美国数学月刊征解题

(2017年10月AMM征解题)求证
[prodlimits_{j ge 1} {{e^{ - 1/j}}left( {1 + frac{1}{j} + frac{1}{{2{j^2}}}} ight)} = frac{{{e^{pi /2}} + {e^{ - pi /2}}}}{{pi {e^gamma }}}.]

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 记

[{x_n} = prodlimits_{k = 1}^n {left( {1 + frac{1}{k} + frac{1}{{2{k^2}}}} ight)} = prodlimits_{k = 1}^n {frac{{{{left( {2k + 1} ight)}^2} + 1}}{{{{left( {2k} ight)}^2}}}} ,]
则
egin{align*}
frac{{prodlimits_{k = 1}^{2n} {left( {1 + frac{1}{{{k^2}}}} ight)} }}{{{x_n}}} &= frac{{prodlimits_{k = 1}^{2n} {frac{{{k^2} + 1}}{{{k^2}}}} }}{{prodlimits_{k = 1}^n {frac{{{{left( {2k + 1} ight)}^2} + 1}}{{{{left( {2k} ight)}^2}}}} }} = frac{{left( {{1^2} + 1} ight)left( {{2^2} + 1} ight)left( {{4^2} + 1} ight) cdots left[ {{{left( {2n} ight)}^2} + 1} ight]}}{{{1^2}{3^2} cdots {{left( {2n - 1} ight)}^2}left[ {{{left( {2n + 1} ight)}^2} + 1} ight]}}\
&= 2prodlimits_{k = 1}^n {left( {1 + frac{1}{{4{k^2}}}} ight) cdot } {left[ {frac{{left( {2n} ight)!!}}{{left( {2n - 1} ight)!!}}} ight]^2}frac{1}{{4{n^2} + 4n + 2}},
end{align*}
由Wallis公式可知
[mathop {lim }limits_{n o infty } {left[ {frac{{left( {2n} ight)!!}}{{left( {2n - 1} ight)!!}}} ight]^2}frac{1}{{2n + 1}} = frac{pi }{2}.]
由$mathrm{sinh} x$的无穷乘积
[frac{{sinh left( {pi x} ight)}}{{pi x}} = prodlimits_{k = 1}^infty {left( {1 + frac{{{x^2}}}{{{k^2}}}} ight)} ]
可知
[prodlimits_{k = 1}^infty {left( {1 + frac{1}{{{k^2}}}} ight)} = frac{{{e^pi } - {e^{ - pi }}}}{{2pi }},quad prodlimits_{k = 1}^infty {left( {1 + frac{1}{{4{k^2}}}} ight)} = frac{{{e^{pi /2}} - {e^{ - pi /2}}}}{pi },]
而调和数列
[{H_n} = sumlimits_{k = 1}^n {frac{1}{k}} = ln n + gamma + oleft( 1 ight),]
故
[mathop {lim }limits_{n o infty } nprodlimits_{k = 1}^n {{e^{ - 1/k}}} = {e^{ - gamma }}.]
因此所求积分为
[frac{{frac{{{e^pi } - {e^{ - pi }}}}{{2pi }}}}{{{e^gamma } imes frac{pi }{2} imes 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
egin{align*}
cosh left( {pi x} ight) &= prodlimits_{n = 1}^infty {left( {1 + frac{{4{x^2}}}{{{{left( {2n - 1} ight)}^2}}}} ight)} ,&frac{{sinh left( {pi x} ight)}}{{pi x}} &= prodlimits_{n = 1}^infty {left( {1 + frac{{{x^2}}}{{{n^2}}}} ight)} ,\
cos left( {pi x} ight) &= prodlimits_{n = 1}^infty {left( {1 - frac{{4{x^2}}}{{{{left( {2n - 1} ight)}^2}}}} ight)} ,&frac{{sin left( {pi x} ight)}}{{pi x}} &= prodlimits_{n = 1}^infty {left( {1 - frac{{{x^2}}}{{{n^2}}}} ight)} ,
end{align*}

