오늘의 계산 42 - A Product related to Dedekind eta function

Problem. Show the identity \begin{equation} \label{pb:wts} \prod_{n=1}^{\infty} \left(1 + e^{- \pi n}\right) = 2^{-\frac{1}{8}} e^{\frac{\pi}{24}}. \end{equation}

Proof. This is related to the famous Dedekind eta function $\eta(s)$, which is defined by \begin{equation*} \eta(\tau) = e^{\frac{\pi i \tau}{12}} \prod_{n=1}^{\infty} \left( 1 - e^{2\pi i n \tau} \right), \end{equation*} although we does not need this relation in this posting. Let us denote \begin{equation} \label{pb:px} P(x) = \prod_{n=1}^{\infty} \left( 1 - e^{-2\pi n x} \right) \end{equation} and prove the identity \begin{equation} \label{pb:iden1} \frac{P(x)}{P\left(\frac{1}{x}\right)} = \frac{e^{\frac{\pi}{12}\left(x - \frac{1}{x}\right)}}{\sqrt{x}}, \end{equation} which holds for $\Re x > 0$.

By analytic continuation, it suffices to prove \eqref{pb:iden1} only for positive $x$. Taking logarithm to \eqref{pb:px}, \begin{align*} \log P(x) & = \sum_{n=1}^{\infty} \log \left( 1 - e^{-2\pi n x} \right) \\ & = - \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{m} e^{-2\pi mn x} \\ & = - \sum_{m=1}^{\infty} \frac{1}{m} \sum_{n=1}^{\infty} e^{-2\pi mn x} \\ & = - \sum_{m=1}^{\infty} \frac{1}{m} \frac{1}{e^{2\pi n x} - 1} \\ & = - \sum_{m=1}^{\infty} \frac{1}{m} \left( \frac{1}{e^{2\pi m x} - 1} + \frac{1}{2} - \frac{1}{2\pi m x} - \frac{1}{2} \right) - \sum_{m=1}^{\infty} \frac{1}{2\pi m^2 x} \\ & = - \sum_{m=1}^{\infty} \frac{1}{m} \left( \frac{1}{2} \frac{e^{2\pi m x} + 1}{e^{2\pi m x} - 1} - \frac{1}{2\pi m x} - \frac{1}{2} \right) - \frac{\zeta(2)}{2\pi x} \\ & = - \frac{1}{2} \sum_{m=1}^{\infty} \frac{1}{m} \left( \coth (\pi m x) - \frac{1}{\pi m x} - 1 \right) - \frac{\pi}{12 x}. \end{align*} Now fix $x$ and define $h(t)$ and $f(t)$ by \begin{equation} \label{pb:hx} h(t) = \begin{cases} 1 & \text{if} \ 1 \leq t \\ t & \text{if} \ -1 < t < 1 \\ -1 & \text{if} \ t \leq -1 \end{cases} \end{equation} and \begin{equation} \label{pb:fx} f(t) = \frac{1}{t} \left( \coth (\pi x t) - \frac{1}{\pi x t} - h(t) \right). \end{equation} Singularity inspection of \eqref{pb:fx} at $t = 0$ and $t = \pm \infty$ shows that \begin{equation*} f(t) = \left(\frac{\pi x}{3}-1\right) + O \left(t^2 \right) \quad \text{as} \ t \to 0 \end{equation*} and \begin{equation*} f(t) = O \left(t^{-2}\right) \quad \text{as} \ |t| \to \infty. \end{equation*} Thus $f$ is an integrable function whose singularity at the origin cancels out to yield $f(0) = \frac{\pi}{3} - 1$. Since $f$ is even, we obtain \begin{align*} \log P(x) & = - \frac{1}{2} \sum_{m=1}^{\infty} f(m) - \frac{\pi}{12 x} \\ & = - \frac{1}{4} \sum_{-\infty}^{\infty} f(m) + \frac{f(0)}{4} - \frac{\pi}{12 x} \\ & = - \frac{1}{4} \sum_{-\infty}^{\infty} f(m) + \frac{\pi}{12} \left( x - \frac{1}{x} \right) - \frac{1}{4}. \end{align*} To apply Poisson summation formula, we evaluate the Fourier transform $\hat{f}(\xi)$ of $f$. Since $f$ is real-valued and even, the same holds for $\hat{f}(\xi)$. Thus we may assume $\xi > 0$ for technical reason. Then \begin{align*} \hat{f}(\xi) & = \int_{-\infty}^{\infty} \frac{1}{t} \left( \coth (\pi x t) - \frac{1}{\pi x t} - h(t) \right) e^{-2\pi i \xi t} \; dt \\ & = 2 \int_{0}^{\infty} \frac{1}{t} \left( \coth (\pi x t) - \frac{1}{\pi x t} - h(t) \right) \cos (2\pi i \xi t) \; dt \\ & = 2 \int_{0}^{\infty} \frac{1}{u} \left( \coth (\pi u) - \frac{1}{\pi u} \right) \cos \left( \frac{2 \pi \xi u}{x} \right) \; du - 2 \int_{0}^{\infty} \frac{h(t)}{t} \cos (2 \pi \xi t) \; dt. \end{align*} Here we note that \begin{equation*} \frac{1}{t} \left( \coth (\pi t)-\frac{1}{\pi t} \right) = \frac{2}{\pi}\sum_{n=1}^{\infty} \frac{1}{t^2 + n^2} \end{equation*} and \begin{equation*} \int_{0}^{\infty} \frac{\cos \alpha u}{u^2 + \beta^2} \; du = \frac{\pi}{2\beta} \, e^{-\alpha \beta} \end{equation*} for $\alpha, \beta > 0$. Then we obtain \begin{align*} 2 \int_{0}^{\infty} \frac{1}{u} \left( \coth (\pi u) - \frac{1}{\pi u} \right) \cos \left( \frac{2 \pi \xi u}{x} \right) \; du \\ & = \frac{4}{\pi} \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{1}{u^2 + n^2} \cos \left( \frac{2 \pi \xi u}{x} \right) \; du \\ & = 2 \sum_{n=1}^{\infty} \frac{1}{n} \exp \left( -\frac{2 \pi \xi n}{x} \right) \\ & = -2 \log\left( 1 - \exp \left( - \frac{2 \pi \xi}{x} \right) \right), \end{align*} hence we have \begin{equation} \label{pb:fhat} \hat{f}(\xi) = -2 \left( \log\left( 1 - \exp \left( - \frac{2 \pi \xi}{x} \right) \right) + \int_{0}^{\infty} \frac{h(t)}{t} \cos (2 \pi \xi t) \; dt \right). \end{equation} Again, inspecting the singularity of \eqref{pb:fhat} at $\xi = 0$ shows that \begin{align*} \hat{f}(\xi) & = -2 \log\left( \frac{1 - e^{- \frac{2 \pi \xi}{x}}}{\frac{2 \pi \xi}{x}} \right) -2 \log \left( \frac{2 \pi \xi}{x} \right) - \frac{\sin 2 \pi \xi}{\pi \xi} - 2 \int_{2 \pi \xi}^{\infty} \frac{\cos u}{u} \; du \\ & = -2 \log \left( \frac{2 \pi \xi}{x} \right) - 2 - 2 \left( - \log \left( 2\pi \xi \right) - \gamma \right) + o(1)\\ & = 2 \log x - 2 + 2 \gamma + o(1), \end{align*} where we used the identity \begin{equation*} \int_{x}^{\infty} \frac{\cos u}{u} \; du = - \log x - \gamma + o(1) \quad \text{as} \ x \to 0^{+}. \end{equation*} Thus the singularity cancels out and we obtain $\hat{f}(0) = 2 \log x - 2 + 2 \gamma$. Then Poisson summation formula yields \begin{align*} \log P(x) & = - \frac{1}{4} \sum_{-\infty}^{\infty} \hat{f}(\xi) + \frac{\pi}{12} \left( x - \frac{1}{x} \right) - \frac{1}{4} \\ & = - \frac{1}{2} \sum_{\xi = 1}^{\infty} \hat{f}(\xi) - \frac{1}{4} \left( 2 \log x - 2 + 2 \gamma \right) + \frac{\pi}{12} \left( x - \frac{1}{x} \right) - \frac{1}{4}, \end{align*} and plugging \eqref{pb:fhat} yields \begin{align} \log P(x) & = \sum_{\xi = 1}^{\infty} \log\left( 1 - \exp \left( - \frac{2 \pi \xi}{x} \right) \right) - \frac{1}{2} \log x + \frac{\pi}{12} \left( x - \frac{1}{x} \right) + C \nonumber \\ & = \log P\left( \tfrac{1}{x} \right) - \frac{1}{2} \log x + \frac{\pi}{12} \left( x - \frac{1}{x} \right) + C, \label{pb:iden2} \end{align} where $C$ is an absolute constant given by \begin{equation} \label{pb:c} C = \frac{1}{4} - \frac{\gamma}{2} + \sum_{\xi = 1}^{\infty} \int_{0}^{\infty} \frac{h(t)}{t} \cos (2 \pi \xi t) \; dt. \end{equation} Now plugging $x = 1$ to \eqref{pb:iden2} shows that $C = 0$, therefore we have proved \eqref{pb:iden1}.

