TITLE: Find values for which the integral converges
QUESTION [0 upvotes]: Find all the values of $\alpha$ and $\beta$ for which the following integral converges:
$$ \int_{0}^{\infty} \bigg(|1-x|^{-\alpha} - (1+x)^{-\alpha}\bigg)x^{-\beta}dx $$
Of course integral converges when $0 < \alpha <1$ and $1 - \alpha < \beta < 1$. Integral also converges when (for example) $\beta = 1.5$ which I checked in Wolfram Alpha.
I would appreciate any tips or hints.

REPLY [1 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\int_{0}^{\infty}\pars{\verts{1 - x}^{-\alpha} - \pars{1 + x}^{-\alpha}}x^{-\beta}\,\dd x
\label{1}\tag{1}
\\[5mm] = &\
\int_{0}^{1}\bracks{\pars{1 - x}^{-\alpha} - \pars{1 + x}^{-\alpha}}x^{-\beta}\,\dd x
\label{2}\tag{2}
\\[2mm] + &
\int_{1}^{\infty}\bracks{\pars{x - 1}^{-\alpha} - \pars{1 + x}^{-\alpha}}x^{-\beta}\,\dd x
\label{3}\tag{3}
\end{align}

When $\ds{x \sim 1}$ $\ds{\pars{~\mbox{see} (\ref{2})\ \mbox{and}\ (\ref{3})~}}$, the convergence is enforced by
\begin{equation}
\Re\pars{-\alpha} > -1 \implies \bbx{\Re\pars{\alpha} < 1} \label{4}\tag{4}\\
\end{equation}
When $\ds{x \gtrsim 0}$, the integrand is $\ds{\propto x^{1 - \beta}.\,\,\,}$ See (\ref{1}).
$$
\mbox{Then,}\quad
\Re\pars{1 - \beta} > - 1 \implies \bbx{\Re\pars{\beta} < 2} \\
$$
When $\ds{x \to \infty}$, the integrand is $\ds{\propto x^{-\alpha -1 - \beta}\,.\,\,}$
See (\ref{3}). Then, $\ds{\Re\pars{-\alpha - \beta} < 0 \implies \bbx{\Re\pars{\beta} > -\Re\pars{\alpha}}}$

Summarizing,
$\ds{-\Re\pars{\alpha} < \Re\pars{\beta} < 2}$
which obviously yields $\ds{\Re\pars{\alpha} > -2}$.

With (\ref{4}):
$$
\left\{\begin{array}{rcccl}
\ds{-\Re\pars{\alpha}} & \ds{<} & \ds{\Re\pars{\beta}} & \ds{<} & \ds{2}
\\[2mm]
\ds{-2} & \ds{<} & \ds{\Re\pars{\alpha}} & \ds{<} & \ds{1}
\end{array}\right.
$$