# Prove $\int_{0}^{\infty}{1\over x}\cdot{1-e^{-\phi{x}}\over 1+e^{\phi{x}}}dx=\ln\left({\pi\over 2}\right)$

Integrate

$$I=\int_{0}^{\infty}{1\over x}\cdot{1-e^{-\phi{x}}\over 1+e^{\phi{x}}}\,dx=\ln\left({\pi\over 2}\right)\tag1,$$

where $\phi={1+\sqrt5\over 2}$.

Recall

$\tanh y=-{1-e^{2y}\over 1+e^{2y}}$, setting $y={\phi{x}\over 2}$ we have

$$\tanh\left({\phi{x}\over 2}\right)=-{1-e^{x\phi}\over 1+e^{x\phi}}\tag2.$$

Sub $(2)$ into $(1)\rightarrow (3)$

$$I=-\int_{0}^{\infty}{\tanh\left({x\phi\over 2}\right)\over x}\,dx\tag4.$$

Recall

$$\tanh x=\sum_{n=1}^{\infty}{(-1)^{n-1}2^{2n}(2^{2n}-1)B_nx^{2n-1}\over (2n)!}\tag5$$

Sub $(5)$ into $(4)\rightarrow (6)$

$$I=\sum_{n=1}^{\infty}{(-1)^{n-1}2^{2n}(2^{2n}-1)B_nx^{2n-1}\over (2n)!}\int_{0}^{\infty}{\left({x\phi\over 2}\right)^{2n-1}\over x}\,dx\tag6.$$

The problem is that

$$\int_{0}^{\infty}{\left({x\phi\over 2}\right)^{2n-1}\over x}dx$$ diverges. I went wrong somewhere, can anyone help please?

• Looks like you can do something like I did here, in the Frullani part. – mickep Jun 8 '16 at 13:13
• Substituting $u=\phi x$ in the first expression, one finds that $\phi$ drops out entirely. So $\phi$ is entirely superfluous and the problem is really to show $\displaystyle \int_0^\infty \frac{1-e^{-u}}{1+e^u}\frac{du}{u}=\ln\frac{\pi}{2}.$ – Semiclassical Jun 8 '16 at 13:31
• in those cases, expand $\frac{1-e^{-u}}{1+e^u}$ in term of $e^{-u}$ : $\frac{1-e^{-u}}{1+e^u} = \frac{e^{-u}-e^{-2u}}{1+e^{-u}} = (e^{-u}-e^{-2u}) \sum_{n=0}^\infty (-1)^n e^{-nu} = \sum_{n=1}^\infty (-1)^{n+1} e^{-nu} -\sum_{n=2}^\infty (-1)^{n} e^{-nu}$ $= -1 +2 \sum_{n=1}^\infty (-1)^{n+1} e^{-nu}$, and $\int_0^\infty e^{-nu} du = \frac{1}{n}$ so that $\int_0^\infty \frac{1-e^{-u}}{1+e^u} = -1 +2 \sum_{n=1}^\infty (-1)^{n+1} \frac{1}{n} = -1 + 2 \eta(1) = -1 + 2 \ln 2$ – reuns Jun 8 '16 at 13:40
• The radius of convergence for the series for $\tanh$ isn't infinite, so you can't use it for $\left({0 \ .. +\infty}\right)$. – GFauxPas Jun 8 '16 at 14:18

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ It's clear that the integral is $\ds{\phi}$-independent. ${\phi \equiv {1 + \root{5} \over 2} > 0}$.

Namely, $\ds{\int_{0}^{\infty}{1 - \expo{-\phi x} \over 1 + \expo{\phi x}} \,{\dd x \over x}\ \stackrel{\phi x\ \to\ x}{=}\ \int_{0}^{\infty}{1 - \expo{-x} \over 1 + \expo{x}}\,{\dd x \over x}}$

\begin{align} \int_{0}^{\infty}{1 - \expo{-x} \over 1 + \expo{x}}\,{\dd x \over x} & = \int_{0}^{\infty}{1 - \expo{-x} \over 1 + \expo{-x}}\,\expo{-x}\,{\dd x \over x} \ \stackrel{\expo{-x}\ =\ t}{=}\ -\int_{0}^{1}{1 - t \over 1 + t}\,{\dd t \over \ln\pars{t}} \\[3mm] & = \int_{0}^{1}{1 - t \over 1 + t}\ \overbrace{\int_{0}^{\infty}t^{\mu}\,\dd\mu}^{\ds{-\,{1 \over \ln\pars{t}}}}\ \,\dd t = \int_{0}^{\infty}\int_{0}^{1}{t^{\mu} - t^{\mu + 1} \over 1 + t}\,\dd\mu\,\dd t \\[3mm] & = \int_{0}^{\infty}\bracks{% 2\int_{0}^{1}{t^{\mu} - t^{\mu + 1} \over 1 - t^{2}}\,\dd\mu - \int_{0}^{1}{t^{\mu} - t^{\mu + 1} \over 1 - t}\,\dd\mu} \\[3mm] & = \int_{0}^{\infty}\bracks{% \int_{0}^{1}{t^{\mu/2 - 1/2} - t^{\mu/2} \over 1 - t}\,\dd\mu - \int_{0}^{1}{t^{\mu} - t^{\mu + 1} \over 1 - t}\,\dd\mu} \\[8mm] & = \int_{0}^{\infty}\left\lbrack% \int_{0}^{1}{1 - t^{\mu/2} \over 1 - t}\,\dd\mu - \int_{0}^{1}{1 - t^{\mu/2 - 1/2} \over 1 - t}\,\dd\mu\right. \\[3mm] & \left.\mbox{} + \int_{0}^{1}{1 - t^{\mu} \over 1 - t}\,\dd\mu - \int_{0}^{1}{1 - t^{\mu + 1} \over 1 - t}\,\dd\mu\right\rbrack \\[8mm] & = \int_{0}^{\infty}\bracks{% \Psi\pars{1 + {\mu \over 2}} - \Psi\pars{\half + {\mu \over 2}} + \Psi\pars{1 + \mu} - \Psi\pars{2 + \mu}}\,\dd\mu \\[3mm] & = \left.% \ln\pars{\Gamma^{2}\pars{1 + \mu/2}\Gamma\pars{1 + \mu} \over \Gamma^{2}\pars{1/2 + \mu/2}\Gamma\pars{2 + \mu}} \right\vert_{\ 0}^{\ \infty} \\[3mm] & = \underbrace{% \lim_{\mu \to \infty}\ln\pars{\Gamma^{2}\pars{1 + \mu/2} \over \Gamma^{2}\pars{1/2 + \mu/2}\pars{1 + \mu}}}_{\ds{-\ln\pars{2}}}\ -\ \underbrace{\ln\pars{\Gamma^{2}\pars{1}\Gamma\pars{1} \over \Gamma^{2}\pars{1/2}\Gamma\pars{2}}}_{\ds{-\ln\pars{\pi}}}\ =\ \color{#f00}{\ln\pars{\pi \over 2}} \end{align}

