole contour as a contour and I call it C.

$$f(z) = \frac{1}{z^p(2+z)}$$ We have only one pole at $z=-2$: $$\text{Res}_{z=-2} f(z) = \frac{1}{(-2)^p}$$

I use the residue theorem. I know that the: $$\oint_C f(z) dz = 2\pi i \frac{1}{(-2)^p}$$

and: $$\oint_C f(z) dz = \int_0^\infty \frac{1}{x^p(2+x)} - e^{-2\pi i p} \int_0^\infty \frac{1}{x^p(2+x)} = I(1-e^{-2\pi i p}) = 2\pi i \frac{1}{(-2)^p}$$

I get that:

$$I = 2\pi i \frac{(-2)^{-p}}{(1-e^{-2\pi i p})} = 2\pi i\frac{(-2)^{-p}}{(e^{\pi i p}-e^{- i \pi p})e^{-i \pi p}} = \frac{\pi (-2)^{-p}}{\sin(\pi p)e^{-i \pi p}}$$

I am afraid there is some mistake. Thank you for your help.

EDIT:

Noting that: $$e^{-\pi i p} = (e^{i\pi})^{-p} = (-1)^{-p} $$ $$I = \frac{\pi}{\sin(\pi p)} \left(\frac{-2}{-1}\right)^{-p} = \frac{\pi 2^{-p}}{\sin(\pi p)}$$

Answer 1

If we wish to use contour analysis, we can write as a consequence of the residue theorem

$$\begin{align} \oint_{C_{\text{keyhole}}}\frac{1}{z^p(2+z)}\,dz&=(1-e^{-i2\pi p})I\\\\ &=2\pi i \frac{1}{2^p\color{red}{e^{i\pi p}}} \end{align}$$

First, note that we have

$$\begin{align} 1-e^{-2\pi p}&=e^{-2\pi p}(e^{i\pi p}-e^{-i\pi p})\\\\ &=2i e^{-2\pi p}\sin(\pi p) \end{align}$$

Second, note that the residue at $z=-2$ is simply $\frac1{2^pe^{i\pi p}}$, given the choice of branch cut along the positive real axis (i.e., $(-2)^p=2^pe^{p\log(-1)}=2^pe^{i\pi p}$, since $\log(-1)=i\pi$).

Solving for $I$ reveals that

$$\begin{align} I&=2\pi i \frac{1}{2^p e^{i\pi p}(1-e^{-i2\pi p})}\\\\ &=2^{-p}\pi \csc(\pi p) \end{align}$$

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Alternatively, using real analysis, we have

$$\begin{align} I&=\int_0^\infty \frac{x^{-p}}{(2+x)}\,dx\\\\ &\overbrace{=}^{x\mapsto 2x}2^{-p}\int_0^{\infty} \frac{x^{-p}}{(1+x)}\,dx\\\\ &=2^{-p}B(1-p,p)\\\\ &=2^{-p}\frac{\Gamma(1-p)\Gamma(p)}{\Gamma(1)}\\\\ &=2^{-p}\pi\csc(\pi p) \end{align}$$

as expected! Note that we used the Euler's reflection formula to arrive at the final result.