Integral using residue theorem complex analysis How to solve this integral using residue theorem?
$$\int_0^∞ \frac{x^{(a-1)}}{(x+b)(x+c)} \, dx $$
$0 < a < 1, \ \ \ b > 0, \ \ \ c > 0$
 A: Herein, we present an approach that exploits the residue theorem.  Let $I(a,b,c)$ be the integral given by
$$\bbox[5px,border:2px solid #C0A000]{I(a,b,c)=\int_0^\infty \frac{x^{a-1}}{(x+b)(x+c)}\,dx} \tag 1$$
where $0<a<1$, $b>0$, and $c>0$.

Next, we move to the complex plane and evaluate the closed contour integral $J(a,b,c)$ given by 
$$\bbox[5px,border:2px solid #C0A000]{J(a,b,c)=\oint_C \frac{z^{a-1}}{(z+b)(z+c)}\,dz} \tag2$$
where we choose the branch cut that emanates from the branch point at $z=0$ and extends along the positive real axis to $z=\infty$.  

In $(2)$, the contour $C$ is the classical "keyhole" contour comprised of $(i)$ the ray from $\epsilon>0$ to $R>\max(b,c)$ on the upper side of the branch cut, $(ii)$ the  counter clockwise circular path on which $z=Re^{i\phi}$, from $\phi =0$ to $\phi =2\pi$, $(iii)$ the ray from $R$ to $\epsilon$ on the lower side of the branch cut, and $(iv)$ the clockwise circular path on which $z=\epsilon e^{i\phi}$, from $\phi=2\pi$ to $\phi=0$.

From the residue theorem, we have
$$\begin{align}J(a,b,c)&=2\pi i \text{Res}\left( \frac{z^{a-1}}{b+c},z=-b,-c\right)\\\\
&=2\pi i \left( \frac{(-b)^{a-1}-(-c)^{a-1}}{c-b}\right)\\\\
&=\bbox[5px,border:2px solid #C0A000]{2\pi i e^{i\pi a}\left(\frac{b^{a-1}-c^{a-1}}{b-c}\right)}\tag 3
\end{align}$$

Next, we express $J(a,b,c)$ as the sum of integrals over the four contours that comprise $C$.  Proceeding, we write 
$$\begin{align}
J(a,b,c)&=\int_\epsilon^R \frac{x^{a-1}}{(x+b)(x+c)}\,dx+\int_0^{2\pi}\frac{iR^{a}e^{ia\phi}}{(Re^{i\phi}+b)(Re^{i\phi}+c)}\,d\phi\\\\
&+\int_R^\epsilon \frac{x^{a-1}e^{i2\pi(a-1)}}{(x+b)(x+c)}\,dx+\int_{2\pi}^0\frac{i\epsilon^{a}e^{ia\phi}}{(\epsilon e^{i\phi}+b)(\epsilon e^{i\phi}+c)}\,d\phi \tag 3
\end{align}$$
As $R\to \infty$ and $\epsilon\to0$, the second and fourth integrals on the right-hand side of $(3)$ approach $0$.  Therefore, we can write
$$\bbox[5px,border:2px solid #C0A000]{\lim_{\epsilon \to 0,R\to \infty}J(a,b,c)=(1-e^{i2\pi (a-1)})\int_0^\infty \frac{x^{a-1}}{(x+b)(x+c)}\,dx }\tag 4$$


Putting together $(3)$ and $(4)$ yields
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^{a-1}}{(x+b)(x+c)}\,dx=-\frac{\pi}{\sin(\pi a)}\,\frac{b^{a-1}-c^{a-1}}{b-c}}$$

A: The branch point at $x= 0$ tell us to use $x = e^u$ :
$$I = \int_0^\infty \frac{x^{a-1}}{(x+b)(x+c)}dx  = \int_{-\infty}^\infty \frac{e^{au}}{(e^u +b)(e^u+c)}du$$
Since $e^{2i \pi a}I  =\int_{2i\pi-\infty}^{2i\pi+\infty} \frac{e^{au}}{(e^u +b)(e^u+c)}du$ you have with $C_R$ the rectangular contour $-R \to R\to R+2i\pi\to R-2i\pi \to-R$ and assuming $b > 0,c> 0$ : 
$$I\frac{1-e^{2i \pi a}}{2i\pi} = \lim_{R \to \infty} \frac{1}{2i\pi}\int_{C_R}
 \frac{e^{au}}{(e^u +b)(e^u+c)}du $$
$$ = Res( \frac{e^{au}}{(e^u +b)(e^u+c)},\log(b)+i\pi)+Res( \frac{e^{au}}{(e^u +b)(e^u+c)},\log(c)+i\pi)$$ $$ = \frac{e^{a(\log(b)+i\pi)}}{e^{\log(b)+i\pi}(e^{\log(b)+i\pi}+c)}+\frac{e^{a(\log(c)+i\pi)}}{e^{\log(c)+i\pi}(e^{\log(c)+i\pi}+b)} = e^{i \pi a}\frac{b^{a-1}-c^{a-1}}{b-c} $$
By analytic continuation it stays true for every $b,c \in \mathbb{C} \setminus (-\infty,0]$ 
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

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

$$\bbox[8px,#efe,border:0.1em groove navy]{\mbox{It's interesting to see that the integral can be performed by 'real methods'}}$$
\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.

A: You can use the result:
$$\int_0^\infty \frac{x^{-p}}{1+x}dx = \frac{\pi}{\sin{\pi p}}\tag{1}$$
To prove this result, you can define $z^{-p}$ as $r^{-p}\exp(-i\pi p\theta)$ where $z = r\exp(i\theta)$ is the polar representation of $z$, where we choose $0\leq\theta<2\pi$ so that the branch cut is put on the positive real axis. Then consider the contour that starts on the real axis at $z = \epsilon$ moves to $z =R$ on the real axis, then it moves via a counterclockwise circle of radius $R$ and center the origin to just below the real axis so that we just don't cross the branch cut, we then move infinitesimally below the positive real axis to just below $z = \epsilon$, and then we close the contour via a clockwise circle of radius $\epsilon$ with center the origin.
There is then one pole at $z = -1$ inside the contour, while the branch point singularity at $z = 0$ is outside the contour, the contour integral can thus be calculated using the residue method. 
To calculate the integral in the question, you can expand
$$\frac{1}{(x+b)(x+c)}$$
in partial fractions and then use (1) for each term (you don't have the re-calculate the contour integral when $x+1$ in the numerator is changed to $x+b$ as you can substitute $x = b t$ and then factor out powers of $b$).
Convergence issues are not a problem either when using (1) term by term for the partial fractions to get the desired result which has a different range of convergence for the parameter $a$. You can invoke the principle of analytic continuation to prove the result for $0<a<1$
