Eufisky - The lost book

## 美国数学月刊无穷乘积与西西新年祝福题

$$\sum_{i=1}^{\infty}{a_{i}^{-6}\left(\frac{\sin a_i-\sinh a_i}{\cos a_i+\cosh a_i}\right)^2}=\frac{1}{80}.$$

1、求无穷积分$$\int_{1/2}^{+\infty}{\int_{1/2}^{+\infty}{\int_{1/2}^{+\infty}{\frac{dxdydz}{\prod\limits_{cyc}{\left[242x^5-\left(y-1\right)^5-\left(z+1\right)^5\right]}}}}}.$$

2、设${\left( {1 + \frac{1}{x}} \right)^x} = e\left( {1 - \sum\limits_{k = 1}^\infty {\frac{{{d_k}}}{{{{\left( {\frac{{11}}{{12}} + x} \right)}^k}}}} } \right),$求
${\sum\limits_{k = 1}^\infty {\frac{1}{{1 + d_k^2}}} }.$

(2017年10月AMM征解题)求证

$\prod\limits_{j \ge 1} {{e^{ - 1/j}}\left( {1 + \frac{1}{j} + \frac{1}{{2{j^2}}}} \right)} = \frac{{{e^{\pi /2}} + {e^{ - \pi /2}}}}{{\pi {e^\gamma }}}.$

${x_n} = \prod\limits_{k = 1}^n {\left( {1 + \frac{1}{k} + \frac{1}{{2{k^2}}}} \right)} = \prod\limits_{k = 1}^n {\frac{{{{\left( {2k + 1} \right)}^2} + 1}}{{{{\left( {2k} \right)}^2}}}} ,$
\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*}

$\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{{2n + 1}} = \frac{\pi }{2}.$

$\frac{{\sinh \left( {\pi x} \right)}}{{\pi x}} = \prod\limits_{k = 1}^\infty {\left( {1 + \frac{{{x^2}}}{{{k^2}}}} \right)}$

$\prod\limits_{k = 1}^\infty {\left( {1 + \frac{1}{{{k^2}}}} \right)} = \frac{{{e^\pi } - {e^{ - \pi }}}}{{2\pi }},\quad \prod\limits_{k = 1}^\infty {\left( {1 + \frac{1}{{4{k^2}}}} \right)} = \frac{{{e^{\pi /2}} - {e^{ - \pi /2}}}}{\pi },$

${H_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} = \ln n + \gamma + o\left( 1 \right),$
$\mathop {\lim }\limits_{n \to \infty } n\prod\limits_{k = 1}^n {{e^{ - 1/k}}} = {e^{ - \gamma }}.$

$\frac{{\frac{{{e^\pi } - {e^{ - \pi }}}}{{2\pi }}}}{{{e^\gamma } \times \frac{\pi }{2} \times \frac{{{e^{\pi /2}} - {e^{ - \pi /2}}}}{\pi }}} = \frac{{{e^{\pi /2}} + {e^{ - \pi /2}}}}{{\pi {e^\gamma }}}.$

\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*}

\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*}

\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*}

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*}

## Problems of the Miklós Schweitzer Memorial Competition

1、AOPS论坛

2、匈牙利语版本

3、英文版本

4、其它试题

Suppose that $f: \mathbb{R}^+ \to \mathbb{R}^+$ is a continuous function such that for all positive real numbers $x,y$ the following is true :
$$(f(x)-f(y)) \left ( f \left ( \frac{x+y}{2} \right ) - f ( \sqrt{xy} ) \right )=0.$$
Is it true that the only solution to this is the constant function ?

For any $y \in {\bf R}^+$, there thus exists $x$ arbitrarily close to $1$ for which $f(x) \neq f(y)$, hence $f((x+y)/2) = f(\sqrt{xy})$.  By continuity, this implies that $f((1+y)/2) = f(\sqrt{y})$ for all $y \in {\bf R}^+$.  Making the substitution $z := (1+y)/2$, we conclude that $f(z) = f(g(z))$ for all $z \in {\bf R}^+$, where $g(z) := \sqrt{2z-1}$.  The function $g$ has the fixed point $z=1$ as an attractor, so on iteration and by using the continuity of $f$ we conclude that $f(z)=f(1)$ for all $z \in {\bf R}^+$, so $f$ is indeed constant.

