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A. Roy
A. Roy
9:15 AM
Does anyone have a quick proof or reference for the identity:
$$2^{1/4}e^{-\pi/24}=(1+e^{-\pi})(1+e^{-3\pi})(1+e^{-5\pi})\dots$$

It is in Fedor Petrov's comment on an answer here: https://mathoverflow.net/questions/321839/double-sum-over-lattice-points-in-circle
 
 
1 hour later…
Martin Sleziak
Martin Sleziak
10:38 AM
I will just add a direct link to the comment in question:
@GHfromMO It is quite interesting (for me) that all the number theory used for example in your solution may be replaced by applying the linear change of variables. By the way, another standard way to calculate this sum compared to the genuine answer gives the nice, of course known, formula $2^{1/4}e^{-\pi/24}=(1+e^{-\pi})(1+e^{-3\pi})(1+e^{-5\pi})\dots$. — Fedor Petrov 15 hours ago
 
Martin Sleziak
Martin Sleziak
10:50 AM
> The obtained results seem numerically correct, however, I couldn't succeed in expressing them as $I=\pi^2/12-\pi/4\ln(2)$, for example showing that
$$\prod_{n\geq 1}\left(1+e^{-(2n-1)\pi} \right)\stackrel{?}{=}2^{1/4}e^{-\pi/24}$$
> Edit: Finally, a proof of this last identity can be found in this article by Xu Ce (expression (6.3)).
> $$\begin{eqnarray*}
\sum_{k=1}^{+\infty}\pi k\int_{(2k-1)\pi}^{(2k+1)\pi}\frac{dx}{1+e^x}&=&
\pi\sum_{k=0}^{+\infty}\log\left(1+e^{-(2k+1)\pi}\right)\\
&=&\pi\log\prod_{k\geq 0}\left(1+e^{-(2k+1)\pi}\right)\\
&=&\pi\log\left(2^{1/4}e^{-\pi/24}\right)\tag{3}\end{eqnarray*}$$
> where the last identity is kindly provided by a special value of the Dedekind eta function.
@A.Roy I am not sure how much the above links help. I have simply searched on Mathematics for posts containing "e^{-\pi/24}".
The same search on MathOverflow only returns one post. I was too lazy to try what happens if I search in comments.
 
A. Roy
A. Roy
11:54 AM
@MartinSleziak thank you very much! Both links are useful. I don't have permission to comment on the post (frustrating) and never thought of searching for that fragment, though I knew that it was an eta identity somehow.
 
Martin Sleziak
Martin Sleziak
I think that asking a question about this identity would be perfectly fine on Mathematics. And if you included link to the comment on question, it would also comply with the requirement to include context.
Maybe somebody will be able to tell you more. I just did the lame thing - let's search/google for this.
BTW the links in this comment seem to be related to the same product. (It is a comment on one of the answers I linked to.)
(+1) This is really nice! Here's with and without poisson summation. — r9m Jan 13 '18 at 20:59
 
A. Roy
A. Roy
12:10 PM
Yes I saw that. Thanks again for such a quick response. Like a dunderhead I was trying to search for the whole identity.
 

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