学习一下那个网站里的自动标号设置，我也尝试一下。
以下内容复制自一楼链接，粹纯测试。
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*}
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发表于 2012-8-17 00:19
发表于 2012-8-17 01:03