The identity \eqref{pb:iden1} can be used to evaluate some exotic products. Plugging $x = \sqrt{2}$, we have \begin{equation*} \prod_{n=1}^{\infty} \left(1 + e^{- \sqrt{2} \pi n}\right) = 2^{-\frac{1}{4}} e^{\frac{\pi}{12\sqrt{2}}}. \end{equation*} Also, plugging $x = 1+i$ gives \begin{equation*} \prod_{n=1}^{\infty} \left(1 + e^{- \pi (2n-1)}\right) = 2^{\frac{1}{4}} e^{-\frac{\pi}{24}}. \end{equation*} Finally, define \begin{equation} \label{pb:qx} Q(x) = \frac{P(2x)}{P(x)} = \prod_{n=1}^{\infty} \left(1 + e^{-2\pi n x}\right). \end{equation} Then \begin{align*} P\left(\tfrac{1}{2}\right) \prod_{n=1}^{\infty} \left(1 + e^{- \pi (2n-1)}\right) & = \left[ \prod_{n=1}^{\infty} \left(1 - e^{- 2 \pi n} \right) \right] \left[ \prod_{n=1}^{\infty} \left(1 - e^{- \pi (4n-2)}\right) \right] \\ & = \frac{P(1)^2}{P(2)} = 2^{\frac{1}{2}} e^{-\frac{\pi}{8}} \frac{P(1)^2}{P\left(\frac{1}{2}\right)} = 2^{\frac{1}{2}} e^{-\frac{\pi}{8}} P\left(\tfrac{1}{2}\right) Q\left(\tfrac{1}{2}\right)^{2}. \end{align*} Therefore, combining these results together, we have \begin{equation*} \prod_{n=1}^{\infty} \left(1 + e^{- \pi n}\right) = Q(\tfrac{1}{2}) = 2^{-\frac{1}{8}} e^{\frac{\pi}{24}}. \end{equation*}