Again, improper integrals involving $\ln(1+x^2)$ How can I check for which values of $\alpha $ this integral $$\int_{0}^{\infty} \frac{\ln(1+x^2)}{x^\alpha}\,dx $$ converges? 
I managed to do this in $0$ because I know $\ln(1+x)\sim x$ near 0. 
I have no idea how to do this in $\infty$ where nothing is known about $\ln(x)$ .
Will someone help me ? 
Thanks! 
 A: $\newcommand{\+}{^{\dagger}}%
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The integral converges whenever $\ds{1 < \Re\alpha < 3}$ as it has been proved above by $\tt\mbox{@Olivier Bégassat}$.

\begin{align}
&\int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over x^{\alpha}}\,\dd x
=
\overbrace{\int_{1}^{\infty}{\ln\pars{x} \over \pars{x - 1}^{\alpha/2}}\,\half\,
\pars{x - 1}^{-1/2}\,\dd x}^{\ds{1 + x^{2} \to x}}
=
\half\int_{1}^{\infty}{\ln\pars{x} \over \pars{x - 1}^{\pars{\alpha + 1}/2}}\,\dd x
\\[3mm]&=
\half\overbrace{\int_{1}^{0}{\ln\pars{1/x} \over \pars{1/x - 1}^{\pars{\alpha + 1}/2}}\,
\pars{-\,{\dd x \over x^{2}}}}^{\ds{x \to {1 \over x}}}
=
-\,\half\int^{1}_{0}
{x^{\pars{\alpha - 3}/2}\ln\pars{x} \over \pars{1 - x}^{\pars{\alpha + 1}/2}}\,\dd x
\\[3mm]&=
-\,\half\lim_{\mu \to 0}\partiald{}{\mu}\int^{1}_{0}
{x^{\mu + \pars{\alpha - 3}/2}\pars{1 - x}^{-\pars{\alpha + 1}/2}}\,\dd x
=
-\,\half\lim_{\mu \to 0}\partiald{{\rm B}\pars{\mu + \alpha/2 - 1/2,1/2 - \alpha/2}}{\mu}
\end{align}
where ${\rm B}\pars{x,y}$ is the $\it Beta$ function:
$$
{\rm B}\pars{a,b} \equiv \int_{0}^{1}t^{a - 1}\pars{1 - t}^{b - 1}\,\dd t\,,\qquad
\Re a >0\,,\quad \Re b > 0
$$

With the identity
$\ds{{\rm B}\pars{a,b} ={\Gamma\pars{a}\Gamma\pars{b} \over \Gamma\pars{a + b}}}$
$\ds{\pars{~\Gamma\pars{z}\ \mbox{is the}\ {\it Gamma}\ \mbox{function}~}}$: 
\begin{align}
&\int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over x^{\alpha}}\,\dd x=
-\,\half\lim_{\mu \to 0}\partiald{}{\mu}\bracks{%
{\Gamma\pars{\mu + \alpha/2 - 1/2}\Gamma\pars{1/2 - \alpha/2}
  \over \Gamma\pars{\mu}}}
\\[3mm]&=-\,\half\,\Gamma\pars{\half - {\alpha \over 2}}\lim_{\mu \to 0}
\partiald{}{\mu}
\bracks{\mu\Gamma\pars{\alpha/2 - 1/2 + \mu} \over \Gamma\pars{1 + \mu}}
\end{align}
where we used the identity $\Gamma\pars{z + 1} = z\Gamma\pars{z}$.

\begin{align}
&\int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over x^{\alpha}}\,\dd x
=
-\,\half\,\Gamma\pars{\half - {\alpha \over 2}}\times\lim_{\mu \to 0}\!\!\!
\braces{\!\!\!%
\bracks{%
\Gamma\pars{\alpha/2 - 1/2 + \mu}
+ \mu\Psi\pars{\alpha/2 - 1/2 + \mu}}\Gamma\pars{1 + \mu}
- \Psi\pars{1 + \mu}\bracks{\mu\Gamma\pars{\alpha/2 - 1/2 + \mu}}
\over \Gamma\pars{1 + \mu}\!\!\!}
\end{align}
$\ds{\Psi\pars{z} \equiv \totald{\ln\pars{\Gamma\pars{z}}}{z}}$ is the $\it digamma$
function.

With $\ds{\Gamma\pars{1} = 1}$ and the identity
$\ds{\Gamma\pars{z}\Gamma\pars{1 - z} = {\pi \over \sin\pars{\pi z}}}$:
\begin{align}
&\int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over x^{\alpha}}\,\dd x\\[3mm]&=
-\,\half\,\Gamma\pars{\half - {\alpha \over 2}}\Gamma\pars{{\alpha \over 2} - \half}
=-\,\half\,
{\pi \over \Gamma\pars{1/2 + \alpha/2}\sin\pars{\pi\bracks{1/2 - \alpha/2}}}
\Gamma\pars{{\alpha \over 2} - \half}
\\[3mm]&=-\,\half\,{\pi \over \cos\pars{\pi\alpha/2}}
{\Gamma\pars{\alpha/2 - 1/2} \over \Gamma\pars{1/2 + \alpha/2}}
=-\half\,\pi\sec\pars{\pi\alpha \over 2}\,
{\Gamma\pars{\alpha/2 - 1/2} \over \pars{\alpha/2 - 1/2}\Gamma\pars{-1/2 + \alpha/2}}
\end{align}

$$\color{#0000ff}{\large%
\int_{0}^{\infty}{\ln\pars{1 + x^{2}} \over x^{\alpha}}\,\dd x
=
-\,{\pi\sec\pars{\pi\alpha/2} \over \alpha - 1}\,,\qquad
1 < \Re\alpha < 3}
$$
A: To check integrability at the vicinity of $0$, use
$$\frac{\ln(1+x^2)}{x^{\alpha}}\sim\frac{1}{x^{\alpha-2}}$$
which is integrable iff $\alpha-2<1$.
At $+\infty$, one has
$$\frac{\ln(1+x^2)}{x^{\alpha}}\sim\frac{2\ln(x)}{x^{\alpha}}.$$
It is not very difficult to prove that

For all $\alpha,\beta\in\Bbb R$,
  $$\frac{1}{x^{\alpha}\ln(x)^{\beta}}$$
  is integrable at the vicinity of $+\infty$ iff either $\alpha>1$ or $\alpha=1$ and $\beta>1$

It follows that your function is integrable on $\Bbb R^*_+$ iff $1<\alpha<3$.
A: For large values of $x$, we can build the expansion
$$  
\log[1 + x^2] = 2 \log[x] + \frac{1}{x^2} - \frac{1}{2 x^4} + \frac{1}{3 x^6} - \frac{1}{4 x^8}   
$$
So, I think that the only problem is now to compute the antiderivative of $\log[x] / x^a$ which is quite simple using integration by parts ($u = \log[x]$ , $v'= 1 / x^a$).  
The result is $\frac{x^{1-a} \log[x]}{1-a} - \frac{x^{1-a}}{1-a)^2}$  
Remember how behave the product of $x$ to power by $\log[x]$. You then arrive to what Olivier was telling.  
I hope that this clarifies. If not, just post and we shall continue. Cheers.
A: HINT 
I suggest you develop $\log(1+x^2)$ as an infinite series (Taylor); divide each term by $x^a$, compute the anti-derivative and look where and when arrive the problems when you compute the integral between zero and infinity.
