How to evaluate $\int_1^{+\infty} \frac{\ln^2(1+x)}{x^{2\alpha}}\mathrm dx$? 
$$\int_1^{+\infty} \frac{\ln^2(1+x)}{x^{2\alpha}}\mathrm dx$$

I suspect that this integral converges for $\alpha \ge 1/2$. However, I do not know how to evaluate this integral. 
 A: The case $ a=1$
Just to give an idea how at least the $n\in\mathbb{N}$ case can be handeled let's have a look at easiest case where the integral converges $a=1$.
We get
$$
I(1)=\int_1^{\infty}\frac{\log^2(1+x)}{x^2}\underbrace{=}_{i.b.p}-\frac{\log(1+x)^2}{x}\big|_{1}^{\infty}+2\underbrace{\int_1^{\infty}\frac{\log(1+x)}{(1+x)x}}_{J}
$$
in accordance with mathematica
now we use a partial fraction composition to further reduce the remaining integral
$$
J=\int_1^{\infty}\frac{\log(1+x)}{(1+x)x}=\int_1^{\infty}\frac{\log(1+x)}{x}-\int_1^{\infty}\frac{\log(1+x)}{(1+x)}
$$
the last integral is trivial and the first one can be evaluated in terms of dilogarithms. We obtain
$$
J=[-\log(1+x)^2+\text{Li}_2(-x)]\big|_{1}^{\infty}
$$
Now using the property $\text{Li}_2(1-y)+\text{Li}_2(1-1/y)=\frac{\log(y)^2}{2}$ for $y=1+x$ we get
$$
J=-\text{Li}_2\left(1-\frac{1}{1+x}\right)\big|_{1}^{\infty}
$$
Together with the values $\text{Li}_2(1)=\zeta(2)$ and $\text{Li}_2(\frac{1}{2})=\frac{\zeta(2)}{2}-\frac{\log^2(2)}{2}$ 
we might condlude that

$$
I(1)=\frac{\pi ^2}{6}+2 \log ^2(2)
$$

The case $a=3/2$
Let's see what happens for The first step is exactly the same, integration b parts
$$
I(3/2)=\int_1^{\infty}\frac{\log^2(1+x)}{x^3}\underbrace{=}_{i.b.p}-\frac{\log(1+x)^2}{2x^2}\big|_{1}^{\infty}+\underbrace{\int_1^{\infty}\frac{\log(1+x)}{(1+x)x^2}}_{J_{3/2}}
$$
Again a partial fraction decomposition will help us 
$\frac{1}{(1+x)x^2}=\frac{1}{x^2}+\frac{1}{x+1}-\frac{1}{x}$: This is awesome! We now 2/3 of the remaining integrals from our calulation of $I(1)$
$$
J_{3/2}=J+\underbrace{\int_1^{\infty}\frac{\log(1+x)}{x^2}}_{K_{3/2}}
$$
and it gets even better, it turns out that $K_{3/2}$ has an elementary antiderivative:
$$
K_{3/2}=\left(\log (x)-\log (x+1)-\frac{\log (x+1)}{x}\right)\big
|_{1}^{\infty}=-2 \log (2)
$$
and therefore

$$
I(3/2)=2\log(2)-\frac{\pi ^2}{12}
$$

The general case $a=n/2$
Following the same steps before, we see that everything boils down to integrate
$$
J_{n/2}=\int_1^{\infty}\frac{\log(1+x)}{(1+x)x^n}
$$
One may prove by induction that
$$
Q_n(x)=\frac{1}{(1+x)x^n}=\frac{1}{x^n}-Q_{n-1}(x)
$$
So we may obtain
$$
J_{n/2}=\int_1^{\infty}\frac{\log(1+x)}{x^n}-J_{(n-1)/2}(x)
$$
And we are down to evaluate $K_{n/2}$ for $n>2$
$$
K_{n/2}=\int_{1}^{\infty}\frac{\log(1+x)}{x^n}
$$
one may prove by induction that the primitive is indeed elementary
$$
K_{n/2}=\sum_{m=3}^n\frac{(-1)^{m-2}}{x^{m-2}m(n-1)}\big|_{1}^{\infty}+\frac{1}{n-1}\left(-\frac{\log(1+x)}{x^{n-1}}+(-1)^n [\log(1+x)-\log(x)]\right)\big|_{1}^{\infty}=
\sum_{m=3}^n\frac{(-1)^{m-2}}{m(n-1)}+\frac{1}{n-1}\left(-\log(2)+(-1)^n \log(2)\right)
$$
This enables us to calculate the integral for every $a\in n/2$ with $n\geq2$
A: At request I put this in an answer, even though I would prefer to have it as a comment. Please don't upvote...
Mathematica gives a result including hypergeometric and polygamma functions (see below, I hope the formatting will turn out OK $(a=\alpha)$. I hope I did not mess the formulas up.). I don't see, at the moment, how it could be obtained by hand.
$$
\begin{gathered}
\frac{1}{(1-2 a)^2 a^2}2^{-2 a-1} \Biggl[(1-2 a) \, _3F_2\left(2 a,2 a,2 a;2 a+1,2
   a+1;\frac{1}{2}\right)\\
+a \, _2F_1\left(2 a,2 a;2 a+1;\frac{1}{2}\right)
   (\log (4)-2 a \log (4))\\
+2^{2 a+1} a^2 \biggl\{2 a \log ^2(2)-2 \pi  \csc (2 \pi 
   a)\\+\psi ^{(0)}(1-a)-\psi ^{(0)}\left(\frac{3}{2}-a\right)-\log ^2(2)+\log
   (4)\biggr\}\Biggr]
\end{gathered}
$$
After a FullSimplify it some parts get turned into HarmonicNumbers instead, but I don't have the guts to format that properly in TeX at the moment...
A: 
I do not know how to evaluate this integral.

No one does. :-$)$ That's because, for the integral to make sense, the lower limit has to be $0$. 
Then, for $1<2a<3,~$ we have $I~=~2\pi~\csc(2\pi a)\cdot\dfrac{H_{2a-2}}{2a-1},~$ which can be shown by twice 
differentiating $J_n(k)~=~\displaystyle\int_0^\infty\frac{x^{k-1}}{(1+x)^n}~dx~=~B(k,~n-k)~$ under the integral sign with 
regard to the parameter k, then letting $k=1-2a$ and $n=0$. Why does $J_n(k)$ evaluate to 
the beta function ? Well, just set $t=\dfrac1{1+x},~$ and see what happens... :-$)$ It goes on without 
saying that knowledge of the reflection formula for the gamma and polygamma functions 
is paramount, as is also the latter's relation to harmonic numbers.
