(2017年10月AMM征解题)求证
\[\prod\limits_{j \ge 1} { {e^{ - 1/j}}\left( {1 + \frac{1}{j} + \frac{1}{ {2{j^2}}}} \right)} = \frac{ { {e^{\pi /2}} + {e^{ - \pi /2}}}}{ {\pi {e^\gamma }}}.\]记
\[{x_n} = \prod\limits_{k = 1}^n {\left( {1 + \frac{1}{k} + \frac{1}{
{2{k^2}}}} \right)} = \prod\limits_{k = 1}^n {\frac{ { { {\left( {2k + 1} \right)}^2} + 1}}{ { { {\left( {2k} \right)}^2}}}} ,\]则\begin{align*}\frac{ {\prod\limits_{k = 1}^{2n} {\left( {1 + \frac{1}{ { {k^2}}}} \right)} }}{ { {x_n}}} &= \frac{ {\prod\limits_{k = 1}^{2n} {\frac{ { {k^2} + 1}}{ { {k^2}}}} }}{ {\prod\limits_{k = 1}^n {\frac{ { { {\left( {2k + 1} \right)}^2} + 1}}{ { { {\left( {2k} \right)}^2}}}} }} = \frac{ {\left( { {1^2} + 1} \right)\left( { {2^2} + 1} \right)\left( { {4^2} + 1} \right) \cdots \left[ { { {\left( {2n} \right)}^2} + 1} \right]}}{ { {1^2}{3^2} \cdots { {\left( {2n - 1} \right)}^2}\left[ { { {\left( {2n + 1} \right)}^2} + 1} \right]}}\\&= 2\prod\limits_{k = 1}^n {\left( {1 + \frac{1}{ {4{k^2}}}} \right) \cdot } {\left[ {\frac{ {\left( {2n} \right)!!}}{ {\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{ {4{n^2} + 4n + 2}},\end{align*}由Wallis公式可知\[\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{ {\left( {2n} \right)!!}}{ {\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{ {2n + 1}} = \frac{\pi }{2}.\]由$\mathrm{sinh} x$的无穷乘积\[\frac{ {\sinh \left( {\pi x} \right)}}{ {\pi x}} = \prod\limits_{k = 1}^\infty {\left( {1 + \frac{ { {x^2}}}{ { {k^2}}}} \right)} \]可知\[\prod\limits_{k = 1}^\infty {\left( {1 + \frac{1}{ { {k^2}}}} \right)} = \frac{ { {e^\pi } - {e^{ - \pi }}}}{ {2\pi }},\quad \prod\limits_{k = 1}^\infty {\left( {1 + \frac{1}{ {4{k^2}}}} \right)} = \frac{ { {e^{\pi /2}} - {e^{ - \pi /2}}}}{\pi },\]而调和数列\[{H_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} = \ln n + \gamma + o\left( 1 \right),\]故\[\mathop {\lim }\limits_{n \to \infty } n\prod\limits_{k = 1}^n { {e^{ - 1/k}}} = {e^{ - \gamma }}.\]因此所求积分为\[\frac{ {\frac{ { {e^\pi } - {e^{ - \pi }}}}{ {2\pi }}}}{ { {e^\gamma } \times \frac{\pi }{2} \times \frac{ { {e^{\pi /2}} - {e^{ - \pi /2}}}}{\pi }}} = \frac{ { {e^{\pi /2}} + {e^{ - \pi /2}}}}{ {\pi {e^\gamma }}}.\]事实上，我们还有
\begin{align*}\cosh \left( {\pi x} \right) &= \prod\limits_{n = 1}^\infty {\left( {1 + \frac{ {4{x^2}}}{ { { {\left( {2n - 1} \right)}^2}}}} \right)} ,&\frac{ {\sinh \left( {\pi x} \right)}}{ {\pi x}} &= \prod\limits_{n = 1}^\infty {\left( {1 + \frac{ { {x^2}}}{ { {n^2}}}} \right)} ,\\\cos \left( {\pi x} \right) &= \prod\limits_{n = 1}^\infty {\left( {1 - \frac{ {4{x^2}}}{ { { {\left( {2n - 1} \right)}^2}}}} \right)} ,&\frac{ {\sin \left( {\pi x} \right)}}{ {\pi x}} &= \prod\limits_{n = 1}^\infty {\left( {1 - \frac{ { {x^2}}}{ { {n^2}}}} \right)} ,\end{align*}另外
\begin{align*}\sqrt 2 \sin \left( {\frac{ {x + 1}}{4}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 + \frac{ { { {\left( { - 1} \right)}^n}x}}{ {2n + 1}}} \right)} ,\\\sqrt {x + 1} \sin \left( {\frac{ {\sqrt {x + 1} }}{2}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 - \frac{x}{ {4{n^2} - 1}}} \right)} ,\\ - \sqrt {x - 1} \mathrm{csch}\left( {\frac{\pi }{2}} \right)\sin \left( {\frac{ {\sqrt {x - 1} }}{2}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 - \frac{x}{ {4{n^2} + 1}}} \right)} ,\\ - \sqrt { - x - 1} \mathrm{csch}\left( {\frac{\pi }{ {\sqrt a }}} \right)\sin \left( {\frac{ {\sqrt { - x - 1} }}{ {\sqrt a }}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 + \frac{x}{ {a{n^2} + 1}}} \right)} ,\\\frac{ { {e^{ - \gamma x}}}}{ {\Gamma \left( {1 + x} \right)}} &= \prod\limits_{n = 1}^\infty {\frac{ {1 + x/n}}{ { {e^{x/n}}}}},\end{align*} 对于求和,我们有\begin{align*}\sum\limits_{n = 1}^\infty {\frac{1}{ { {n^2} - {x^2}}}} &= \frac{1}{ {2{x^2}}} - \frac{\pi }{ {2x}}\cot \left( {\pi x} \right),&&\left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{ { { {\left( { {n^2} - {x^2}} \right)}^2}}}} &= - \frac{1}{ {2{x^4}}} - \frac{ { {\pi ^2}}}{ {4{x^2}}}\mathrm{csc}^2\left( {\pi x} \right) + \frac{\pi }{ {4{x^3}}}\cot \left( {\pi x} \right),&&\left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{ { { {\left( {2n - 1} \right)}^2} - {x^2}}}} &= \frac{\pi }{ {4x}}\tan \left( {\frac{\pi }{2}x} \right),&& \left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{ { { {\left[ { { {\left( {2n - 1} \right)}^2} - {x^2}} \right]}^2}}}} &= \frac{ { {\pi ^2}}}{ {16{x^2}}}\sec \left( {\frac{\pi }{2}x} \right) - \frac{\pi }{ {8{x^3}}}\tan \left( {\frac{\pi }{2}x} \right),&&\left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{ { {n^2} + {x^2}}}} &= \frac{\pi }{ {2x}}\coth \left( {\pi x} \right) - \frac{1}{ {2{x^2}}}, &&\left| x \right| < \infty\\\sum\limits_{n = 1}^\infty {\frac{1}{ { { {\left( {2n - 1} \right)}^2} + {x^2}}}} &= \frac{\pi }{ {4x}}\tanh \left( {\frac{\pi }{2}x} \right),&&\left| x \right| < \infty\end{align*}其中$\mathrm{sinh}x=\frac{e^x-e^{-x}}2,\mathrm{cosh}x=\frac{e^x+e^ {-x}}2,\mathrm{csch}x=\frac2{e^x-e^ {-x}},\mathrm{tanh}x=\frac{e^x-e^ {-x}}{e^x+e^ {-x}},\mathrm{coth}x=\frac{e^x+e^ {-x}}{e^x-e^ {-x}}$.The Weierstrass factorization theorem. Sometimes called the Weierstrass product/factor theorem.
Let $f$ be an entire function, and let $\{a_n\}$ be the non-zero zeros of $ƒ$ repeated according to multiplicity; suppose also that $ƒ''$ has a zero at $z= 0$ of order $m\geq 0$ (a zero of order $m=0$ at $z=0$ means $f(0)\neq 0$.
Then there exists an entire function $g$ and a sequence of integers $\{p_n\}$ such that\[f(z)=z^m e^{g(z)} \prod_{n=1}^\infty E_{p_n}\left(\frac{z}{a_n}\right).\]
====Examples of factorization====
\begin{align*}\sin \pi z &= \pi z \prod_{n\neq 0} \left(1-\frac{z}{n}\right)e^{z/n} = \pi z\prod_{n=1}^\infty \left(1-\left(\frac{z}{n}\right)^2\right)\\\cos \pi z &= \prod_{q \in \mathbb{Z}, \, q \; \text{odd} } \left(1-\frac{2z}{q}\right)e^{2z/q} = \prod_{n=0}^\infty \left( 1 - \left(\frac{z}{n+\tfrac{1}{2}} \right)^2 \right) \end{align*}
The cosine identity can be seen as special case of
\[\frac{1}{\Gamma(s-z)\Gamma(s+z)} = \frac{1}{\Gamma(s)^2}\prod_{n=0}^\infty \left( 1 - \left(\frac{z}{n+s} \right)^2 \right)\]for $s=\tfrac{1}{2}$.Mittag-Leffler's theorem.
== Pole expansions of meromorphic functions ==
Here are some examples of pole expansions of meromorphic functions:\begin{align*}\frac{1}{\sin(z)}&= \sum_{n \in \mathbb{Z}} \frac{(-1)^n}{z-n\pi}= \frac{1}{z} + 2z\sum_{n=1}^\infty (-1)^n \frac{1}{z^2 - (n\,\pi)^2},\\\cot(z) &\equiv \frac{\cos (z)}{\sin (z)}= \sum_{n \in \mathbb{Z}} \frac{1}{z-n\pi}= \frac{1}{z} + 2z\sum_{k=1}^\infty \frac{1}{z^2 - (k\,\pi)^2},\\\frac{1}{\sin^2(z)} &= \sum_{n \in \mathbb{Z}} \frac{1}{(z-n\,\pi)^2},\\
\frac{1}{z \sin(z)}&= \frac{1}{z^2} + \sum_{n \neq 0} \frac{(-1)^n}{\pi n(z-\pi n)}= \frac{1}{z^2} + \sum_{n=1}^\infty \frac{(-1)^n}{n\,\pi} \frac{2z}{z^2 - (n\,\pi)^2}.\end{align*}