Is this identity or an equivalent one already referenced in the litterature? In particular, is it even true?


${\frac{\left ( -1 ; e^{-4\pi} \right) ^2\_{\infty}}{\left ( e^{-2\pi} ; e^{-2\pi} \right) ^4\_{\infty}} = \frac{32 \pi^3 \left(\sqrt{2}-1 \right )\sqrt{2^\frac{1}{4}+2^\frac{3}{4}}}{\Gamma^4{\left(\frac{1}{4} \right )}} \simeq 4,030103529...}$


It appears in relation to a [particular elliptic function](https://mathoverflow.net/questions/91851/elliptic-function-with-constant-real-part-on-the-unit-square-diagonals).


Similar identities also arise in this [post](https://mathoverflow.net/questions/91763/special-values-of-a-doubly-periodic-meromorphic-function)


Thanks in advance, 



q-Pochhammer Symbol Identity
The equality is indeed correct. It follows from identities in Ramanujan's notebook.


First notice that 
$\left(-1;e^{-4\pi}\right)\_{\infty}^{2}=2\left(-e^{-4\pi};e^{-4\pi}\right)\_{\infty}^{2},$ 
so we are trying to prove the identity


$$\frac{\left(-e^{-4\pi};e^{-4\pi}\right)\_{\infty}^{2}}{\left(e^{-2\pi};e^{-2\pi}\right)\_{\infty}^{4}}=\frac{16\pi^{3}\left(\sqrt{2}-1\right)\sqrt{2^{\frac{1}{4}}+2^{\frac{3}{4}}}}{\Gamma^{4}\left(\frac{1}{4}\right)}.$$


The right hand side many be cleaned up further, and written as 


$$\frac{32\pi^{3}2^{\frac{1}{8}}\sqrt{\sqrt{2}-1}}{\Gamma^{4}\left(\frac{1}{4}\right)}.$$ 


Now, since $1+q^{4}=\frac{1-q^{8}}{1-q^{4}},$ the left hand side is


$$\frac{\left(e^{-8\pi};e^{-8\pi}\right)\_{\infty}^{2}}{\left(e^{-4\pi};e^{-4\pi}\right)\_{\infty}^{2}\left(e^{-2\pi};e^{-2\pi}\right)\_{\infty}^{4}}=\frac{\left(e^{-8\pi}\right)\_{\infty}^{2}}{\left(e^{-4\pi}\right)\_{\infty}^{2}\left(e^{-2\pi}\right)\_{\infty}^{4}}.\ \ \ \ \ \ \ \ \ \ (1)$$ 


On page 326 of Bruce C Brendts “Ramanujan's Notebook Part V” he shows that


$$\left(e^{-2\pi}\right)\_{\infty}=\frac{\Gamma\left(\frac{1}{4}\right)}{2\pi^{\frac{3}{4}}}e^{\frac{\pi}{12}}$$


$$\left(e^{-4\pi}\right)\_{\infty}=\frac{2^{-\frac{3}{8}}\Gamma\left(\frac{1}{4}\right)}{2\pi^{\frac{3}{4}}}e^{\frac{\pi}{6}}$$ and 


$$\left(e^{-8\pi}\right)\_{\infty}=\frac{2^{-\frac{13}{16}}\Gamma\left(\frac{1}{4}\right)}{2\pi^{\frac{3}{4}}}\left(\sqrt{2}-1\right)^{\frac{1}{4}}e^{\frac{\pi}{3}}.$$


Combining these three together in equation (1) yields the desired result.


**Remark:** You could have proceeded in a different manner by noticing that the denominator is (almost) the [Dedkind eta function](http://mathworld.wolfram.com/DedekindEtaFunction.html). The product can be written as 


$$\frac{\left(-e^{-4\pi};e^{-4\pi}\right)\_{\infty}^{2}}{\left(e^{-2\pi};e^{-2\pi}\right)\_{\infty}^{4}}=\frac{\eta(4i)^2}{\eta(i)^4\eta(2i)^2}.$$ 


Indeed there are many ways to write this product, another I stumbled across is $$\frac{\left(-e^{-4\pi};e^{-4\pi}\right)\_{\infty}^{2}}{\left(e^{-2\pi};e^{-2\pi}\right)\_{\infty}^{4}}=\left(\frac{1}{\vartheta\_{4}\left(e^{-4\pi}\right)\vartheta\_{4}\left(e^{-2\pi}\right)}\right)^{2},$$ where $\vartheta\_4(q)$ is a [Jacobi theta function.](http://mathworld.wolfram.com/JacobiThetaFunctions.html)


