Does $\int_{-\infty}^\infty \frac{\mathrm dx}{(1+x^2)^\alpha}$ converge? I'm wondering when the integral 
$$
\int_{-\infty}^\infty \frac{\mathrm dx}{(1+x^2)^\alpha}
$$
converges for the real number $\alpha$.
 A: The integral is evaluated in terms of the Gamma function as follows
$${\frac {\sqrt {\pi }\Gamma  \left( -1/2+\alpha \right) }{\Gamma 
 \left( \alpha \right) }}
$$
Then $\alpha >1/2$ is domain with convergence.  When $\alpha =1/2$ the integral does not converge.  When $-1/2 + \alpha = -n$ where $n$ is a positive integer, the integral does not converge.
Do you agree?
A: Outline: Use Comparison. Break up into two parts at $0$. By symmetry the two integrals both exist or both fail to exist, and if they exist they have the same value.
Since our function is nicely behaved on the interval $[-1,1]$, we only need to worry about $\int_1^\infty \frac{1}{(1+x^2)^\alpha}\,dx$. 
For $\alpha \gt 1/2$, we have convergence, by comparison with $\int_1^\infty \frac{1}{x^{2\alpha}}\,dx$.
If $\alpha\le 1/2$, we have divergence. For in the interval from $1$ to $\infty$, we have $(1+x^2)^\alpha \le (2x^2)^\alpha$. 
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{}$
\begin{align}
&\mbox{When}\ \verts{x}\to\infty,\ \mbox{the integrand}\ \sim x^{-2\alpha}
\\[3mm]&\mbox{Its antiderivative}\ \sim x^{-2\alpha + 1}
\\[3mm]&\mbox{Convergengence requires}\ -2\,\Re\pars{\alpha} + 1 < 0
\end{align}

It converges whenever $\ds{\color{#66f}{\large\Re\pars{\alpha} > \half}}$.

A: Note that for large $x$
$$ \frac{1}{(1+x^2)^\alpha} \sim \frac{1}{x^{2\alpha}} $$
Now check for what values of alpha the last integrand converge.
