$\mathrm G$ is Catalan's constant.

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

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 $$P_{4,4}\left(\frac12,\frac7{12},\frac1{20},\frac{13}{20};\frac14,\frac1{12},\frac3{20},\frac{11}{20}\right)=\alpha$$

  • $\begingroup$ Were you ask a question? or just write a proof? $\endgroup$ Mar 13, 2019 at 21:26
  • $\begingroup$ See the edit. My question is highlighted in yellow $\endgroup$
    – clathratus
    Mar 13, 2019 at 21:33
  • $\begingroup$ @Mr.N what are you asking? $$\mathrm L(x)=\frac1\pi\int_0^{\pi x}\log(\sin t)dt...$$ $\endgroup$
    – clathratus
    Jan 23, 2020 at 2:37
  • $\begingroup$ @Mr.N yes. This is all my own work $\endgroup$
    – clathratus
    Jan 23, 2020 at 16:01

3 Answers 3


Here's a partial answer. Let me start with expressing my astonishment about the construction of such a complicated product, and even more about the closed expression you have found. Truely Ramanujan-like!

§1 Definitions (in Mathematica)

The auxiliary function

fe[n_, x_] := (n + x)^(n + x)/((E n)^(2 x) (n - x)^(n - x))

The partial product (up to the nn-th factor)

p[nn_] := 
 Product[(fe[n, 1/2] fe[n, 7/12] fe[n, 1/20] fe[n, 13/20])/(
  fe[n, 1/4] fe[n, 1/12] fe[n, 3/20] fe[n, 11/20]), {n, 1, nn}]

The closed expression result given in the OP

pr = Exp[(47 Catalan)/(30 \[Pi]) + 3/4] Sqrt[
   33/(91 \[Pi]) Sqrt[2/\[Pi] 11^(1/5)/7^(1/3) (3^3/13^3)^(1/5)]];

N[pr] = 0.780459...

§2 Check of numerics

Numerically, the product is slowly going down to pr

For instance N[p[100]/pr-1] = 0.0013684

§3 Sum over $\log(E_n)$

This is the elementary constituent of the ($\log$) of the product.

$$\sum_{n=1}^\infty \log(E_n(x)) = \zeta ^{(1,0)}(-1,1-x)-\zeta ^{(1,0)}(-1,x+1)+x+x (-(\log (2)+\log (\pi )))$$

Here $\zeta ^{(1,0)}(s,z) = \frac{\partial}{\partial s}\zeta(s,z)$.


I asked the same question on Math Overflow, and I got an answer in the affirmative. Check it out.



$\displaystyle U(-x)=\left(\frac{2\pi |x|}{e}\right)^x \exp\left(\frac{Cl_2(2\pi x)}{2\pi}\right)~~$ for $~~|x|<1~~$ , $~~Cl_2(z) $ is the Clausen function

$\displaystyle \alpha = \prod\limits_{x\in\{-\frac{1}{2},-\frac{7}{12},-\frac{1}{20},-\frac{13}{20},\\\hspace{5mm}+\frac{1}{4},+\frac{1}{12},+\frac{3}{20},+\frac{11}{20}\}} U(-x)~~~~$ with $~~~~\displaystyle \sum\limits_{x\in\{-\frac{1}{2},-\frac{7}{12},-\frac{1}{20},-\frac{13}{20},\\\hspace{5mm}+\frac{1}{4},+\frac{1}{12},+\frac{3}{20},+\frac{11}{20}\}} Cl_2(2\pi x) = \frac{47}{15}G$

The rest of the product is just simple multiplication.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.