A conjectured closed form of $\int\limits_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx$

Question

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 1

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.

Answer 2

$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\expo}[1]{{\rm e}^{#1}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\pp}{{\cal P}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert} \newcommand{\yy}{\Longleftrightarrow}$ $\ds{% {\cal I} \equiv \int\limits_{0}^{\infty} {x - 1 \over \sqrt{\vphantom{\large A}2^{x} - 1\,}\,\ln\pars{2^{x} - 1}}\,\dd x:\ {\large ?}}$

With the change of variables $z \equiv 2^{x} - 1\yy x = \ln\pars{1 + z}/\ln\pars{2},\ {\cal I}$ is reduced to $$ {\cal I} = {1 \over \ln^{2}\pars{2}} \int\limits_{0}^{\infty}{\ln\pars{1 + z} - \ln\pars{2} \over z^{1/2}\,\pars{1 + z}\,\ln\pars{z}} \,\dd z $$ Now, we split the integral from $\pars{0, 1}$ and from $\pars{1, \infty}$. In the second one, we makes the change $z \to 1/z$ such that we are left with an integration over $\pars{0, 1}$: \begin{align} {\cal I} &= {1 \over \ln^{2}\pars{2}} \int\limits_{0}^{1}{\ln\pars{1 + z} - \ln\pars{2} \over z^{1/2}\,\pars{1 + z}\,\ln\pars{z}} \,\dd z + {1 \over \ln^{2}\pars{2}}\int\limits_{0}^{1} {\ln\pars{1 + 1/z} - \ln\pars{2} \over z^{-1/2}\,\pars{1 + 1/z}\,\bracks{-\ln\pars{z}}} \,{\dd z \over z^{2}} \\[3mm]&= {1 \over \ln^{2}\pars{2}} \int\limits_{0}^{1}{\ln\pars{1 + z} - \ln\pars{2} \over z^{1/2}\,\pars{1 + z}\,\ln\pars{z}} \,\dd z - {1 \over \ln^{2}\pars{2}}\int\limits_{0}^{1} {\ln\pars{1 + z} - \ln\pars{z} - \ln\pars{2} \over z^{1/2}\,\pars{1 + z}\,\ln\pars{z}}\,\dd z \\[3mm]&= {1 \over \ln^{2}\pars{2}} \int\limits_{0}^{1}{1 \over z^{1/2}\,\pars{1 + z}} \,\dd z\,, \quad \pars{~\mbox{Let's}\quad r \equiv z^{1/2}\yy\ z = r^{2}~} \\[3mm]&= {2 \over \ln^{2}\pars{2}} \underbrace{\quad\int\limits_{0}^{1}{\dd r \over r^{2} + 1}\quad} _{\ds{\arctan\pars{1}\ =\ {\pi \over 4}}} = \color{#ff0000}{\Large{\pi \over 2\ln^{2}\pars{2}}} \end{align}