Solve the improper integral $\int_{-\infty}^{+\infty}{\frac{1}{|x^2-1|^\alpha}dx}$ at the variation of the value of $\alpha$ 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?
 A: 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 $1/2<\alpha<1$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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
$\bbx{\ds{\Gamma\pars{1 - \alpha}\Gamma\pars{\alpha - 1/2} \over \root{\pi}}}$.
