Crthuky

$mathrm G$ is Catalan's constant.

I recently found the product
$$
alpha=prod_n=1^inftyfracE_n(frac12)E_n(frac712)E_n(frac120)E_n(frac1320)E_n(frac14)E_n(frac112)E_n(frac320)E_n(frac1120)=\
expleft[frac47mathrm G30pi+frac34right]sqrtfrac3391pisqrtfrac2pifracsqrt[5]11sqrt[3]7sqrt[5]frac3^313^3$$

Where $$E_n(x)=fracj(n+x)(en)^2xj(n-x)qquad xin(0,1)$$
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 $$mathrm L(x)=frac1piint_0^pi xlog(sin t)dt$$
And we use $$sin t=tprod_ngeq1left(1-fract^2pi^2 n^2right)$$
To see that $$log(sin t)=log(t)+sum_ngeq1logfracpi^2n^2-t^2pi^2n^2$$
Then integrate both sides over $[0,x]$ to get
$$pimathrm L(x/pi)=x(log x-1)+sum_ngeq1xlogbigg(1-fracx^2pi^2n^2bigg)-2x+pi nlogfracpi n+xpi n-x$$
$$pimathrm L(x/pi)=logleft[fracj(x)e^xright]+sum_ngeq1logleft[fracj(pi n+x)(epi n)^2xj(pi n-x)right]$$
$xmapsto pi x$:
$$pimathrm L(x)=logleft[fracj(pi x)e^pi xright]+sum_ngeq1logleft[fracj(pi n+pi x)(epi n)^2pi xj(pi n-pi x)right]$$
$$mathrm L(x)=logleft[left(fracpieright)^xj(x)right]+sum_ngeq1log E_n(x)$$
Then we define $$U(x)=prod_ngeq1E_n(x)$$
To see that $$U(x)=left(fracepi xright)^xexpmathrm L(x)$$
Where we used $$sum_nlog(a_n)=logleft[prod_na_nright]$$
and the neat rules $$log(a^b)=log(e^blog a)=blog a$$
$$log(a)pm b=logleft(e^pm baright)$$
to simplify the expressions. Next, we define
$$P_mu,nu(a_1,a_2,dots,a_mu;b_1,b_2,dots,b_nu)=fracprod_i=1^mu U(a_i)prod_i=1^nu U(b_i)$$
And we see that
$$P_mu,nu(a_1,dots,a_mu;b_1,dots,b_nu)=prod_ngeq1fracprod_i=1^mu E_n(a_i)prod_i=1^nu E_n(b_i)$$
This gives $$P_1,1(x_1;x_2)=left(fracepiright)^x_1-x_2fracj(x_2)j(x_1)expleft[mathrm L(x_1)-mathrm L(x_2)right]$$
Then we define $$mathrmT(x)=frac1piint_0^pi xlog(tan t)dt=mathrm L(x)-mathrm L(x+1/2)-frac12log2$$
To get that
$$P_1,1left(x;x+frac12right)=sqrtfrac2pie,fracj(x+1/2)j(x)expmathrm T(x)$$
So we have
$$P_2,2left(x_1,x_2+frac12 ;x_2,x_1+frac12right)=fracj(x_1+1/2)j(x_2)j(x_2+1/2)j(x_1)expleft[mathrm T(x_1)-mathrm T(x_2)right]$$
Then using the identities
$$mathrm L(1/2)=-frac12log2$$
$$mathrm L(1/4)=fracmathrm G2pi-frac14log2$$
We get $$P_1,1left(frac12;frac14right)=frac1(2pi)^1/4expleft[fracmathrm G2pi+frac14right]tag1$$
From here, the identity
$$-mathrm T(1/12)=frac2mathrm G3pi$$
which gives
$$P_1,1left(frac712;frac112right)=sqrtfrac67pisqrt[6]7expleft[frac2mathrm G3pi+frac12right]tag2$$
Then from here, the identity
$$mathrm T(1/20)-mathrm T(3/20)=frac2mathrm G5pi$$
gives $$P_2,2left(frac120,frac1320;frac320,frac1120right)=left(fracj(11)j(3)j(13)right)^1/20expfrac2mathrm G5pitag3$$
Then multiplying $(1),(2),$ and $(3)$, we have the desired result, namely
$$P_4,4left(frac12,frac712,frac120,frac1320;frac14,frac112,frac320,frac1120right)=alpha$$

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.

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.