Integral

[Tolaso J Kos]

Evaluate the integral:

$$\mathcal{J}= \int_{0}^{\infty }\frac{x-1}{\sqrt{2^x-1}\ln \left ( 2^x-1 \right )}\, {\rm d}x$$

[Tsakanikas Nickos]

Thank you, Apostolos, for your hint!
Here is the requested solution:

\begin{align*}
\int_{0}^{+\infty} \frac{x-1}{ \sqrt{2^{x}-1} \log(2^{x}-1) } \mathrm{d}x
&\overset{2^{x}-1=u \, , \, dx = \frac{du}{(\log2)(u+1)}}{\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=} \int_{0}^{+\infty} \frac{ \frac{1}{\log2}\left(\log(u+1) -\log2\right)}{(\log2)(u+1)\sqrt{u}\log{u}} \mathrm{d}u \\
&= \frac{1}{\log^{2}2} \int_{0}^{+\infty} \frac{ \log(u+1) - \log2 }{(u+1)\sqrt{u}\log{u}} \mathrm{d}u \\
&= \frac{1}{\log^{2}2} \left[ \int_{0}^{1}\frac{ \log(u+1) - \log2 }{(u+1)\sqrt{u}\log{u}} \mathrm{d}u + \int_{1}^{+\infty} \frac{\log(u+1) -\log2}{(\log2)(u+1)\sqrt{u}\log{u}} \mathrm{d}u \right] \\
&\overset{u=\frac{1}{t} \, , \, du=-\frac{1}{t^{2}}dt}{\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=} \frac{1}{\log^{2}2} \left[ \int_{0}^{1} \frac{ \log(u+1) - \log2 }{(u+1)\sqrt{u}\log{u}} \mathrm{d}u + \int_{1}^{0} \frac{ \log(\frac{1}{t}+1) -\log 2 }{(\frac{1}{t}+1)\sqrt{\frac{1}{t}}\log(\frac{1}{t})} \left( -\frac{1}{t^2} \right) \mathrm{d}t \right] \\
&= \frac{1}{\log^{2}2} \left[ \int_{0}^{1} \frac{ \log(u+1) - \log2 }{(u+1)\sqrt{u}\log{u}} \mathrm{d}u - \int_{0}^{1} \frac{ \log(t+1) - \log{t} - \log2 }{ t^{2} \frac{t+1}{t}\frac{1}{\sqrt{t}}\log{t} } \mathrm{d}t \right] \\
&= \frac{1}{\log^{2}2} \left[ \int_{0}^{1} \frac{ \log(u+1) - \log2 }{(u+1)\sqrt{u}\log{u}} \mathrm{d}u - \int_{0}^{1} \frac{ \log(t+1) - \log{t} - \log2 }{ (t+1)\sqrt{t}\log{t} } \mathrm{d}t \right] \\
&\overset{t=u \text{ in the second integral (dummy variable) } }{\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=} \frac{1}{\log^{2}2} \int_{0}^{1} \frac{1}{ (u+1)\sqrt{u} } \mathrm{d}u \\
&\overset{ \sqrt{u} = \xi \implies u=\xi^2 \, , \, du = 2\xi d\xi }{\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=} \frac{1}{\log^{2}2} \int_{0}^{1} \frac{2\xi}{(\xi^2+1)\xi} \mathrm{d}\xi \\
&= \frac{2}{\log^{2}2} \arctan{\xi} \Big|_{0}^{1} \\
&= \frac{\pi}{2\log^{2}2}
\end{align*}