I want to prove the variation from here

$$\int_{0}^{1}{(x-1)^2\over \sqrt{2^x-1}\ln(2^x-1)}dx=\color{blue}{\pi\over \ln^2{2}}\tag1$$

Sub $u=2^x-1\rightarrow du=2^x\ln{2}dx$

$x=1\rightarrow u=1$ and $x=0\rightarrow u=0$

$(x-1)^2={1\over \ln^2{2}}\ln^2\left({1+u\over 2}\right)$

$$I={1\over \ln{2}}\int_{0}^{1}{(x-1)^2\over \sqrt{u}(u+1)\ln{u}}du\tag2$$

$$I={1\over \ln^3{2}}\int_{0}^{1}{\ln^2\left({u+1\over 2}\right)\over \sqrt{u}(u+1)\ln{u}}du\tag3$$

Sub $u=e^z\rightarrow du=e^zdz$

$u=1\rightarrow z=0$ and $u=0\rightarrow z=-\infty$

${e^z+1\over 2}=e^{z/2}\cosh(z/2)$

$$I=-{1\over \ln^3{2}}\int_{0}^{\infty}{\ln^2(e^{z/2}\cosh(z/2))\over ze^{-z/2}(1+e^z)}dz\tag4$$


$$I=-{1\over \ln^3{2}}\int_{0}^{\infty}{\ln^2(e^{z/2}\cosh(z/2))\over 2z\cosh(z/2)}dz\tag5$$

I am stuck! Can anyone demonstrate how to prove I in step by step manner please, thank.

I try to understand their proofs From here, but couldn't follow.


marked as duplicate by Jack D'Aurizio integration Jun 10 '16 at 11:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Since when $(x-1)=(x-1)^2$. If that is true than I agree it is a duplicate. The closed form is also different. $\endgroup$ – gymbvghjkgkjkhgfkl Jun 10 '16 at 13:54
  • $\begingroup$ @JackD'Aurizio: At a superficial look, at least, the two integrals do not seem identical: this one has a $(x-1)^2$, while the other one only a $x-1$. The upper bound here is $1$, while it is $\infty$ there. Could you give a hint regarding why you consider this one to be a duplicate, please? $\endgroup$ – Alex M. Jun 10 '16 at 19:04
  • $\begingroup$ @AlexM.: Ron Gordon's technique applies here, too, with minor changes. Probably not an exact duplicate, but an abstract duplicate for sure. Honestly, I voted for closing in order to push the OP towards the study of the solutions already on MSE, opposed to the tendency to collect answers as soon as possible, without reading them too carefully. Anyway, I might be wrong, and you (and others) are free to vote for reopening. $\endgroup$ – Jack D'Aurizio Jun 10 '16 at 21:33
  • $\begingroup$ @JackD'Aurizio: The thing is that the OP himself links to that other page, and begins the post by stating that his post is a variation of the other. To me, this is a sign that he has already studied that page ("due dilligence"). $\endgroup$ – Alex M. Jun 10 '16 at 22:07