a closed form for the following product.

Question

while solving some infinite products I recently encountered a terribly hard infinite product which is $$\prod_{n=0}^\infty 1-e^{-(4n+1)\pi}$$ I tried many methods theorems but still can't obtain a closed-form or a perfect answer, but what I did obtain is an approximation which is as follows $$\approx \frac{4^{1/16}}{e^{\pi/24}}$$ Any help would be appreciated.

Answer 1

$$\prod _{n=0}^{\infty } \left(1-e^{-(4n+1)\pi }\right)=\left(1-e^{-\pi }\right)\prod _{n=1}^{\infty } \left(1-e^{-(4n+1)\pi }\right)$$ $$\log \left(\prod _{n=1}^{\infty } \left(1-e^{-(4n+1)\pi}\right)\right)=\sum _{n=1}^{\infty } \log \left(1-\frac 1{e^{(4n+1)\pi}}\right)$$ $$\log \left(1-\frac 1{e^{(4n+1)\pi}}\right)=-\sum_{p=1}^\infty \frac 1 p e^{-(4n+1)p\pi }$$ Reverse the summations and you face a geometric progression $$\sum_{n=1}^\infty \frac 1 p e^{-(4n+1)p\pi }=\frac{e^{-p\pi }}{\left(e^{4 \pi p}-1\right) p}\sim \frac 1p e^{-5p\pi }$$ $$\sum_{p=1}^\infty \frac 1p e^{-5p\pi }= -\log \left(1-e^{-5 \pi }\right)$$ $$\prod _{n=0}^{\infty } \left(1-e^{-(4n+1)\pi }\right)\sim \left(1-e^{-\pi }\right)\left(1-e^{-5\pi }\right)=0.95678593754691\cdots$$ while, computed, the decimal representation of the infinite product is $0.95678593754641$ (the result is then in an absolute error of $5.03 \times 10^{-12}$).

Your approximation $\sqrt[8]{2} e^{-\pi /24}$ is in an absolute error of $7.72 \times 10^{-5}$