Find values for which the integral converges

calculusconvergence-divergenceintegration

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.

Best Answer

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}): $$ \left\{\begin{array}{rcccl} \ds{-\Re\pars{\alpha}} & \ds{<} & \ds{\Re\pars{\beta}} & \ds{<} & \ds{2} \\[2mm] \ds{-2} & \ds{<} & \ds{\Re\pars{\alpha}} & \ds{<} & \ds{1} \end{array}\right. $$
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