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?
 A: Hint:
First, set $\phi{x}\mapsto x$. We have
\begin{equation}
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
\end{equation}
Consider the parametric integral
\begin{equation}
I(a):=\int_{0}^{\infty}{e^{-(a-1)x}\over x}\cdot{e^{-{x}}-e^{-2{x}}\over 1+e^{-{x}}}dx
\end{equation}
and
\begin{equation}
I'(a)=\int_{0}^{\infty}{{e^{-(a+1)x}-e^{-ax}}\over 1+e^{-{x}}}dx
\end{equation}
Now use the geometric expansion
\begin{equation}
{1\over 1+e^{-{x}}}=\sum_{k=0}^\infty\,(-1)^ke^{-k{x}}
\end{equation}
and the following relation
\begin{equation}
\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)
\end{equation}
then do as shown in this answer.
A: By Frullani's theorem:
$$ I = \sum_{n\geq 1}(-1)^{n+1}\int_{0}^{+\infty}\frac{e^{-n\varphi x}-e^{-(n+1)\varphi x}}{x}\,dx =\sum_{n\geq 1}(-1)^{n+1}\log\left(\frac{n+1}{n}\right)\tag{1}$$
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.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\imp}{\Longrightarrow}
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 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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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.
A: 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.