另外
egin{align*}
sqrt 2 sin left( {frac{{x + 1}}{4}pi } ight) &= prodlimits_{n = 0}^infty {left( {1 + frac{{{{left( { - 1} ight)}^n}x}}{{2n + 1}}} ight)} ,\
sqrt {x + 1} sin left( {frac{{sqrt {x + 1} }}{2}pi } ight) &= prodlimits_{n = 0}^infty {left( {1 - frac{x}{{4{n^2} - 1}}} ight)} ,\
- sqrt {x - 1} mathrm{csch}left( {frac{pi }{2}} ight)sin left( {frac{{sqrt {x - 1} }}{2}pi } ight) &= prodlimits_{n = 0}^infty {left( {1 - frac{x}{{4{n^2} + 1}}} ight)} ,\
- sqrt { - x - 1} mathrm{csch}left( {frac{pi }{{sqrt a }}} ight)sin left( {frac{{sqrt { - x - 1} }}{{sqrt a }}pi } ight) &= prodlimits_{n = 0}^infty {left( {1 + frac{x}{{a{n^2} + 1}}} ight)} ,\
frac{{{e^{ - gamma x}}}}{{Gamma left( {1 + x} ight)}} &= prodlimits_{n = 1}^infty {frac{{1 + x/n}}{{{e^{x/n}}}}},
end{align*}

对于求和,我们有
egin{align*}
sumlimits_{n = 1}^infty {frac{1}{{{n^2} - {x^2}}}} &= frac{1}{{2{x^2}}} - frac{pi }{{2x}}cot left( {pi x} ight),&&left| x ight| < infty \
sumlimits_{n = 1}^infty {frac{1}{{{{left( {{n^2} - {x^2}} ight)}^2}}}} &= - frac{1}{{2{x^4}}} - frac{{{pi ^2}}}{{4{x^2}}}mathrm{csc}^2left( {pi x} ight) + frac{pi }{{4{x^3}}}cot left( {pi x} ight),&&left| x ight| < infty \
sumlimits_{n = 1}^infty {frac{1}{{{{left( {2n - 1} ight)}^2} - {x^2}}}} &= frac{pi }{{4x}} an left( {frac{pi }{2}x} ight),&& left| x ight| < infty \
sumlimits_{n = 1}^infty {frac{1}{{{{left[ {{{left( {2n - 1} ight)}^2} - {x^2}} ight]}^2}}}} &= frac{{{pi ^2}}}{{16{x^2}}}sec left( {frac{pi }{2}x} ight) - frac{pi }{{8{x^3}}} an left( {frac{pi }{2}x} ight),&&left| x ight| < infty \
sumlimits_{n = 1}^infty {frac{1}{{{n^2} + {x^2}}}} &= frac{pi }{{2x}}coth left( {pi x} ight) - frac{1}{{2{x^2}}}, &&left| x ight| < infty\
sumlimits_{n = 1}^infty {frac{1}{{{{left( {2n - 1} ight)}^2} + {x^2}}}} &= frac{pi }{{4x}} anh left( {frac{pi }{2}x} ight),&&left| x ight| < 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 $mgeq 0$ (a zero of order $m=0$ at $z=0$ means $f(0) eq 0$.
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} ight).]

====Examples of factorization====

egin{align*}sin pi z &= pi z prod_{n eq 0} left(1-frac{z}{n} ight)e^{z/n} = pi zprod_{n=1}^infty left(1-left(frac{z}{n} ight)^2 ight)\cos pi z &= prod_{q in mathbb{Z}, \, q ; ext{odd} } left(1-frac{2z}{q} ight)e^{2z/q} = prod_{n=0}^infty left( 1 - left(frac{z}{n+ frac{1}{2}} ight)^2 ight) 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} ight)^2 ight)]
for $s= frac{1}{2}$.

Mittag-Leffler's theorem.

== Pole expansions of meromorphic functions ==
Here are some examples of pole expansions of meromorphic functions:

egin{align*}frac{1}{sin(z)}&= sum_{n in mathbb{Z}} frac{(-1)^n}{z-npi}= frac{1}{z} + 2zsum_{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-npi}= frac{1}{z} + 2zsum_{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 eq 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*}