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$$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right)=\ln\left(\sqrt{2\over \pi}\cdot{2\over \Gamma^2\left({3\over 4}\right)}\right)\tag1$$

Where $\beta(n)$ is Beta dirichlet function

$(1)$ becomes

$$ln\left({2\over \pi}\right)+{1\over n\Gamma(n)}\sum_{n=1}^{\infty}(-1)^{n-1}\int_{0}^{\infty}{x^{n-1}\over e^x+e^{-x}}\mathrm dx\tag2$$

How can we show that the closed form for $(1)$ is correct?

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    $\begingroup$ And why would anybody give a damn, just because it's non-trivial, so you can't solve it? If that were a criterion, most people would work on Riemann Hypothesis, now. So why this? Is it necessary for anything important you're working upon, or just for the lolz? $\endgroup$
    – user436658
    Jun 24 '17 at 19:59
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    $\begingroup$ @ProfessorVector Sometimes it's just fun to work on a hard problem. $\endgroup$ Jun 24 '17 at 20:00
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    $\begingroup$ @Will Fisher Granted, but I haven't seen any work of the OP on that problem. $\endgroup$
    – user436658
    Jun 24 '17 at 20:10
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    $\begingroup$ @ProfessorVector: it's a common trend here on MSE. Questions involving tough integrals and series are always appreciated and no one demand any work from OP. The only reason for this behavior is that most people here love solving these problems and are ready for any question which gives opportunity to show their skills. Work from OP is expected mostly for problems which are not too difficult. $\endgroup$ Jun 25 '17 at 3:58
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    $\begingroup$ @Paramanand Singh You mean that everybody who had a chance to show off upvotes questions, no matter how useless they are? That would explain a lot of what I've seen at MSE, indeed. $\endgroup$
    – user436658
    Jun 25 '17 at 5:28
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Let us consider the two sums separately: $$ S_1 = \sum_{n=1}^\infty (-1)^{n-1}\frac{\beta(n)}{n} \\ S_2 = \sum_{n=1}^\infty (-1)^{n-1}\ln\frac{n+1}{n} $$

For $S_1$, let us first consider the related sum, for which $S_1$ is the limiting value $S(1)$: $$ S(x) = \sum_{n=1}^\infty (-1)^{n-1}\frac{\beta(n)x^n}{n} \\ S'(x) = \sum_{n=1}^\infty (-1)^{n-1}\beta(n)x^{n-1} = \sum_{n=1}^\infty (-1)^{n-1}x^{n-1}\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)^n} \\ = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1} \sum_{n=1}^\infty \left( \frac{-x}{2k+1} \right)^{n-1} = \sum_{k=0}^\infty \frac{1}{2k+1+x} = \frac{1}{2} \Phi \left(-1, 1, \frac{1+x}{2} \right) $$ Where $\Phi$ is the Lerch Transcendent, and $|x|<1$. Now, upon integration, we can recover $S_1$: $$ S_1 - S(0) =\int_0^1 S'(x) dx = \int_0^1 \frac{1}{2} \Phi \left(-1, 1, \frac{1+x}{2} \right)dx = \int_\frac{1}{2}^1 \Phi (-1, 1, x)dx \\ = \lim_{s\to0}\int_\frac{1}{2}^1 \Phi (-1, 1+s, x)dx = \lim_{s\to0} \left[ \frac{-1}{s}\Phi (-1, s, x) \right|_\frac{1}{2}^1 \\ = \lim_{s\to0} \frac{\Phi \left(-1, s, \frac{1}{2}\right) - \Phi(-1, s, 1)}{s} = \lim_{s\to0} \frac{2^s \beta(s) - \eta(s)}{s} \\ =^H \lim_{s\to0} \ \ln(2) 2^s \beta(s) + 2^s \beta'(s) - \eta'(s) = \frac{1}{2} \ln(2) + \ln \frac{\Gamma^2(\frac{1}{4})}{2 \pi \sqrt{2}} - \frac{1}{2} \ln \frac{\pi}{2} = \ln \frac{\sqrt{2 \pi}}{\Gamma^2(\frac{3}{4})} $$ And as $S(0) = 0$, this is all $S_1$. For $S_2$: $$ S_2 = \sum_{n=1}^\infty (-1)^{n-1}\ln\frac{n+1}{n} = \ln \prod_{n=1}^\infty \left( \frac{n+1}{n} \right)^{(-1)^{n-1}} \\ = \lim_{N \to \infty} \ln \prod_{n=1}^N \frac{n^2}{(n - \frac{1}{2})(n + \frac{1}{2})} = \lim_{N \to \infty} \ln\frac{\pi \Gamma^2(N+1)}{2\Gamma(N + \frac{1}{2})\Gamma(N + \frac{3}{2})} = \ln \frac{\pi}{2} $$ The sum therefore equals: $$ S_1-S_2=\ln \frac{2\sqrt2}{\sqrt{\pi} \ \Gamma^2(\frac{3}{4})} $$ as predicted.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\substack{\ds{\sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\bracks{{\beta\pars{n} \over n} -\ln\pars{n + 1 \over n}} = \ln\pars{\root{2 \over \pi}\,{2 \over \Gamma^{\,2}\pars{3/4}}}:\ {\large ?}.} \\[3mm] \ds{\beta:\texttt{Dirichlet Beta Function}.}}$

