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$\mathrm G$ is Catalan's constant.

I recently found the product $$ \alpha=\prod_{n=1}^{\infty}\frac{E_n(\frac12)E_n(\frac7{12})E_n(\frac1{20})E_n(\frac{13}{20})}{E_n(\frac14)E_n(\frac1{12})E_n(\frac3{20})E_n(\frac{11}{20})}=\\ \exp\left[\frac{47\mathrm G}{30\pi}+\frac34\right]\sqrt{\frac{33}{91\pi}\sqrt{\frac2\pi\frac{\sqrt[5]{11}}{\sqrt[3]{7}}\sqrt[5]{\frac{3^3}{13^{3}}}}}$$

Where $$E_n(x)=\frac{j(n+x)}{(en)^{2x}j(n-x)}\qquad x\in(0,1)$$ and $j(x)=x^x$.

Could I have some numerical evidence, or better yet an alternate proof? My tools are limited to desmos, which cannot really handle infinite products. Thanks.


My Proof:

We define $$\mathrm L(x)=\frac1\pi\int_0^{\pi x}\log(\sin t)dt$$ And we use $$\sin t=t\prod_{n\geq1}\left(1-\frac{t^2}{\pi^2 n^2}\right)$$ To see that $$\log(\sin t)=\log(t)+\sum_{n\geq1}\log\frac{\pi^2n^2-t^2}{\pi^2n^2}$$ Then integrate both sides over $[0,x]$ to get $$\pi\mathrm L(x/\pi)=x(\log x-1)+\sum_{n\geq1}x\log\bigg(1-\frac{x^2}{\pi^2n^2}\bigg)-2x+\pi n\log\frac{\pi n+x}{\pi n-x}$$ $$\pi\mathrm L(x/\pi)=\log\left[\frac{j(x)}{e^x}\right]+\sum_{n\geq1}\log\left[\frac{j(\pi n+x)}{(e\pi n)^{2x}j(\pi n-x)}\right]$$ $x\mapsto \pi x$: $$\pi\mathrm L(x)=\log\left[\frac{j(\pi x)}{e^{\pi x}}\right]+\sum_{n\geq1}\log\left[\frac{j(\pi n+\pi x)}{(e\pi n)^{2\pi x}j(\pi n-\pi x)}\right]$$ $$\mathrm L(x)=\log\left[\left(\frac\pi{e}\right)^xj(x)\right]+\sum_{n\geq1}\log E_n(x)$$ Then we define $$U(x)=\prod_{n\geq1}E_n(x)$$ To see that $$U(x)=\left(\frac{e}{\pi x}\right)^x\exp\mathrm L(x)$$ Where we used $$\sum_{n}\log(a_n)=\log\left[\prod_{n}a_n\right]$$ and the neat rules $$\log(a^b)=\log(e^{b\log a})=b\log a$$ $$\log(a)\pm b=\log\left(e^{\pm b}a\right)$$ to simplify the expressions. Next, we define $$P_{\mu,\nu}(a_1,a_2,\dots,a_\mu;b_1,b_2,\dots,b_\nu)=\frac{\prod_{i=1}^\mu U(a_i)}{\prod_{i=1}^\nu U(b_i)}$$ And we see that $$P_{\mu,\nu}(a_1,\dots,a_\mu;b_1,\dots,b_\nu)=\prod_{n\geq1}\frac{\prod_{i=1}^\mu E_n(a_i)}{\prod_{i=1}^\nu E_n(b_i)}$$ This gives $$P_{1,1}(x_1;x_2)=\left(\frac{e}{\pi}\right)^{x_1-x_2}\frac{j(x_2)}{j(x_1)}\exp\left[\mathrm L(x_1)-\mathrm L(x_2)\right]$$ Then we define $$\mathrm{T}(x)=\frac{1}{\pi}\int_0^{\pi x}\log(\tan t)dt=\mathrm L(x)-\mathrm L(x+1/2)-\frac12\log2$$ To get that $$P_{1,1}\left(x;x+\frac12\right)=\sqrt{\frac{2\pi}e}\,\frac{j(x+1/2)}{j(x)}\exp\mathrm T(x)$$ So we have $$P_{2,2}\left(x_1,x_2+\frac12 ;x_2,x_1+\frac12\right)=\frac{j(x_1+1/2)j(x_2)}{j(x_2+1/2)j(x_1)}\exp\left[\mathrm T(x_1)-\mathrm T(x_2)\right]$$ Then using the identities $$\mathrm L(1/2)=-\frac12\log2$$ $$\mathrm L(1/4)=\frac{\mathrm G}{2\pi}-\frac14\log2$$ We get $$P_{1,1}\left(\frac12;\frac14\right)=\frac1{(2\pi)^{1/4}}\exp\left[\frac{\mathrm G}{2\pi}+\frac14\right]\tag{1}$$ From here, the identity $$-\mathrm T(1/12)=\frac{2\mathrm G}{3\pi}$$ which gives $$P_{1,1}\left(\frac7{12};\frac1{12}\right)=\sqrt{\frac6{7\pi\sqrt[6]{7}}}\exp\left[\frac{2\mathrm G}{3\pi}+\frac12\right]\tag{2}$$ Then from here, the identity $$\mathrm T(1/20)-\mathrm T(3/20)=\frac{2\mathrm G}{5\pi}$$ gives $$P_{2,2}\left(\frac1{20},\frac{13}{20};\frac3{20},\frac{11}{20}\right)=\left(\frac{j(11)j(3)}{j(13)}\right)^{1/20}\exp\frac{2\mathrm G}{5\pi}\tag{3}$$ Then multiplying $(1),(2),$ and $(3)$, we have the desired result, namely $$P_{4,4}\left(\frac12,\frac7{12},\frac1{20},\frac{13}{20};\frac14,\frac1{12},\frac3{20},\frac{11}{20}\right)=\alpha$$

