TITLE: Integral using residue theorem complex analysis
QUESTION [6 upvotes]: How to solve this integral using residue theorem?
$$\int_0^∞ \frac{x^{(a-1)}}{(x+b)(x+c)} \, dx $$
$0 < a < 1, \ \ \ b > 0, \ \ \ c > 0$

REPLY [3 votes]: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{\infty}{x^{a - 1} \over \pars{x + b}\pars{x + c}}
\,\dd x:\ ?.\qquad 0 < a <1\,,\quad b >0\,,\quad c>0}$.

$$\bbox[8px,#efe,border:0.1em groove navy]{\mbox{It's interesting to see that the integral can be performed by 'real methods'}}$$
\begin{align}
&\color{#f00}{\int_{0}^{\infty}{x^{a - 1} \over \pars{x + b}\pars{x + c}}\,\dd x} =
{1 \over c - b}\pars{\int_{0}^{\infty}{x^{a - 1} \over x + b}\,\dd x -
\int_{0}^{\infty}{x^{a - 1} \over x + c}\,\dd x}\label{1}\tag{1}
\end{align}

The LHS integral converges whenever $\ds{0 < \Re\pars{a} < \color{#f00}{2}}$ albeit the OP asked for the condition $\ds{0 < a < \color{#f00}{1}}$.
The RHS integrals converge whenever $\ds{0 < \Re\pars{a} < \color{#f00}{1}}$ which coincides with the OP condition $\ds{0 < a < 1}$  whenever $\ds{a \in {\mathbb R}}$.
$\bbox[8px,#ffe,border:0.1em groove navy]{\mbox{Then,}\ \eqref{1}\ \mbox{is evaluated with}\ \ds{0 < \Re\pars{a} < \color{#f00}{1}}\ \mbox{which is more general that the above OP condition}}$


In the first integral we make $\ds{x/b \mapsto x}$ while in the second we make $\ds{x/c \mapsto x}$: 
\begin{align}
&\color{#f00}{\int_{0}^{\infty}{x^{a - 1} \over \pars{x + b}\pars{x + c}}\,\dd x} =
{b^{a - 1} - c^{a - 1} \over c - b}\int_{0}^{\infty}{x^{a - 1} \over x + 1}
\,\dd x\label{2}\tag{2}
\end{align}

Now, $\ds{t \equiv {1 \over x + 1}\implies x = {1 \over t} - 1\implies
\totald{x}{t} = -\,{1 \over t^{2}}}$:
\begin{align}
&\color{#f00}{\int_{0}^{\infty}{x^{a - 1} \over \pars{x + b}\pars{x + c}}\,\dd x} =
{b^{a - 1} - c^{a - 1} \over c - b}\int_{1}^{0}t\pars{{1 \over t} - 1}^{a - 1}
\,\pars{-\,{\dd t \over t^{2}}}
\\[5mm] = &\
{b^{a - 1} - c^{a - 1} \over c - b}\int_{0}^{1}t^{-a}\pars{1 - t}^{a - 1}\,\dd t 
\label{3.a}\tag{3.a}
=
{b^{a - 1} - c^{a - 1} \over c - b}\,
{\Gamma\pars{-a + 1}\Gamma\pars{a} \over \Gamma\pars{1}}
\\[5mm] = &\
\color{#f00}{{b^{a - 1} - c^{a - 1} \over c - b}\,{\pi \over \sin\pars{\pi a}}}
\label{3.b}\tag{3.b}
\end{align}

Note that the integral involved in \eqref{3.a} $\pars{~the\ Beta\ function~}$
  converges whenever
  $$
\Re\pars{-a} > -1\,,\quad\Re\pars{a - 1} > -1\qquad\implies\qquad
0 < \Re\pars{a} < 1
$$
  which defines the general condition for the integrals evaluation.