(1950年)令$a>0,d>0$,设$$f(x)=\frac{1}{a}+\frac{x}{a(a+d)}+\cdots+\frac{x^n}{a(a+d)\cdots(a+nd)}+\cdots$$给出$f(x)$的封闭解.

\begin{align*}\sum_{n=0}^{\infty}{\frac{x^n}{\prod\limits_{k=0}^n{\left( a+kd \right)}}}&=\frac{\Gamma \left( \frac{a}{d} \right)}{d}\sum_{n=0}^{\infty}{\frac{\left( x/d \right) ^n}{\Gamma \left( \frac{a}{d}+n+1 \right)}}\\&=\frac{\Gamma \left( \frac{a}{d} \right)}{d}\gamma \left( \frac{a}{d},\frac{x}{d} \right) \left( \frac{d}{x} \right) ^{a/d}\frac{e^{x/d}}{\Gamma \left( \frac{a}{d} \right)}=\left( \frac{d}{x} \right) ^{a/d}\frac{e^{x/d}}{d}\gamma \left( \frac{a}{d},\frac{x}{d} \right) ,\end{align*}

$g(x) = x^a f(x^d)$ satifies $g'(x) = x^{a-1} + x^{d-1} g(x)$. Solve the associated differential equation and conclude.

$$\int_0^1{\frac{\arctan x}{x\sqrt{1-x^2}}dx}.$$

Let $n$ be a positive integer. Prove that, for $0<x<\frac\pi{n+1}$,
$$\sin{x}-\frac{\sin{2x}}{2}+\cdots+(-1)^{n+1}\frac{\sin{nx}}{n}-\frac{x}{2}$$
is positive if $n$ is odd and negative if $n$ is even.

Since

\begin{align*}f_n(x) &= \sin{x} - \frac {\sin{2x}}{2} + \cdots + ( - 1)^{n + 1}\frac {\sin{nx}}{n} - \frac {x}{2},\\f_n'(x) &= - \mbox{Re}\left(\sum_{n = 1}^{n}z^n\right) - \frac12.\end{align*}

After some simplifications we get
$$f_n'(x) = \frac {( - 1)^{n + 1}}{2}((1 - \cos(x))\frac {\sin((n + 1)x)}{\sin(x)} + \cos((n + 1)x))$$
and $$f_n''(x) = \frac {( - 1)^{n}}{2}\frac {(n + 1)\sin(nx) + n\sin((n + 1)x)}{1 + \cos(x)}.$$
The formula for $f_n''$ shows that $( - 1)^n f$ is convex for $0 < x < \frac {\pi}{n + 1}$. Since $f_n(0) = 0$ and $f_n'(0) = \frac {( - 1)^{n + 1}}{2}$.We are ready when we can show that $( - 1)^{n + 1}f_n(\frac {\pi}{n + 1}) > 0$.

We have to distinct between two different, but very similar cases, namely $n$ is odd, and $n$ is even.
Let's restrict to the case $n$ is even.
We prove $f_{2n}(\frac {\pi}{2n + 1}) < 0$.

\begin{align*}f_{2n}\left( \frac{\pi}{2n+1} \right) &=\sum_{k=1}^{2n}{\left( -1 \right)}^{k+1}\frac{\sin \left( \frac{k\pi}{2n+1} \right)}{k}-\frac{\pi}{2\left( 2n+1 \right)}\\&=\frac{\pi}{2n+1}\left( \sum_{k=1}^n{\frac{\sin \left( \frac{\left( 2k-1 \right) \pi}{2n+1} \right)}{\frac{\left( 2k-1 \right) \pi}{2n+1}}}-\sum_{k=1}^n{\frac{\sin \left( \frac{2k\pi}{2n+1} \right)}{\frac{2k\pi}{2n+1}}} \right) -\frac{\pi}{2\left( 2n+1 \right)}.\end{align*}

The function $x \mapsto \frac {\sin(x)}{x}$ is descending on $[0,\pi]$, thus
both sums lay between $a$ and $a + \frac {2\pi}{2n + 1}$, where $a = \int_0^{\pi}\frac {\sin(x)}{x}\,dx$.

