Change of variables in $\int_{0}^{+\infty} \frac{x^{-\beta}}{1+x} \, dx$

Question

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.

Answer 1

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,

$$\int_0^1 dx \frac{x^{-\beta}}{1+x} \left ( 1+ x^{2 \beta-1}\right ) $$

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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 $\ds{\Gamma\pars{1} = 1}$.