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the OP asks for some numerical evidence: plotted below is the constant $\alpha$ minus the $\prod_{n=1}^N$ of the expression in OP, as a function of $N$; so at least within 1 part in 1000 the infinite product does seem to converge from above to the stated constant.

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    $\begingroup$ This is perfect, thank you. What software did you use to plot this? $\endgroup$ – clathratus Mar 23 at 18:44
  • $\begingroup$ The downvote is mine. $\endgroup$ – user64494 Mar 23 at 18:57
  • $\begingroup$ oh, this is just Mathematica output. $\endgroup$ – Carlo Beenakker Mar 23 at 20:24
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Using the Maple code from user64494's answer (but using mul(evalf(...)) instead of evalf(product(...)) for greater efficiency, and 20 digits), I computed the product for 1000,2000,3000,4000 terms. It took about four minutes.

The answers were: 0.78056641010997748236, 0.78056641010997748236, 0.78056641010997748236, 0.78056641010997748236.

Now assuming that the product to $N$ terms has an asymptotic expansion $$a_0+a_1/N+a_2/N^2+a_3/N^3+\cdots,$$ which can surely be proved but I didn't, extrapolation of these 4 values gives $$a_0\approx 0.7804591974129376479,a_1\approx 0.107, a_2\approx -0.0463, a_3\approx 0.0151.$$ The value of $a_0$ agrees with the proposed infinite product 0.7804591974129374862 to 15 digits.

ADDED: By evaluating $x^x$ as evalf(x)^x, the same computation runs in about 20 seconds even with twice the precision. Adding the product with 5000 terms, agreement to 20 digits is obtained.

ADDED[2]: I don't know how it does it, but Maple's built-in code for numerical evaluation of infinite products takes mere seconds to confirm this to 200 digits.

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  • $\begingroup$ Also really great! Thanks for all the hard effort! 20 digits of precision is good enough for me. $\endgroup$ – clathratus Mar 24 at 2:58
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The following Mathematica code

NProduct[(1/2 + n)^(1/2 + n)/(Exp[1]*n)^(2*1/2)/(n - 1/2)^(n - 
  1/2)*(7/12 + n)^(7/12 + n)/(Exp[1]*n)^(2*7/12)/(n - 7/12)^(n - 
  7/12)*(1/20 + n)^(1/20 + n)/(Exp[1]*n)^(2*1/20)/(n - 1/20)^(n - 
  1/20)*(13/20 + n)^(13/20 + n)/(Exp[1]*n)^(2*13/20)/(n -13/20)^(n -13/20)/
((1/4 + n)^(1/4 + n)/(Exp[1]*n)^(2*1/4)/(n - 
       1/4)^(n - 1/4))/((1/12 + n)^(1/12 + n)/(Exp[1]*
       n)^(2*1/12)/(n - 1/12)^(n - 1/12))/((11/20 + n)^(11/20 + 
      n)/(Exp[1]*n)^(2*11/20)/(n - 11/20)^(n - 11/20))/((3/20 + 
     n)^(3/20 + n)/(Exp[1]*n)^(2*3/20)/(n - 3/20)^(n - 3/20)),
{n,1,Infinity}, AccuracyGoal -> 3, WorkingPrecision -> 15]

performs

$0.78046 $

If somebody verifies the above code, it would be kind of her/him.

Addition. The Maple command for the product up to $100$

Digits:=15:evalf(product((1/2+n)^(1/2+n)*(7/12+n)^(7/12+n)*(1/20+n)^(1/20+n)*(13/20+n)^(13/20+n)*(n-1/4)^(n-1/4)*(n-1/12)^(n-1/12)*(n-11/20)^(n-11/20)*(n-3/20)^(n-3/20)/(exp(1)*n*(n-1/2)^(n-1/2)*sqrt(exp(1)*n)*(n-7/12)^(n-7/12)*(n-1/20)^(n-1/20)*(n-13/20)^(n-13/20)*(1/4+n)^(1/4+n)*(1/12+n)^(1/12+n)*(11/20+n)^(11/20+n)*(3/20+n)^(3/20+n)), n = 1 .. 100));

produces $0.781527175985084 $.

