Evaluating $\int_{0}^{\infty}\frac{\mathrm{d}x}{x^{\alpha}(x + 1)}$ I need some help to evaluate the following integral.

$$\int_{0}^{\infty}{\mathrm{d}x \over x^{\alpha}\left(x + 1\right)}$$ where $\alpha \in \left(0,1\right)$

I've tried many ways (the best one seems to be developing by Taylor series) but actually I have no solution.
Some ideas?
Thank you.
 A: I think we will need some complex analysis here. 
Take a branch of $1/z^a$ defined in $\mathbb{C}\setminus [0,+\infty)$ and consider the integral
$$\int_{\gamma}\frac{dz}{z^a(z+1)}$$
where $\gamma$ is a close path composed by an arc of a inner circle of radius $0<r<1$, an arc of an outer circle of radius $R>1$  and two parallel segments over and under the segment $[r,R]$. 
Then by the residue theorem
$$\int_{\gamma}\frac{dz}{z^a(z+1)}=2\pi i\mbox{Res}\left(\frac{1}{z^a(z+1) },-1\right)=2\pi i e^{-i\pi a}.$$
Now we take the limit as $R\to+\infty$ and $r\to 0^+$.
It is easy to see that the integrals along the arcs of the circles goes to $0$. Hence
$$\int_0^{+\infty}\frac{dx}{x^a(x+1)}-\int_0^{+\infty}\frac{dx}{x^ae^{2\pi ia}(x+1)}=2\pi i e^{-i\pi a}$$
which implies that
$$\int_0^{+\infty}\frac{dx}{x^a(x+1)}=\frac{2\pi i e^{-i\pi a}}{1-e^{-2i\pi a}}=\frac{\pi}{\sin (\pi a)}.$$
A: Recalling that the Beta function can be written in the form $$B\left(a,b\right)=\int_{0}^{\infty}\frac{x^{a-1}}{\left(1+x\right)^{a+b}}dx,\,\textrm{Re}\left(a\right)>0,\,\textrm{Re}\left(b\right)>0
 $$ we have that $$\begin{align} 
\int_{0}^{\infty}\frac{x^{-\alpha}}{1+x}dx=
  & \int_{0}^{\infty}\frac{x^{1-\alpha-1}}{\left(1+x\right)^{1-\alpha+\alpha}}dx
  \\=
  &B\left(1-\alpha,\alpha\right)
  \\ =
 & \Gamma\left(1-\alpha\right)\Gamma\left(\alpha\right)
  \\ =
 & \color{red}{\frac{\pi}{\sin\left(\pi\alpha\right)}}
 \end{align}$$ where the last identity follows from the Gamma reflection formula. 
A: There is a way using complex analysis. You make a cut along the positive real axis. Then define for $z=re^{i\phi}$, $r>0$, $\phi \in (0,2\pi)$:
$$ z^a = r^a e^{i\phi a}$$
Make a closed contour that consists of  twice going along the positive real axis (once above, the other below)
$ z(t) = t+i\epsilon , 0<t\leq R$,  $z(t)=t-i\epsilon, R\geq t>0$,
a small halfcircle $ z(t)=\epsilon e^{it}, 3\pi/2>t>\pi/2$ and a large circle $z(t) = R e^{it}, \ 0<t<2\pi$. (A drawing would obviously help). Show that the circle contributions go to zero when $\epsilon\rightarrow 0$ and $R\rightarrow \infty$ and that there is a residue for $z=-1$, to obtain:
$$ 2\pi i e^{-i a\pi} = \int_0^\infty \frac{1}{x^a(1+x)} dx \left( 1- e^{-2\pi i a}\right) $$ from which you get the result $\frac{\pi}{\sin(\pi a)}$, if I didn't mess up somewhere.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{0}^{\infty}{\dd x \over x^{\alpha}\pars{1 + x}} =
\int_{0}^{\infty}{x^{\pars{\color{red}{1 - \alpha}} - 1} \over
1 + x}\,\dd x
\\[2mm] & \mbox{is a nice example of}\
\underline{Ramanujan's\ Master\ Theorem}\, :
\\[3mm] &\
\begin{array}{ll}
{\Large \bullet} & \ds{1 \over 1 + x}\
\mbox{expansion in powers of}\ \ds{x}\ \mbox{is given by}
\\ & \ds{\sum_{k = 0}^{\infty}\pars{-x}^{k} =
\sum_{k = 0}^{\infty}\Gamma\pars{\color{red}{k} + 1}
\,{\pars{-x}^{k} \over k!}}  
\\[5mm]
{\Large \bullet} & \mbox{The}\ \underline{integral\ value}\
\mbox{is given by}\
\\
& \ds{\Gamma\pars{\color{red}{1 - \alpha}}
\color{red}{\Gamma\pars{-\bracks{1 - \alpha} + 1}}} =
\bbx{\pi \over \sin\pars{\pi\alpha}} \\ &
\end{array}
\end{align}
Ramanujan's Master Theorem.
