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A tough product

Calculus (Integrals, Series)
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Tolaso J Kos
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A tough product

#1

Post by Tolaso J Kos » Tue Feb 09, 2016 9:10 am

Prove that:

$$\prod_{n=1}^{\infty}\left ( 1+e^{-\pi n} \right )=\frac{e^{\pi/24}}{\sqrt[8]{2}}$$
Imagination is much more important than knowledge.
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Re: A tough product

#2

Post by r9m » Sun Feb 14, 2016 11:12 pm

There'd be a way with Mellin Transform and residue theorem, but I suppose we could start with, \begin{align*}\sum\limits_{n=1}^{N} \log (1+e^{-2\pi n}) &= \sum\limits_{n=1}^{N-1} n\left(\log (1+e^{-2\pi n}) - \log (1+e^{-2\pi (n+1)}) \right) + N\log (1+e^{-2\pi N})\\&= \sum\limits_{n=1}^{N-1}\int_n^{n+1} \frac{2\pi n}{1+e^{2\pi x}}\,dx + N \log (1+e^{-2\pi N})\\&= 2\pi\sum\limits_{n=1}^{N-1}\int_n^{n+1} \frac{x - \{x\}}{1+e^{2\pi x}}\,dx + N \log (1+e^{-2\pi N})\\&= 2\pi \int_{0}^{N} \frac{x - \{x\}}{1+e^{2\pi x}}\,dx + N \log (1+e^{-2\pi N})\end{align*}

Letting $N \to \infty$, and recalling the fourier series of: $\displaystyle \{x\} = \frac{1}{2} - \frac{1}{\pi}\sum\limits_{j=1}^{\infty} \frac{\sin 2\pi j x}{j}$ we have:

\begin{align*}\sum\limits_{n=1}^{\infty} \log (1+e^{-2\pi n}) &= 2\pi \int_{0}^{\infty} \frac{x - \{x\}}{1+e^{2\pi x}}\,dx \\&= 2\pi \int_0^{\infty} \frac{x - \frac{1}{2}}{1+e^{2\pi x}}\,dx + 2\sum\limits_{j=1}^{\infty} \frac{1}{j} \int_0^{\infty} \frac{\sin 2\pi j x}{1+e^{2\pi x}}\,dx \\&= 2\pi \int_0^{\infty} \frac{x - \frac{1}{2}}{1+e^{2\pi x}}\,dx + 2\sum\limits_{k=1}^{\infty}\sum\limits_{j=1}^{\infty} (-1)^{k-1}\frac{1}{j} \int_0^{\infty} e^{-2\pi kx}\sin 2\pi j x\,dx \\&= 2\pi \int_0^{\infty} \frac{x - \frac{1}{2}}{1+e^{2\pi x}}\,dx + \frac{1}{\pi}\sum\limits_{k=1}^{\infty}\sum\limits_{j=1}^{\infty} \frac{(-1)^{k-1}}{k^2+j^2} \\&= \frac{1}{2\pi}\eta(2) - \frac{1}{2} \log 2 - \frac{1}{4\pi}\left(\sum\limits_{(j,k) \in \mathbb{Z^{2}}\setminus \{(0,0)\}} \frac{(-1)^{k}}{j^2+k^2} - 2\zeta(2) + 2\eta(2)\right)\\&= \frac{\pi}{12} - \frac{1}{2}\log 2 - \frac{1}{4\pi}\sum\limits_{(j,k) \in \mathbb{Z^{2}}\setminus \{(0,0)\}} \frac{(-1)^{k}}{j^2+k^2}\\&= \frac{\pi}{12} - \frac{1}{2}\log 2 + \frac{1}{2\pi}\eta(1)\beta(1) = \frac{\pi}{12} - \frac{3}{8}\log 2\end{align*}

In a similar manner we have:

\begin{align*}\sum\limits_{n=0}^{\infty} \log (1+e^{-(2n+1)\pi}) = 2\pi\sum\limits_{n=1}^{\infty} \int_{n-\frac{1}{2}}^{n+\frac{1}{2}}\frac{n}{1+e^{2\pi x}}\,dx &= \frac{\pi}{24} - \frac{1}{\pi}\sum\limits_{j,k \ge 1} \frac{(-1)^{j+k}}{j^2+k^2}\\&= \frac{\pi}{24} - \frac{1}{4\pi}\left(\sum\limits_{(j,k) \in \mathbb{Z^{2}}\setminus \{(0,0)\}} \frac{(-1)^{j+k}}{j^2+k^2} + 4\eta(2)\right)\\&= -\frac{\pi}{24} - \frac{1}{4\pi}\sum\limits_{(j,k) \in \mathbb{Z^{2}}\setminus \{(0,0)\}} \frac{(-1)^{j+k}}{j^2+k^2}\\&= -\frac{\pi}{24} + \frac{1}{\pi}\eta(1)\beta(1) = -\frac{\pi}{24} + \frac{1}{4}\log 2\end{align*}

Thus, $$\prod_{n=1}^{\infty}\left(1+e^{-\pi n} \right) = \prod_{n=1}^{\infty}\left(1+e^{-2\pi n} \right) \prod_{n=0}^{\infty}\left(1+e^{-(2n+1)\pi } \right) = \frac{e^{\pi/24}}{2^{1/8}}$$
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Re: A tough product

#3

Post by galactus » Sun Feb 14, 2016 11:20 pm

Wow, rd, you're just showin' off now :clap2:
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Re: A tough product

#4

Post by whitexlotus » Sun Sep 04, 2016 9:59 pm

http://zerocollar.blogspot.cl/" onclick="window.open(this.href);return false;
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