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

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?

• After a substitution of y = 2^x-1, does this not reduce to knowledge about identities of polylogaritms ? – mick Oct 9 '13 at 21:41

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.

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

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}

Let $\tan^2t=2^x-1\;\Rightarrow\;x=\dfrac{\ln(1+\tan^2t)}{\ln 2}\;\Rightarrow\;dx=\dfrac{2\tan t\sec^2t\ dt}{(1+\tan^2t)\ln 2}$ and the corresponding region is $0<t<\dfrac\pi2$. Using identity $\sec^2t=1+\tan^2t$, then the integral turns out to be $$\mathcal{I}=\frac{1}{\ln^22}\int_0^{\Large\frac\pi2}\frac{2\ln(\sec t)-\ln2}{\ln (\tan t)}\ dt.\tag1$$ Now, using property $$\int_b^af(x)\ dx=\int_b^af(a+b-x)\ dx$$ equation $(1)$ becomes $$\mathcal{I}=\frac{1}{\ln^22}\int_0^{\Large\frac\pi2}\frac{2\ln(\csc t)-\ln2}{\ln (\cot t)}\ dt.\tag2$$ Adding $1$ and $2$ yields \begin{align} 2\mathcal{I}&=\frac{1}{\ln^22}\int_0^{\Large\frac\pi2}\left(\frac{2\ln(\sec t)-\ln2}{\ln (\tan t)}+\frac{2\ln(\csc t)-\ln2}{\ln (\cot t)}\right)\ dt\\ &=\frac{1}{\ln^22}\int_0^{\Large\frac\pi2}\left(\frac{2\ln\left(\dfrac{1}{\cos t}\right)-\ln2}{\ln (\tan t)}+\frac{2\ln\left(\dfrac{1}{\sin t}\right)-\ln2}{\ln \left(\dfrac{1}{\tan t}\right)}\right)\ dt\\ &=\frac{1}{\ln^22}\int_0^{\Large\frac\pi2}\left(\frac{-2\ln(\cos t)-\ln2}{\ln (\tan t)}+\frac{2\ln(\sin t)+\ln2}{\ln (\tan t)}\right)\ dt\\ &=\frac{2}{\ln^22}\int_0^{\Large\frac\pi2}\frac{\ln(\sin t)-\ln(\cos t)}{\ln \left(\dfrac{\sin t}{\cos t}\right)}\ dt\\ \mathcal{I}&=\frac{1}{\ln^22}\int_0^{\Large\frac\pi2}\ dt\\\\ \int_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx&=\large\color{blue}{\frac{\pi}{2\ln^22}}.\qquad\qquad\qquad\blacksquare \end{align}

• Wow, this is remarkably brilliant! I'm impressed. – Sophie Agnesi Jun 10 '16 at 12:06

Substitute $(2^x-1) = t^2$ to get,

$\text{I} = \displaystyle \dfrac{1}{\ln^2 2} \int_{0}^{\infty} \left( \dfrac{\ln (t^2+1) - \ln 2}{(t^2 + 1) \ln t} \right) \mathrm{d}t$

Substitute $t \mapsto \dfrac{1}{t}$

$\implies \text{I} = -\displaystyle \dfrac{1}{\ln^2 2} \int_{0}^{\infty} \left( \dfrac{\ln (t^2+1) - \ln 2}{(t^2 + 1) \ln t} \right) \mathrm{d}t + \dfrac{2}{\ln^2 2} \int_{0}^{\infty} \dfrac{\mathrm{d}t}{t^2+1}$

$\implies \text{I} = -\text{I} + \dfrac{\pi}{\ln^2 2}$

$\implies \text{I} = \dfrac{\pi}{2 \ln^2 2}$

• good one $(+1)$ – tired Jul 9 '17 at 20:48
• And with the right substitution the integral just dissolves away. Nice work. – omegadot Jan 25 at 1:04