Also

N[Exp[47*Catalan/30/Pi + 3/4]*  Sqrt[33/91/Pi*Sqrt[2/Pi*11^(1/5)/7^(1/3)*3^(3/5)/13^(3/5)]], 15]

$0.780459197412937 $

Edit. A typo in the codes ($(n-1/2)^{n-1/2}$ instead of $(n-1/2)^{n-1}$) is corrected. That typo leads to incorrect results.

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Following off of @user64494, the Mathematica code

Product[(1/2 + n)^(1/2 + n)/(Exp[1]*n)^(2*1/2)/(n - 1/2)^(n - 1/2)*(7/12 + n)^(7/12 + n)/(Exp[1]*n)^(2*7/12)/(n - 7/12)^(n - 7/12)*(1/20 + n)^(1/20 + n)/(Exp[1]*n)^(2*1/20)/(n - 1/20)^(n - 1/20)*(13/20 + n)^(13/20 + n)/(Exp[1]*n)^(2*13/20)/(n - 13/20)^(n - 13/20)/((1/4 + n)^(1/4 + n)/(Exp[1]*n)^(2*1/4)/(n - 1/4)^(n - 1/4))/((1/12 + n)^(1/12 + n)/(Exp[1]* n)^(2*1/12)/(n - 1/12)^(n - 1/12))/((11/20 + n)^(11/20 + n)/(Exp[1]*n)^(2*11/20)/(n - 11/20)^(n - 11/20))/((3/20 + n)^(3/20 + n)/(Exp[1]*n)^(2*3/20)/(n - 3/20)^(n - 3/20)), {n, 1, Infinity}]

gives the closed form output

$$\frac{{2}^{1/4} 3^{13/20} 11^{11/20} 5^{\frac{1}{5} \zeta \left(-1,-\frac{1}{4}\right)-\frac{1}{480}} \exp \left(\frac{1}{5} \zeta ^{(1,0)}\left(-1,-\frac{1}{4}\right)+\zeta ^{(1,0)}\left(-1,\frac{19}{20}\right)-\zeta ^{(1,0)}\left(-1,-\frac{1}{20}\right)+\frac{97 C}{60 \pi }+\frac{361}{480}\right)}{7^{7/12} 13^{13/20} \pi ^{3/4} {Glaisher}^{1/40}}$$

where Glaisher's constant is approximately 1.28243 and $\zeta$ is the (generalized) Riemann zeta function. Evaluating the above mess to 20 decimal places gives

$$0.780~459~197~412~937~486~21.$$

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  • $\begingroup$ Woah! I had no idea that Glaisher's constant was in here! Thanks. $\endgroup$ – clathratus Mar 24 at 15:44
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Each $P(x)=\sum_{n=1}^{\infty} \log E_n(x)$ is convergent and can be calculated using Mathematica, $$ P(x) = x [1-\log(2\pi x)] +\zeta^{(1,0)}(-1,1-x) -\zeta^{(1,0)}(-1,x), $$ as $\sum_n(n+a)\log(n+a) "=" -\zeta^{(1,0)}(-1,a)$. Further, the following identity holds (see, e.g., Zeta2, 10.02.20.0043.01): $$ \zeta^{(1,0)}(-1,x) - \zeta^{(1,0)}(-1,1-x) = \frac{1}{2\pi\mathrm{i}} \mathrm{Li}_2(\mathrm{e}^{2 \pi \mathrm{i} x})+ \frac{\pi \mathrm{i}}{2}\left(\frac{1}{6}-x-x^2\right), $$ with polylog $\mathrm{Li}$, such that $$ P(x) = x [1-\log(2\pi x)] -\frac{1}{2\pi\mathrm{i}} \mathrm{Li}_2(\mathrm{e}^{2 \pi \mathrm{i} x}) - \frac{\pi \mathrm{i}}{2}\left(\frac{1}{6}-x-x^2\right). $$ The non-trivial polylogs cancel in the final sum of the $P$'s, with the result $$ \log \alpha = \frac{47 C}{30 \pi }+\frac{3}{4}+\frac{\log 2}{4}-\frac{13}{20} \log \frac{13}{3}-\frac{7}{12} \log 7 + \frac{11}{20}\log 11-\frac{3 }{4}\log \pi. $$

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  • $\begingroup$ Woah! An alternate proof! Fantastic! Thanks:) $\endgroup$ – clathratus Mar 27 at 15:06

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