$\int_{0}^{\infty} {x^{\alpha} \over \left(x + 1\right)}dx$ I need some help to evaluate the following integral.

$$\int_{0}^{\infty} {x^{\alpha} \over \left(x + 1\right)}dx$$ 

I know you need to use a branch cut but not sure how to start. 
Any help is always appreciated. 
Edit: 
Need Help to Establish the Cauchy Principle for the improper integral if that provides any direction. 
 A: Assuming $\alpha\in\mathbb{R}$, the given integral is convergent iff $\alpha\in(-1,0)$.
With such assumption we have
$$ \int_{0}^{+\infty}\frac{x^\alpha}{x+1}\,dx = \int_{-\infty}^{+\infty}\frac{e^{\alpha z}}{1+e^{-z}}\,dz $$
and the last integral can be computed by considering the integral of $f(z)=\frac{e^{\alpha z}}{1+e^{-z}}$ over the rectangle contour (counter-clockwise oriented) having vertices at $-R,R,R+2\pi i,-R+2\pi i$. 
By computing the residue at the pole $z=\pi i$ it follows that
$$ (1-e^{2\pi i\alpha})\int_{-\infty}^{+\infty}f(z)\,dz = 2\pi i e^{\pi i\alpha} $$
from which:
$$ \int_{-\infty}^{+\infty}f(z)\,dz = \color{red}{-\frac{\pi}{\sin(\pi \alpha)}}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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Some alternatives for
$\ds{\,\mc{I} \equiv \int_{0}^{\infty}{x^{\alpha} \over x + 1}\,\dd x}$ !!!.
$\ds{\qquad \Re\pars{\alpha} \in \pars{-1,0}}$.

$\ds{\Large\left.a\right)}$
\begin{align}
2\pi\ic\pars{1^{\alpha}} &
\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,
-\int_{-\infty}^{-\epsilon}
{\pars{-x}^{\alpha}\expo{-\pi\alpha\ic} \over x - 1}\,\dd x -
\int_{\pi}^{-\pi}{\epsilon^{\alpha}\expo{\alpha\theta\,\ic} \over \epsilon\expo{\theta\,\ic} - 1}\,\epsilon\expo{\theta\,\ic}\ic\,\dd\theta -
\int_{-\epsilon}^{-\infty}
{\pars{-x}^{\alpha}\expo{\pi\alpha\ic} \over x - 1}\,\dd x
\\[5mm] &
\,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{=}\,\,\,
\int_{0}^{\infty}
{x^{\alpha}\expo{-\pi\alpha\ic} \over x + 1}\,\dd x -
\int_{0}^{\infty}
{x^{\alpha}\expo{\pi\alpha\ic} \over x + 1}\,\dd x =
-2\ic\sin\pars{\pi\alpha}\int_{0}^{\infty}{x^{\alpha} \over x + 1}\,\dd x
\\[5mm] & \implies
\bbx{\int_{0}^{\infty}{x^{\alpha} \over x + 1}\,\dd x = -\pi\csc\pars{\pi a}}
\end{align}

$\ds{\Large\left.b\right)}$
\begin{align}
\mc{I} & = \int_{1}^{\infty}{\pars{x - 1}^{\alpha} \over x}\,\dd x = \int_{1}^{0}{\pars{1/x - 1}^{\alpha} \over 1/x}\,\pars{-\,{1 \over x^{2}}}
\,\dd x =
\int_{0}^{1}x^{-\alpha - 1}\pars{1 - x}^{\alpha}\,\dd x
\\[5mm] & =
{\Gamma\pars{-\alpha}\Gamma\pars{\alpha + 1} \over \Gamma\pars{1}} =
{\pi \over \sin\pars{-\pi\alpha}} \implies
\bbx{\int_{0}^{\infty}{x^{\alpha} \over x + 1}\,\dd x = -\pi\csc\pars{\pi a}}
\end{align}
This one is equivalent to the initial change of variables
$\ds{{1 \over x + 1} \mapsto x}$.

$\ds{\Large\left.c\right)}$
\begin{align}
\mc{I} & =
\int_{0}^{1}{x^{\alpha} \over 1 + x}\,\dd x +
\int_{1}^{\infty}{x^{\alpha} \over x + 1}\,\dd x =
\int_{0}^{1}{x^{\alpha} \over 1 + x}\,\dd x -
\int_{1}^{0}{x^{-\alpha} \over x\pars{1 + x}}\,\dd x
\\[5mm] & =
\int_{0}^{1}{x^{\alpha} + x^{-\alpha - 1} \over 1 + x}\,\dd x =
\int_{0}^{1}{x^{\alpha} + x^{-\alpha - 1} - x^{\alpha + 1} - x^{-\alpha}
\over 1 - x^{2}}\,\dd x
\\[5mm] & =
{1 \over 2}\int_{0}^{1}{x^{\alpha/2 - 1/2} + x^{-\alpha/2 - 1} - x^{\alpha/2} -
x^{-\alpha/2 - 1/2} \over 1 - x}\,\dd x
\\[5mm] & =
{1 \over 2}\bracks{%
-\Psi\pars{{\alpha \over 2} + {1 \over 2}} - \Psi\pars{-\,{\alpha \over 2}} +
\Psi\pars{{\alpha \over 2} + 1} +
\Psi\pars{-\,{\alpha \over 2} + {1 \over 2}}}
\\[5mm] & =
{1 \over 2}\braces{%
\bracks{\Psi\pars{-\,{\alpha \over 2} + {1 \over 2}} -
\Psi\pars{{\alpha \over 2} + {1 \over 2}}} +
\bracks{\Psi\pars{{\alpha \over 2} + 1} - \Psi\pars{-\,{\alpha \over 2}}}}
\\[5mm] & =
{1 \over 2}\bracks{%
\pi\cot\pars{\pi\bracks{{\alpha \over 2} + {1 \over 2}}} +
\pi\cot\pars{\pi\bracks{-{\alpha \over 2}}}} =
-\,{1 \over 2}\,\pi\bracks{%
\tan\pars{\pi\bracks{\alpha \over 2}} +
\cot\pars{\pi\bracks{\alpha \over 2}}}
\\[5mm] & =
-\,{1 \over 2}\,\pi\,{1 \over \sin\pars{\pi\alpha/2}\cos\pars{\pi\alpha/2}}
\implies
\bbx{\int_{0}^{\infty}{x^{\alpha} \over x + 1}\,\dd x = -\pi\csc\pars{\pi a}}
\end{align}
