Ggtkuk

$mathrm G$ is Catalan's constant.

I recently found the product
$$
alpha=prod_{n=1}^{infty}frac{E_n(frac12)E_n(frac7{12})E_n(frac1{20})E_n(frac{13}{20})}{E_n(frac14)E_n(frac1{12})E_n(frac3{20})E_n(frac{11}{20})}=\
expleft[frac{47mathrm G}{30pi}+frac34right]sqrt{frac{33}{91pi}sqrt{frac2pifrac{sqrt[5]{11}}{sqrt[3]{7}}sqrt[5]{frac{3^3}{13^{3}}}}}$$

Where $$E_n(x)=frac{j(n+x)}{(en)^{2x}j(n-x)}qquad xin(0,1)$$
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 $$mathrm L(x)=frac1piint_0^{pi x}log(sin t)dt$$
And we use $$sin t=tprod_{ngeq1}left(1-frac{t^2}{pi^2 n^2}right)$$
To see that $$log(sin t)=log(t)+sum_{ngeq1}logfrac{pi^2n^2-t^2}{pi^2n^2}$$
Then integrate both sides over $[0,x]$ to get
$$pimathrm L(x/pi)=x(log x-1)+sum_{ngeq1}xlogbigg(1-frac{x^2}{pi^2n^2}bigg)-2x+pi nlogfrac{pi n+x}{pi n-x}$$
$$pimathrm L(x/pi)=logleft[frac{j(x)}{e^x}right]+sum_{ngeq1}logleft[frac{j(pi n+x)}{(epi n)^{2x}j(pi n-x)}right]$$
$xmapsto pi x$:
$$pimathrm L(x)=logleft[frac{j(pi x)}{e^{pi x}}right]+sum_{ngeq1}logleft[frac{j(pi n+pi x)}{(epi n)^{2pi x}j(pi n-pi x)}right]$$
$$mathrm L(x)=logleft[left(fracpi{e}right)^xj(x)right]+sum_{ngeq1}log E_n(x)$$
Then we define $$U(x)=prod_{ngeq1}E_n(x)$$
To see that $$U(x)=left(frac{e}{pi x}right)^xexpmathrm L(x)$$
Where we used $$sum_{n}log(a_n)=logleft[prod_{n}a_nright]$$
and the neat rules $$log(a^b)=log(e^{blog a})=blog a$$
$$log(a)pm b=logleft(e^{pm b}aright)$$
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
$$P_{mu,nu}(a_1,dots,a_mu;b_1,dots,b_nu)=prod_{ngeq1}frac{prod_{i=1}^mu E_n(a_i)}{prod_{i=1}^nu E_n(b_i)}$$
This gives $$P_{1,1}(x_1;x_2)=left(frac{e}{pi}right)^{x_1-x_2}frac{j(x_2)}{j(x_1)}expleft[mathrm L(x_1)-mathrm L(x_2)right]$$
Then we define $$mathrm{T}(x)=frac{1}{pi}int_0^{pi x}log(tan t)dt=mathrm L(x)-mathrm L(x+1/2)-frac12log2$$
To get that
$$P_{1,1}left(x;x+frac12right)=sqrt{frac{2pi}e},frac{j(x+1/2)}{j(x)}expmathrm T(x)$$
So we have
$$P_{2,2}left(x_1,x_2+frac12 ;x_2,x_1+frac12right)=frac{j(x_1+1/2)j(x_2)}{j(x_2+1/2)j(x_1)}expleft[mathrm T(x_1)-mathrm T(x_2)right]$$
Then using the identities
$$mathrm L(1/2)=-frac12log2$$
$$mathrm L(1/4)=frac{mathrm G}{2pi}-frac14log2$$
We get $$P_{1,1}left(frac12;frac14right)=frac1{(2pi)^{1/4}}expleft[frac{mathrm G}{2pi}+frac14right]tag{1}$$
From here, the identity
$$-mathrm T(1/12)=frac{2mathrm G}{3pi}$$
which gives
$$P_{1,1}left(frac7{12};frac1{12}right)=sqrt{frac6{7pisqrt[6]{7}}}expleft[frac{2mathrm G}{3pi}+frac12right]tag{2}$$
Then from here, the identity
$$mathrm T(1/20)-mathrm T(3/20)=frac{2mathrm G}{5pi}$$
gives $$P_{2,2}left(frac1{20},frac{13}{20};frac3{20},frac{11}{20}right)=left(frac{j(11)j(3)}{j(13)}right)^{1/20}expfrac{2mathrm G}{5pi}tag{3}$$
Then multiplying $(1),(2),$ and $(3)$, we have the desired result, namely
$$P_{4,4}left(frac12,frac7{12},frac1{20},frac{13}{20};frac14,frac1{12},frac3{20},frac{11}{20}right)=alpha$$

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.

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
$$a_0approx 0.7804591974129376479,a_1approx 0.107, a_2approx -0.0463, a_3approx 0.0151.$$
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.

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

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