Prove ${\Gamma^2(1/4)\over 4\sqrt{2\pi}}=\prod_{n=1}^{\infty}\left({2n+1\over 2n}\right)^{(-1)^{n+1}}$

Question

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)?

Answer 1

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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}

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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 $$ \color{#f00}{\int_{0}^{\infty}{1 - \expo{-x} \over 1 + \expo{2x}}\, {\dd x \over x}} = \color{#f00}{\ln\pars{{\Gamma^{2}\pars{1/4} \over 4\root{2\pi}}}} \approx 0.2708 $$

Answer 2

$$\prod_{n=1}^{2N}\left(\frac{2n+1}{2n}\right)^{(-1)^{n+1}} = \prod_{m=1}^{N}\frac{(4m)(4m-1)}{(4m+1)(4m-2)}=\frac{\sqrt{\pi }\, \Gamma\left(\frac{5}{4}\right) \Gamma\left(\frac{3}{4}+N\right) \Gamma(1+N)}{\Gamma\left(\frac{3}{4}\right) \Gamma\left(\frac{1}{2}+N\right) \Gamma\left(\frac{5}{4}+N\right)} $$

now it is enough to consider the limit as $N\to +\infty$ to recover the wanted expression.