Lets \begin{equation} \sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\bracks{{\beta\pars{n} \over n} -\ln\pars{n + 1 \over n}} = \mc{S}_{1} - \mc{S}_{2}\,,\quad \left\{\begin{array}{l} \ds{\mc{S}_{1} \equiv \sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,{\beta\pars{n} \over n}} \\[2mm] \ds{\mc{S}_{2} \equiv \sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,\ln\pars{n + 1 \over n}} \end{array}\right.\label{1}\tag{1} \end{equation}


$\ds{\Large\mc{S}_{1}:\ ?.}$ With the $\ds{\beta}$ Integral Representation: \begin{align} \mc{S}_{1} & \equiv \sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,{1 \over n}\ \overbrace{{1 \over \Gamma\pars{n}}\int_{0}^{\infty} {x^{n - 1}\expo{-x} \over 1 + \expo{-2x}}\,\dd x}^{\ds{\beta\pars{n}}}\ =\ -\int_{0}^{\infty} {\bracks{\sum_{n = 1}^{\infty}\pars{-x}^{n}/n!}\expo{-x} \over 1 + \expo{-2x}}\,{\dd x \over x} \\[5mm] & = -\int_{0}^{\infty} {\pars{\expo{-x} - 1}\expo{-x} \over 1 + \expo{-2x}}\,{\dd x \over x} \,\,\,\stackrel{\exp\pars{-2x}\ \mapsto\ x}{=}\,\,\, -\int_{1}^{0}{x - x^{1/2} \over 1 + x}\,{\dd x/\pars{-2x} \over \ln\pars{x}/\pars{-2}} \\[5mm] & = \int_{0}^{1}{x^{-1/2} - 1 \over 1 - x^{2}}\,{x - 1 \over \ln\pars{x}}\,\dd x = \int_{0}^{1}{x^{-1/2} - 1 \over 1 - x^{2}}\ \overbrace{\int_{0}^{1}x^{t}\,\dd t}^{\ds{x - 1 \over \ln\pars{x}}}\ \,\dd x \\[5mm] & = \int_{0}^{1}\int_{0}^{1}{x^{t - 1/2} - x^{t} \over 1 - x^{2}}\,\dd x\,\dd t \,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\, {1 \over 2}\int_{0}^{1}\int_{0}^{1}{x^{t/2 - 3/4} - x^{t/2 - 1/2} \over 1 - x}\,\dd x\,\dd t \\[5mm] & = {1 \over 2}\int_{0}^{1}\bracks{% \int_{0}^{1}{1 - x^{t/2 - 1/2} \over 1 - x}\,\dd x - \int_{0}^{1}\int_{0}^{1}{1 - x^{t/2 - 3/4} \over 1 - x}\,\dd x}\dd t \\[5mm] & = {1 \over 2}\int_{0}^{1} \bracks{\Psi\pars{{t \over 2} + {1 \over 2}} - \Psi\pars{{t \over 2} + {1 \over 4}}}\dd t \qquad\pars{~\Psi:\ Digamma\ Function~} \\[5mm] & = \left. \ln\pars{\Gamma\pars{t/2 + 1/2} \over \Gamma\pars{t/2 + 1/4}} \right\vert_{\ 0}^{\ 1} = \ln\pars{{\Gamma\pars{1} \over \Gamma\pars{3/4}}\,{\Gamma\pars{1/4} \over \Gamma\pars{1/2}}} \\[5mm] & = \ln\pars{{1 \over \Gamma^{\,2}\pars{3/4}}\,{\Gamma\pars{3/4}\Gamma\pars{1/4} \over \root{\pi}}}\quad \pars{~\substack{\mbox{Note that}\\[2mm] \ds{\Gamma\pars{1 \over 2} = \root{\pi}\,,\ \Gamma\pars{1} = 1}}~} \\[5mm] & = \ln\pars{{1 \over \Gamma^{\,2}\pars{3/4}\root{\pi}} \,{\pi \over \sin\pars{\pi/4}}}\qquad\pars{~Euler\ Reflection\ Formula~} \\[5mm] & \implies \bbx{\mc{S}_{1} \equiv \sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,{\beta\pars{n} \over n} = \ln\pars{\root{2\pi} \over \Gamma^{\,2}\pars{3/4}}}\label{2}\tag{2} \end{align}
$\ds{\Large\mc{S}_{2}:\ ?.}$ \begin{align} \mc{S}_{2} & \equiv \sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,\ln\pars{n + 1 \over n} = \sum_{n = 0}^{\infty}\pars{-1}^{n}\int_{0}^{1}{\dd t \over t + n + 1} = \int_{0}^{1}\sum_{n = 0}^{\infty}{\pars{-1}^{n} \over n + t + 1}\,\dd t \\[5mm] & = {1 \over 2}\int_{0}^{1}\sum_{n = 0}^{\infty} \bracks{{1 \over n + \pars{t + 1}/2} - {1 \over n + t/2 + 1}}\,\dd t \\[5mm] & = {1 \over 2}\int_{0}^{1}\bracks{% \Psi\pars{{t \over 2} + 1} - \Psi\pars{t + 1 \over 2}}\,\dd t = \left.\ln\pars{\Gamma\pars{t/2 + 1} \over \Gamma\pars{t/2 + 1/2}} \right\vert_{\ 0}^{\ 1} \\[5mm] & = \ln\pars{{\Gamma\pars{3/2} \over \Gamma\pars{1}}\,{\Gamma\pars{1/2} \over \Gamma\pars{1}}} = \ln\pars{{1 \over 2}\,\Gamma^{\,2}\pars{1 \over 2}} \qquad\pars{~\Gamma-Recursive Property~} \\[5mm] & \implies \bbx{\mc{S}_{2} \equiv \sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\,\ln\pars{n + 1 \over n} = \ln\pars{\pi \over 2}}\label{3}\tag{3} \end{align}
With \eqref{1}, \eqref{2} and \eqref{3}: $$ \bbx{\sum_{n = 1}^{\infty}\pars{-1}^{n - 1}\bracks{{\beta\pars{n} \over n} -\ln\pars{n + 1 \over n}} = \ln\pars{\root{2 \over \pi}\,{2 \over \Gamma^{\,2}\pars{3/4}}}} $$

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Here is a fairly elementary proof of this result. I derive a closed form equivalent to the OP's.

We begin with the following Lemma:


Lemma $1$: $\sum_{k=1}^p \ln(r*k+n) = \ln\left[r^n\left(\frac{n+r}{r}\right)_p\right] \quad \forall p\in\mathbb{N}$

Proof of Lemma $1$: $\sum _{k=1}^p\ln \left(rk+n\right) = \ln \left(\prod _{k=1}^p\left(rk+n\right)\right) = \ln \left(r^p\prod _{k=0}^{p-1}\left(k+\frac{n}{r}\right)\right) = \log\left[r^n\left(\frac{n+r}{r}\right)_p\right]$


