Prove ${\Gamma^2(1/4)\over 4\sqrt{2\pi}}=\prod_{n=1}^{\infty}\left({2n+1\over 2n}\right)^{(-1)^{n+1}}$ Integrate
$$I=\int_{0}^{\infty}{1-e^{-x}\over 1+e^{2x}}\cdot{dx\over x}={\ln{\Gamma^2(1/4)}\over 4\sqrt{2\pi}}\tag1$$
By Frullani's theorem
$$I=\sum_{n=1}^{\infty}(-1)^{n+1}\int_{0}^{\infty}{e^{-2nx}-e^{-(2n+1)x}\over x}dx\tag2$$
$$I=\sum_{n=1}^{\infty}(-1)^{n+1}\ln\left({2n+1\over 2n}\right)\tag3$$
$${\Gamma^2(1/4)\over 4\sqrt{2\pi}}=\prod_{n=1}^{\infty}\left({2n+1\over 2n}\right)^{(-1)^{n+1}}\tag4$$
How can we prove (4)?
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\color{#f00}{\int_{0}^{\infty}{1 - \expo{-x} \over 1 + \expo{2x}}\,
{\dd x \over x}} =
\int_{0}^{\infty}{\expo{-x} - \expo{-2x} \over 1 + \expo{-2x}}\,\expo{-x}\,
{\dd x \over x}\ \stackrel{\phantom{AA}t\ =\ \expo{-x}}{=}\
-\int_{0}^{1}{t - t^{2} \over 1 + t^{2}}\,{\dd t \over \ln\pars{t}}
\\[3mm] = &\
\int_{0}^{1}{t - t^{2} \over 1 + t^{2}}\int_{0}^{\infty}t^{\mu}\,\dd\mu\,\dd t =
\int_{0}^{\infty}\int_{0}^{1}{t^{\mu + 1} - t^{\mu + 2} \over 1 + t^{2}}
\,\dd t\,\dd\mu =
\half\int_{0}^{\infty}\int_{0}^{1}{t^{\mu/2} - t^{\mu/2 + 1/2} \over 1 + t}
\,\dd t\,\dd\mu
\\[3mm] = &\
\half\int_{0}^{\infty}\bracks{%
2\int_{0}^{1}{t^{\mu/2} - t^{\mu/2 + 1/2} \over 1 - t^{2}}\,\dd t -
\int_{0}^{1}{t^{\mu/2} - t^{\mu/2 + 1/2} \over 1 - t}\,\dd t}\,\dd\mu
\\[3mm] = &\
\half\int_{0}^{\infty}\bracks{%
\int_{0}^{1}{t^{\mu/4 - 1/2} - t^{\mu/4 - 1/4} \over 1 - t}\,\dd t -
\int_{0}^{1}{t^{\mu/2} - t^{\mu/2 + 1/2} \over 1 - t}\,\dd t}\,\dd\mu
\\[3mm] = &
\half\int_{0}^{\infty}\bracks{%
\Psi\pars{{\mu \over 4} + {3 \over 4}} -
\Psi\pars{{\mu \over 4} + \half} +
\Psi\pars{{\mu \over 2} + 1} -
\Psi\pars{{\mu \over 2} + {3 \over 2}}}\,\dd\mu
\\[3mm] = &
\left.\half
\ln\pars{{\Gamma^{4}\pars{\mu/4 + 3/4}\Gamma^{2}\pars{\mu/2 + 1}
\over \Gamma^{4}\pars{\mu/4 + 1/2}\Gamma^{2}\pars{\mu/2 + 3/2}}}
\right\vert_{\ 0}^{\infty} =
\bracks{-\,\half\ln\pars{2}} - \bracks{%
-\,\half\,\ln\pars{2\pi} + \ln\pars{{\Gamma\pars{3/4} \over \Gamma\pars{5/4}}}}
\\[3mm] = &\
\ln\pars{\root{\pi}\,{\Gamma\pars{5/4} \over \Gamma\pars{3/4}}}
\end{align}

However,
$$
\root{\pi}\,{\Gamma\pars{5/4} \over \Gamma\pars{3/4}} =
\root{\pi}\,{1 \over 4}\,\Gamma\pars{1/4}\,
{1 \over \pi/\bracks{\Gamma\pars{1/4}\sin\pars{\pi/4}}}
={\Gamma^{2}\pars{1/4} \over 4\root{2\pi}}
$$
and
$$
\color{#f00}{\int_{0}^{\infty}{1 - \expo{-x} \over 1 + \expo{2x}}\,
{\dd x \over x}} =
\color{#f00}{\ln\pars{{\Gamma^{2}\pars{1/4} \over 4\root{2\pi}}}}
\approx 0.2708
$$
A: $$\prod_{n=1}^{2N}\left(\frac{2n+1}{2n}\right)^{(-1)^{n+1}} = \prod_{m=1}^{N}\frac{(4m)(4m-1)}{(4m+1)(4m-2)}=\frac{\sqrt{\pi }\, \Gamma\left(\frac{5}{4}\right) \Gamma\left(\frac{3}{4}+N\right) \Gamma(1+N)}{\Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{1}{2}+N\right) \Gamma\left(\frac{5}{4}+N\right)} $$
now it is enough to consider the limit as $N\to +\infty$ to recover the wanted expression.
