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

I suspect that this integral converges for $\alpha \ge 1/2$. However, I do not know how to evaluate this integral.

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    $\begingroup$ This is going off the screen for me. You should put this in an answer. $\endgroup$ – Adam Jan 8 '16 at 14:18
  • $\begingroup$ @mickep hey can you put that as an answer? I can't read half of your sentence. Cheers! $\endgroup$ – M.S.E Jan 8 '16 at 14:21
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    $\begingroup$ Convergence for $a>1/2$ I believe, but not $a=1/2$. $\endgroup$ – GEdgar Jan 8 '16 at 14:25
  • $\begingroup$ It seems that for $\alpha$ rational the integral can somehow expanded in polylogs values. The method outlined above works at least also for $2\alpha=n/2$ even if things gets a little bit more messy $\endgroup$ – tired Jan 8 '16 at 18:44
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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 $n\geq2$

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  • $\begingroup$ @M.S.E Thanks, please double-check everything for computation errors, i was writing this quiet fast, so no guarantee that everything is 100% correct. $\endgroup$ – tired Jan 8 '16 at 18:57
  • $\begingroup$ nice................+1 $\endgroup$ – Bhaskara-III Jan 12 '16 at 21:34
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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...

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  • $\begingroup$ So: "show the integral converges" is a much more sensible question than "evaluate the integral". $\endgroup$ – GEdgar Jan 8 '16 at 14:24
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    $\begingroup$ the hypergeometrics look if they can somehow simplified... $\endgroup$ – tired Jan 8 '16 at 14:39
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    $\begingroup$ It looks like that there are closed forms at least for $a=1+n/2$ in terms of $\pi,\log[2] $ and $K$ (Catalan's constant) $\endgroup$ – tired Jan 8 '16 at 14:43
  • $\begingroup$ "I don't have the guts to format that properly in TeX at the moment..." I think if you right click you can copy as LaTex just in case somebody reading wants to know. Only in this case after $4 \frac{1}{2}$ years that might be a bit too late! $\endgroup$ – onepound Aug 5 '20 at 9:30
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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.

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