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Firstly the function is even so I can consider the integral $\int_{1}^{+\infty}{\frac{1}{|x^2-1|^\alpha}dx}$. So I have to solve $lim_{c \to 1^+}\int_{c}^{+\infty}{\frac{1}{|x^2-1|^\alpha}dx}.$ At this point I thought to put $t^2 = x^2 -1$ so as to obtain $x=\sqrt{t^2-1}$ and $x'=\frac{2t}{\sqrt{t^2-1}}$ so the integral becames $lim_{c \to 2^+}2\int_{c}^{+\infty}{\frac{1}{|t|^{\alpha-1} \sqrt{t^2-1}}}dt$ (the lower bound of the integral is obtained as following: put $g(t) = \sqrt{t^2-1}$ so $g^{-1(t)}=t^2+1$; now substituting $1^+$ in the prior formula the lower bound is equal to $2^+$ ). But now using the asymptotic comparison test I obtain that improper integral converges for $\alpha>1$. But this result is wrong. Can somebody help me?

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  • $\begingroup$ Are you trying to value the integral when/if it exists? Or are you trying to determine the values of $\alpha$ for which the integral exists as an improper Riemann integral? $\endgroup$ – Mark Viola May 7 '18 at 15:39
  • $\begingroup$ I'm trying to determine the values of α for which the integral exists as an improper Riemann integral $\endgroup$ – user515933 May 7 '18 at 15:40
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First note that the improper integral, if it exists, can be written

$$\begin{align} \int_{0}^\infty \frac{1}{|x^2-1|^\alpha}\,dx&=\lim_{\epsilon_1\to 0^+}\int_0^{1-\epsilon_1}\frac{1}{(1-x^2)^\alpha}\,dx\\\\ &+\lim_{\epsilon_2\to 0^+}\int_{1+\epsilon_2}^{2}\frac{1}{(x^2-1)^\alpha}\,dx\\\\ &+\lim_{L\to\infty}\int_2^L \frac{1}{(x^2-1)^\alpha }\,dx\tag1 \end{align}$$

The first and second integrals on the right-hand side of $(1)$ converge if and only if $\alpha<1$ due to the singularity at $x=1$ since the integrand is $O(|x-1|^{-\alpha})$ as $x\to 1^\pm$.

The third integral on the right-hand side of $(1)$ converges if and only if $\alpha>1/2$ since the integrand is $O(x^{-2\alpha})$ as $x\to\infty$.

Hence, the integral $\int_{0}^\infty \frac{1}{|x^2-1|^\alpha}\,dx$ converges if and only if $1/2<\alpha<1$.

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

The integral behaves, with $\ds{\Re\pars{\alpha} >0}$ as $\ds{x^{-\alpha}}$ and $\ds{x^{-\alpha - 1/2}}$ as $\ds{x \to 0^{+}}$ and $\ds{x \to \infty}$, respectively. Then, it converges whenever $\ds{-\Re\pars{\alpha} > - 1}$ and $\ds{-\Re\pars{\alpha} + 1/2 < 0}$

$\ds{\implies \bbx{1/2 < \Re\pars{\alpha} < 1}}$.

The answer is given by $\bbx{\ds{\Gamma\pars{1 - \alpha}\Gamma\pars{\alpha - 1/2} \over \root{\pi}}}$.

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