We are now ready to prove the result $$\begin{align} \sum_{n=1}^{p}(-1)^{n-1}&\left({\beta(n)\over n}-\ln{n+1\over n}\right)\\ &= \sum_{n=1}^{p}\left(-1\right)^{n-1}\left(\frac{1}{n}\sum _{k=0}^{\infty}\frac{\left(-1\right)^k}{\left(2k+1\right)^n}-\ln \left(\frac{n+1}{n}\right)\right)\\ &=\sum _{n=1}^p\left(-1\right)^n\ln \left(\frac{n+1}{n}\right)-\sum _{n=1}^{p}(-1)^n\sum _{k=0}^{\infty}\frac{(-1)^k}{n\left(2k+1\right)^n}\\ &\stackrel{*}{=} \sum _{n=1}^p\left(-1\right)^n\ln \left(\frac{n+1}{n}\right)-\sum _{k=0}^p\left(-1\right)^k\sum _{n=1}^\infty\frac{\left(-1\right)^n}{n\left(2k+1\right)^n}\\ &\stackrel{**}{=}\sum _{n=1}^p\left(-1\right)^n\ln \left(\frac{n+1}{n}\right)-\sum _{k=0}^p\left(-1\right)^n\ln \left(\frac{2k+1}{2k+2}\right)\\ &=\sum _{n=1}^p\left(-1\right)^n\ln \left(\frac{n+1}{n}\right)-\sum _{k=1}^p\left(-1\right)^k\ln \left(\frac{2k+1}{2k+2}\right)+\ln(2)\\ &=\sum _{k=1}^p\left(-1\right)^k\left[\ln \left(\frac{k+1}{k}\right)-\ln \left(\frac{2k+1}{2k+2}\right)\right]+\ln(2)\\ &=\sum _{k=1}^p\left[\ln \left(\frac{2k+1}{2k}\right)-\ln \left(\frac{4k+1}{4k+2}\right)-\ln \left(\frac{2k}{2k-1}\right)+\ln \left(\frac{4k-1}{4k}\right)\right]+\ln(2)\\ &= \sum _{k=1}^p\left(\ln \left(2k+1\right)-2\ln \left(2k\right)-\ln \left(4k+1\right)+\ln \left(4k+2\right)+\ln \left(2k-1\right)+\ln \left(4k-1\right)-\ln \left(4k\right)\right)+\ln(2)\\ &=\ln \left(2\left(3/2\right)_p\right)-2\ln \left(\left(1\right)_p\right)-\ln \left(4\left(5/4\right)_p\right)+\ln \left(16\left(3/2\right)_p\right)+\ln \left(\frac{1}{2}\left(1/2\right)_p\right)+\ln \left(\frac{1}{4}\left(3/4\right)_p\right)-\ln \left(\left(1\right)_p\right)+\ln(2)\\ &= \ln \left(\frac{2\cdot \left(1/2\right)_p\left(3/4\right)_p\left(3/2\right)_p\left(3/2\right)_p}{\left(1\right)_p\left(1\right)_p\left(1\right)_p\left(5/4\right)_p}\right)\\ &\stackrel{***}{=} \ln \left(\frac{2\cdot \Gamma\left(1/4\right)^2}{\pi ^2\sqrt{2\pi }}\right)+\ln \left(\frac{\Gamma\left(p+1/2\right)\Gamma\left(p+3/4\right)\Gamma\left(p+3/2\right)^2}{\Gamma\left(p+1\right)^3 \Gamma\left(p+5/4\right)}\right) \end{align}$$ Luckily, as can be checked using Stirling's Formula, the second logarithm goes to $0$ as $p$ goes to infinity, i.e. $$\sum_{n=1}^{\infty}(-1)^{n-1}\left({\beta(n)\over n}-\ln{n+1\over n}\right) = \ln \left(\frac{2\cdot \Gamma\left(1/4\right)^2}{\pi ^2\sqrt{2\pi }}\right) + \ln \left(\lim_{p \to \infty}\frac{\Gamma\left(p+1/2\right)\Gamma\left(p+3/4\right)\Gamma\left(p+3/2\right)^2}{\Gamma\left(p+1\right)^3 \Gamma\left(p+5/4\right)}\right) = \color{red}{\ln \left(\frac{2\cdot \Gamma\left(1/4\right)^2}{\pi ^2\sqrt{2\pi }}\right)}$$


Footnotes:

$*\;\;\;\;\;\;$ Interchange double summations
$**\;\;\;\,$ Taylor Series for Natural Logarithm
$***\;$ Gamma representation for Pochhammer Symbol

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