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valid for all $s\ge 1$ $$\beta(s)=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^s}$$

The particular value of $\Gamma\left(\frac{1}{4}\right)=3.6256099...$

Euler's constant is defined by $$\lim_{n \to \infty}\left[H_n-\ln(n)\right]=\gamma$$

Euler also showed that $$\gamma=\sum_{n=1}^{\infty}\left(\frac{1}{n}-\ln\left(\frac{n+1}{n}\right)\right)$$

Where $H_n=\sum_{k=1}^{n}\frac{1}{k}$

Show that,

$$\sum_{n=1}^{\infty}\left[\frac{\beta(2n)}{n}-\ln\left(\frac{n+1}{n}\right)\right] =\gamma+\ln\left(\frac{16\pi^2}{\Gamma^4\left(\frac{1}{4}\right)}\right)$$

We are greatly appreciated if anyone can answer this identity.

  • What have you attempted? Did you try to exploit the integral representation for the $\beta$ function and the fact that $$\log\frac{n+1}{n}=\int_{0}^{+\infty}\frac{e^{-nx}-e^{-(n+1)x}}{x},dx $$ ? – Jack D'Aurizio May 03 '16 at 15:28
  • No I haven't, but I still couldn't figure it out –  May 03 '16 at 15:32
  • Then what have you tried? – Jack D'Aurizio May 03 '16 at 15:32
  • I haven't try any yet, I just took this formula out from my formula book list. I don't know how I got this formula since a few years back I can't remember, that is why I post it for someone help me to answer it so I can learn from it. –  May 03 '16 at 15:39
  • I just hope my answer won't be downvoted like the answer I gave to your other question. It is the same approach, indeed. – Jack D'Aurizio May 03 '16 at 15:48

1 Answers1

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With the same approach to a recent question of yours,

$$ \beta(2n)-1=\frac{1}{\Gamma(2n)}\int_{0}^{+\infty}\frac{x^{2n-1} e^{-3x}}{1+e^{-2x}}\,dx $$ hence: $$ \sum_{n\geq 1}\frac{\beta(2n)-1}{n} = 2\int_{0}^{+\infty}\frac{e^{-3x}}{1+e^{-2x}}\cdot\frac{\cosh x-1}{x}\,dx.$$ By Frullani's theorem, or differentiation under the integral sign (Feynman's trick): $$ 2\int_{0}^{+\infty}\frac{\cosh x-1}{x}e^{-(3+2m)x}\,dx =-\log(m+1)+2\log\left(m+\frac{3}{2}\right)-\log(m+2),$$ hence: $$\sum_{n\geq 1}\frac{\beta(2n)-1}{n} = \sum_{m\geq 0}(-1)^m\left(-\log(m+1)+2\log\left(m+\frac{3}{2}\right)-\log(m+2)\right)$$ is the logarithm of an infinite product that can be computed from the limit product representation for the $\Gamma$ function. Namely: $$ \sum_{m\geq 0}\left[\log\left(2m+\frac{3}{2}\right)-\frac{1}{2}\log(2m+1)-\log\left(2m+\frac{5}{2}\right)+\frac{1}{2}\log(2m+3)\right]\\=\frac{\log 2}{2}-\log\Gamma\left(\frac{3}{4}\right)+\log\Gamma\left(\frac{5}{4}\right).$$ Now it is enough to use the $\Gamma$ reflection formula and the identity $\Gamma(z+1)=z\,\Gamma(z)$ to express your sum in terms of $\gamma$ and $\Gamma\left(\frac{1}{4}\right)$. The original identity is, in fact, equivalent to:

$$ \prod_{m\geq 0}\frac{(4m+3)(4m+3)(4m+6)}{(4m+2)(4m+5)(4m+5)}=\color{red}{\frac{1}{16\pi^2}\,\Gamma\left(\frac{1}{4}\right)^4}.$$

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