Consider the following integral:
$$\mathcal{I}=\int\limits_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx.$$
I tried to evaluate $\mathcal{I}$ in a closed form (both manually and using Mathematica), but without success.
However, if WolframAlpha is provided with a numerical approximation $\,\mathcal{I}\approx 3.2694067500684...$, it returns a possible closed form:
$$\mathcal{I}\stackrel?=\frac\pi{2\,\ln^2 2}.$$
Further numeric caclulations show that this value is correct up to at least $10^3$ decimal digits. So, I conjecture that this is the exact value of $\mathcal{I}$.
Question: Is this conjecture correct?
Answer
Sub $u=\log{(2^x-1)}$. Then $x=\log{(1+e^u)}/\log{2}$, $dx = (1/\log{2}) (du/(1+e^{-u})$. The integral then becomes
$$\begin{align}\frac{1}{\log{2}} \int_{-\infty}^{\infty} \frac{du}{1+e^{-u}} e^{-u/2} \frac{\frac{\log{(1+e^u)}}{\log{2}}-1}{u} = \frac{1}{2\log^2{2}} \int_{-\infty}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}\\ = \underbrace{\frac{1}{2\log^2{2}} \int_{-\infty}^{0} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}}_{u\rightarrow -u} \\+ \frac{1}{2\log^2{2}} \int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}\\ = \underbrace{-\frac{1}{2\log^2{2}} \int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^{-u})}-\log{2}}{u}}_{\log{(1+e^{-u})} = \log{(1+e^u)}-u}\\+ \frac{1}{2\log^2{2}} \int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} \frac{\log{(1+e^u)}-\log{2}}{u}\\ \end{align}$$
The nasty pieces of the integral cancel, and we are left with
$$ \frac{1}{2\log^2{2}}\int_{0}^{\infty} \frac{du}{\cosh{(u/2)}} = \frac{\pi}{2 \log^2{2}} $$
as correctly conjectured.
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