# Consider the following integral: I=\int_0^\infty \frac{x-1}{\sqrt{2^x-1}\ln(2^x-1)}dx

Consider the following integral:
$I={\int }_{0}^{\mathrm{\infty }}\frac{x-1}{\sqrt{{2}^{x}-1}\mathrm{ln}\left({2}^{x}-1\right)}dx$
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Ronnie Schechter
Sub $u=\mathrm{log}\left({2}^{x}-1\right)$. Then . The integral then becomes
$\frac{1}{\mathrm{log}2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{du}{1+{e}^{-u}}\frac{\frac{\mathrm{log}\left(1+{e}^{u}\right)}{\mathrm{log}2}-1}{u}=\frac{1}{2{\mathrm{log}}^{2}2}$
${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{du}{\text{cosh}\left(\frac{u}{2}\right)}\frac{\mathrm{log}\left(1+{e}^{u}\right)-\mathrm{log}2}{u}$
$=\frac{1}{2{\mathrm{log}}^{2}2}{\int }_{-\mathrm{\infty }}^{0}\frac{du}{\text{cosh}\left(\frac{u}{2}\right)}\frac{\mathrm{log}\left(1+{e}^{u}\right)-\mathrm{log}2}{u}+\frac{1}{2{\mathrm{log}}^{2}2}{\int }_{0}^{\mathrm{\infty }}\frac{du}{\text{cosh}\left(\frac{u}{2}\right)}\frac{\mathrm{log}\left(1+{e}^{u}\right)-\mathrm{log}2}{u}$
The nasty pieces of the integral cancel, and we are left with
$\frac{1}{2{\mathrm{log}}^{2}2}{\int }_{0}^{\mathrm{\infty }}\frac{du}{\text{cosh}\left(\frac{u}{2}\right)}=\frac{\pi }{2{\mathrm{log}}^{2}2}$
as correctly conjured.
###### Not exactly what you’re looking for?
enhebrevz
$I={\int }_{0}^{\mathrm{\infty }}\frac{x-1}{\sqrt{{2}^{x}-1}\mathrm{ln}\left({2}^{x}-1\right)}dx$
With the change of variables $z\equiv {2}^{x}-1\to x=\frac{\mathrm{ln}\left(1+z\right)}{\mathrm{ln}\left(2\right)}$, I is reduced to
$I=\frac{1}{{\mathrm{ln}}^{2}\left(2\right)}{\int }_{0}^{\mathrm{\infty }}\frac{\mathrm{ln}\left(1+z\right)-\mathrm{ln}\left(2\right)}{{z}^{\frac{1}{2}}\left(1+z\right)\mathrm{ln}\left(z\right)}dz$
Now, we split the integral from (0,1) and from $\left(1,\mathrm{\infty }\right)$. In the second one, we makes the change $z\to \frac{1}{z}$ such that we are left with an integration over $\left(0,1\right)$:
$I=\frac{1}{{\mathrm{ln}}^{2}\left(2\right)}{\int }_{0}^{1}\frac{\mathrm{ln}\left(1+z\right)-\mathrm{ln}\left(2\right)}{{z}^{\frac{1}{2}}\left(1+z\right)\mathrm{ln}\left(z\right)}dz+\frac{1}{{\mathrm{ln}}^{2}\left(2\right)}{\int }_{0}^{1}\frac{\mathrm{ln}\left(1+\frac{1}{z}\right)-\mathrm{ln}\left(2\right)}{{z}^{-\frac{1}{2}}\left(1+\frac{1}{z}\right)\left[-\mathrm{ln}\left(z\right)\right]}\frac{dz}{{z}^{2}}$
$=\frac{1}{{\mathrm{ln}}^{2}\left(2\right)}{\int }_{0}^{1}\frac{\mathrm{ln}\left(1+z\right)-\mathrm{ln}\left(2\right)}{{z}^{\frac{1}{2}}\left(1+z\right)\mathrm{ln}\left(z\right)}dz-\frac{1}{{\mathrm{ln}}^{2}\left(2\right)}$
${\int }_{0}^{1}\frac{\mathrm{ln}\left(1+z\right)-\mathrm{ln}\left(z\right)-\mathrm{ln}\left(2\right)}{{z}^{\frac{1}{2}}\left(1+z\right)\mathrm{ln}\left(z\right)}dz$
###### Not exactly what you’re looking for?
karton

Substitute $\left({2}^{x}-1\right)={t}^{2}$ to get,
$I=\frac{1}{{\mathrm{ln}}^{2}2}{\int }_{0}^{\mathrm{\infty }}\left(\frac{\mathrm{ln}\left({t}^{2}+1\right)-\mathrm{ln}2}{\left({t}^{2}+1\right)\mathrm{ln}t}\right)dt$
Substitute $t↦\frac{1}{t}$
$\to I=-\frac{1}{{\mathrm{ln}}^{2}2}{\int }_{0}^{\mathrm{\infty }}\left(\frac{\mathrm{ln}\left({t}^{2}+1\right)-\mathrm{ln}2}{\left({t}^{2}+1\right)\mathrm{ln}t}dt+\frac{2}{{\mathrm{ln}}^{2}2}{\int }_{0}^{\mathrm{\infty }}\frac{dt}{{t}^{2}+1}\phantom{\rule{0ex}{0ex}}\to I=-I+\frac{\pi }{{\mathrm{ln}}^{2}2}\phantom{\rule{0ex}{0ex}}\to I=\frac{\pi }{2{\mathrm{ln}}^{2}2}$