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I saw the following infinite product on Wikipedia: $$\Gamma\left(\tfrac14\right)=\mathrm{A}^3e^{-\mathrm{G}/\pi}2^{1/6}\sqrt{\pi}\prod_{k\ge1}\left(1-\frac1{2k}\right)^{(-1)^k k}\tag{1}$$ where $\mathrm A$ is the Glaisher constant, and $\mathrm G$ is Catalan's constant. I am looking for a proof of this product.

I haven't gotten very far with this product, other than noting that if $$\zeta_*(s)=\sum_{k\ge1}\frac1{(1-\frac1{2k})^{(-1)^k ks}}$$ then $$\zeta_*'(0)=3\ln\mathrm A+\frac16\ln2+\frac12\ln\pi-\frac{\mathrm{G}}{\pi}-\ln\Gamma\left(\tfrac14\right),\tag2$$ of course, assuming that $(1)$ is true. Perhaps $(2)$ is easier to prove. Could I have some help? Thanks.

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  • $\begingroup$ Related problem: here --- you can use the definition of $A$ for your proof (it's only a hint) $\endgroup$
    – user90369
    Commented Sep 25, 2019 at 17:42
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    $\begingroup$ You can find a solution with the Borwein-Dykshoorn function, e.g. here (whole script) and here, page 3 . $~~$ N O T E: $~\prod_{k\ge1}\left(1-\frac1{2k}\right)^{(-1)^k k}$ (<- not defined !) has two solutions, $\lim\limits_{n\to\infty}\prod_{k=1}^{2n}\left(1-\frac1{2k}\right)^{(-1)^k k}$ and $\lim\limits_{n\to\infty}\prod_{k=1}^{2n+1}\left(1-\frac1{2k}\right)^{(-1)^k k}$. $\endgroup$
    – user90369
    Commented Sep 26, 2019 at 9:15

2 Answers 2

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A short proof.

$\displaystyle Q_0(x) :=\Gamma(x+1)=\lim_{n\to\infty}\frac{n^x}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)}~~ , ~~~~ Q_1(x) :=\lim_{n\to\infty}\frac{e^{xn}n^{-x^2/2}}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)^k}$

$\displaystyle \pi^{1/2} = Q_0\left(-\frac{1}{2}\right)~~ , ~~ A^3 = 2^{7/12}Q_1\left(-\frac{1}{2}\right)^2 ~~ , ~~ e^{G/\pi} = 2^{3/4}\left(\frac{ Q_0\left(-\frac{1}{2}\right) Q_1\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{3}{4}\right) Q_1\left(-\frac{1}{4}\right) }\right)^2$

It follows:

$\displaystyle \prod\limits_{k=1}^{2n}\left(1-\frac{1}{2k}\right)^{(-1)^k k} = \prod\limits_{k=1}^n \frac{\left(1-\frac{1}{4k}\right)^{2k}}{\left(1-\frac{1}{4k-2}\right)^{2k-1}} = \prod\limits_{k=1}^n \frac{\left(1-\frac{1}{2k}\right)^{2k-1} \left(1-\frac{1}{4k}\right)^{2k}}{ \left(1-\frac{3}{4k}\right)^{2k-1} } = $

$\displaystyle = \frac{ \prod\limits_{k=1}^n \left(1-\frac{3}{4k}\right) n^{-1/2} }{ n^{-3/4} \prod\limits_{k=1}^n \left(1-\frac{1}{2k}\right) } \left(\frac{ e^{-3n/4} n^{-9/32} \prod\limits_{k=1}^n \left(1-\frac{1}{2k}\right)^k \prod\limits_{k=1}^n \left(1-\frac{1}{4k}\right)^k }{\prod\limits_{k=1}^n \left(1-\frac{3}{4k}\right)^k e^{-n/2} n^{-1/8} e^{-n/4} n^{-1/32} }\right)^2$

$\displaystyle \to ~\frac{ Q_0\left(-\frac{1}{2}\right) }{ Q_0\left(-\frac{3}{4}\right) } \left(\frac{ Q_1\left(-\frac{3}{4}\right) }{ Q_1\left(-\frac{1}{2}\right) Q_1\left(-\frac{1}{4}\right) }\right)^2$

$\displaystyle = 2^{-1/6} \frac{ Q_0\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{1}{2}\right) } \cdot 2^{-7/12} Q_1\left(-\frac{1}{2}\right)^{-2} \cdot 2^{3/4} \left(\frac{ Q_0\left(-\frac{1}{2}\right) Q_1\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{3}{4}\right) Q_1\left(-\frac{1}{4}\right)}\right)^2$

$\displaystyle = \Gamma\left(\frac{1}{4}\right) 2^{-1/6} \pi^{-1/2} A^{-3} e^{G/\pi}$

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  • $\begingroup$ Fantastic! Thank you (+1) $\endgroup$
    – clathratus
    Commented Sep 27, 2019 at 19:44
  • $\begingroup$ @clathratus: You are welcome. ;) $\endgroup$
    – user90369
    Commented Sep 27, 2019 at 19:56
  • $\begingroup$ do you have any references for the $e^{\mathrm G/\pi}$ identity? $\endgroup$
    – clathratus
    Commented Sep 27, 2019 at 20:31
  • $\begingroup$ E.g. here, page 195, first formula. But this doesn't look understandable if one doesn't deal in detail with the derivation of the zeta function. Use Catalan, section "Relation to other special functions" (Barnes), with more details Clausen, section "Relation to the Barnes' G-function" . $\endgroup$
    – user90369
    Commented Sep 28, 2019 at 9:02
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    $\begingroup$ @reuns If you hover your cursor over the downvote button, it says "This answer is not useful." In your answer, you yourself say "I don't think giving more details is useful." Twice, I asked for more details, and twice, you refused. Frankly, I find that completely useless. $\endgroup$
    – clathratus
    Commented Sep 29, 2019 at 20:10
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The log of the product can be expressed in term of $\frac12\beta'(-1)-\frac12 \beta'(0)$ and $\frac12\eta'(-1)$

where $\beta(s) = \sum_{k\ge 0} (2k+1)^{-s} (-1)^k$ , it is a L-function with a functional equation.

$G =\beta(2)$, $\log A = 1/12-\zeta'(-1)$

Finally $e^{\beta'(0)}$ can be expressed in term of $\Gamma(1/4)$ and $\Gamma(1/2)$ infinite products.

I don't think giving more details is useful.

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  • $\begingroup$ I think more details would be useful. For instance, what are the relations between $\zeta_*(s)$ and $\beta(s)$ and $\eta(s)$? $\endgroup$
    – clathratus
    Commented Sep 25, 2019 at 22:47
  • $\begingroup$ Your $\zeta_*(s)$ doesn't make any sense. No more details won't be useful, I made clear the needed theorems $\endgroup$
    – reuns
    Commented Sep 25, 2019 at 22:49
  • $\begingroup$ Could you at least show the relation between the log of the product and the various constants you mentioned? What you have as an 'answer' barely gives me any information. $\endgroup$
    – clathratus
    Commented Sep 25, 2019 at 22:55
  • $\begingroup$ Formally the relation is obvious $\endgroup$
    – reuns
    Commented Sep 25, 2019 at 22:56

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