where $\Psi$ is the Digamma function and, by definition, $\ds{\Psi\pars{z} = \totald{\ln\pars{\Gamma\pars{z}}}{z}}$. In the above calculation we used the well known identities ( $\gamma$ is the Euler-Mascheroni constant ): \begin{align} \Psi\pars{z} + \gamma & = \int_{0}^{1}{1 - t^{z - 1} \over 1 - t}\,\dd t\,,\qquad\Re\pars{z} > 0 \\[3mm] \Gamma\pars{1} & = \Gamma\pars{2} = 1\,,\quad\Gamma\pars{\half} = \root{\pi} \,,\quad\Gamma\pars{z + 1} = z\,\Gamma\pars{z} \end{align}

The last $\ds{\mu \to \infty}$ limit can be evaluated with Stirling Asymptotic Formula.

Hint:

First, set $\phi{x}\mapsto x$. We have

$$I(a):=\int_{0}^{\infty}{1\over x}\cdot{1-e^{-{x}}\over 1+e^{{x}}}dx=\int_{0}^{\infty}{1\over x}\cdot{e^{-{x}}-e^{-2{x}}\over 1+e^{-{x}}}dx$$

Consider the parametric integral

$$I(a):=\int_{0}^{\infty}{e^{-(a-1)x}\over x}\cdot{e^{-{x}}-e^{-2{x}}\over 1+e^{-{x}}}dx$$

and

$$I'(a)=\int_{0}^{\infty}{{e^{-(a+1)x}-e^{-ax}}\over 1+e^{-{x}}}dx$$

Now use the geometric expansion

$${1\over 1+e^{-{x}}}=\sum_{k=0}^\infty\,(-1)^ke^{-k{x}}$$

and the following relation

$$\sum_{k=0}^\infty\frac{(-1)^k}{(z+k)^{m+1}}=\frac1{(-2)^{m+1}m!}\!\left(\psi_m\left(\frac{z}{2}\right)-\psi_m\!\left(\frac{z+1}{2}\right)\right)$$ then do as shown in this answer.

$$I = \sum_{n\geq 1}(-1)^{n+1}\int_{0}^{+\infty}\frac{e^{-n\varphi x}-e^{-(n+1)\varphi x}}{x}\,dx =\sum_{n\geq 1}(-1)^{n+1}\log\left(\frac{n+1}{n}\right)\tag{1}$$ and the RHS is the logarithm of Wallis' product, hence $\color{red}{\log\frac{\pi}{2}}$.

You may also notice that the $\varphi$ constant is irrelevant here, since it can be eliminated by the substitution $x=\frac{z}{\varphi}$ in the original integral.

You may like this answer. Noting $$\lim_{n\to\infty}n(t^{1/n}-1)=\ln t$$ then from @Felix Marin's solution, one has \begin{eqnarray} &&\int_{0}^{\infty}{1\over x}\cdot{1-e^{-\phi{x}}\over 1+e^{\phi{x}}}\,dx\\ &=&-\int_0^1\frac{1-t}{1+t}\frac{1}{\ln t}dt\\ &=&-\lim_{n\to\infty}\int_0^1\frac{1-t}{1+t}\frac{1}{n(t^{1/n}-1)}dt\\ &=&\lim_{n\to\infty}\frac{1}{n}\int_0^1\frac{\sum_{k=0}^{n-1}t^{\frac{k}{n}}}{1+t}dt\\ &=&\lim_{n\to\infty}\frac{1}{n}\int_0^1\sum_{j=0}^\infty\sum_{k=0}^{n-1}(-1)^jt^{j+\frac{k}{n}}dt\\ &=&\lim_{n\to\infty}\frac{1}{n}\sum_{j=0}^\infty\sum_{k=0}^{n-1}(-1)^j\frac{1}{j+\frac{k}{n}+1}\\ &=&\sum_{j=0}^\infty(-1)^j\int_0^1\frac{1}{j+t+1}dt\\ &=&\sum_{j=0}^\infty(-1)^j\ln\frac{j+2}{j+1}\\ &=&\ln\prod_{j=1}^\infty\left(\frac{2j}{2j-1}\cdot\frac{2j}{2j+1}\right)\\ &=&\ln\left(\frac{\pi}{2}\right) \end{eqnarray} by the Wallis product.