Change of variables in $\int_{0}^{+\infty} \frac{x^{-\beta}}{1+x} \, dx$ Let $0 < \beta < \frac{1}{2}$. I cannot figure out which change of variable to use in order to prove that :
$$ \int_{0}^{+\infty} \frac{x^{-\beta}}{1+x} \; dx = \int_{0}^{1} \frac{x^{-\beta}}{1+x} (1+x^{2\beta-1}) \; dx $$
I have tried : $t=\frac{1}{1+x}$ and $t=\frac{x}{1+x}$ but it didn't work. 
 A: Split into two pieces; the integral is equal to
$$\int_0^1 dx \frac{x^{-\beta}}{1+x} + \underbrace{\int_1^{\infty} dx \frac{x^{-\beta}}{1+x}}_{x\mapsto 1/x} $$
which is then
$$\int_0^1 dx \frac{x^{-\beta}}{1+x} + \int_0^1 \frac{dx}{x^2} \frac{x^{\beta}}{1+(1/x)}  = \int_0^1 dx \frac{x^{-\beta}}{1+x} + \int_0^1 dx \frac{x^{\beta-1}}{1+x}$$
which is, combining,
$$\int_0^1 dx  \frac{x^{-\beta}}{1+x} \left ( 1+ x^{2 \beta-1}\right ) $$
A: $\newcommand{\+}{^{\dagger}}%
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With $\ds{t \equiv {1 \over 1 + x}\quad\iff\quad x = {1 - t \over t}}$:
\begin{align}
\int_{0}^{\infty}{x^{-\beta} \over 1 + x}\,\dd x&=
\int_{1}^{0}t\pars{1 - t \over t}^{-\beta}\pars{-\,{\dd t \over t^{2}}}
=
\int_{0}^{1}t^{\beta - 1}\pars{1 - t}^{-\beta}\,\dd t
={\rm B}\pars{\beta,1 - \beta}
\end{align}
where ${\rm B}\pars{x,y}$ is the Beta Function which satisfies
$\ds{{\rm B}\pars{x,y} = {\Gamma\pars{x}\Gamma\pars{y} \over \Gamma\pars{x + y}}}$. Then,
\begin{align}
\color{#00f}{\large\int_{0}^{\infty}{x^{-\beta} \over 1 + x}\,\dd x}&=
{\Gamma\pars{\beta}\Gamma\pars{1 - \beta} \over \Gamma\pars{\beta + \bracks{1 - \beta}}} = \color{#00f}{\large{\pi \over \sin\pars{\pi\beta}}}
\end{align}
Here we used the identities
$\ds{\Gamma\pars{z}\Gamma\pars{1 - z} = {\pi \over \sin\pars{\pi z}}}$ and $\ds{\Gamma\pars{1} = 1}$.