Thus $$f_{2n}\left(\frac {\pi}{2n + 1}\right) < \frac {\pi}{2n + 1}\cdot\frac {2\pi}{2n + 1} - \frac {\pi}{2(2n + 1)} < 0.$$

## Euler-Maclaurin求和公式估计梯形积分公式的误差

$\mathop {\lim }\limits_{n \to \infty } {n^4}\left( {\frac{1}{{24}} - n\left( {n\left( {\frac{\pi }{4} - {A_n}} \right) - \frac{1}{4}} \right)} \right).$

Euler-Maclaurin求和公式

\begin{align*}\frac{{b - a}}{n}\sum\limits_{i = 1}^n {\frac{1}{2}\left[ {f\left( {{x_{i - 1}}} \right) + f\left( {{x_i}} \right)} \right]} - \int_a^b {f\left( x \right)dx} = &\sum\limits_{k = 1}^m {\frac{{{B_{2k}}}}{{\left( {2k} \right)!}}{h^{2k}}\left[ {{f^{\left( {2k - 1} \right)}}\left( b \right) - {f^{\left( {2k - 1} \right)}}\left( a \right)} \right]} \\&+ \frac{{{B_{2m + 2}}}}{{\left( {2m + 2} \right)!}}{h^{2m + 2}}{f^{\left( {2m + 2} \right)}}\left( \xi \right)\left( {b - a} \right),\end{align*}

\begin{align*}&{A_n} + \frac{1}{{4n}} - \frac{\pi }{4} = \frac{1}{2}\left[ {\left( {{A_n} - \frac{1}{{2n}} + \frac{1}{n}} \right) + {A_n}} \right] - \frac{\pi }{4} = \frac{{{B_2}}}{{2!}} \cdot \frac{1}{{{n^2}}}\left[ {f'\left( 1 \right) - f'\left( 0 \right)} \right]\\+ &\frac{{{B_4}}}{{4!}} \cdot \frac{1}{{{n^4}}}\left[ {f'''\left( 1 \right) - f'''\left( 0 \right)} \right] + \frac{{{B_6}}}{{6!}} \cdot \frac{1}{{{n^6}}}\left[ {{f^{\left( 5 \right)}}\left( 1 \right) - {f^{\left( 5 \right)}}\left( 0 \right)} \right] + \frac{{{B_8}}}{{8!}} \cdot \frac{1}{{{n^8}}}{f^{\left( 8 \right)}}\left( \xi \right),\end{align*}

${n^4}\left( {\frac{1}{{24}} - n\left( {n\left( {\frac{\pi }{4} - {A_n}} \right) - \frac{1}{4}} \right)} \right) = \frac{1}{{2016}} + \frac{{{B_8}}}{{8!}} \cdot \frac{1}{{{n^2}}}{f^{\left( 8 \right)}}\left( \xi \right),$

## 西西爆难积分题求解

1. $\int_0^1 {\frac{{\ln \left( {1 + {x^{2 + \sqrt 3 }}} \right)}}{{1 + x}}dx} = \frac{{{\pi ^2}}}{{12}}\left( {1 - \sqrt 3 } \right) + \ln 2 \cdot \ln \left( {1 + \sqrt 3 } \right)$
2. $\int_0^1 {\frac{{\ln \left( {1 + {x^{4 + \sqrt {15} }}} \right)}}{{1 + x}}dx} = \frac{{{\pi ^2}}}{{12}}\left( {2 - \sqrt {15} } \right) + \ln \frac{{1 + \sqrt 5 }}{2} \cdot \ln \left( {2 + \sqrt 3 } \right) + + \ln 2 \cdot \ln \left( {\sqrt 3 + \sqrt 5 } \right)$
3. $\int_0^1 {\frac{{\ln \left( {1 + {x^{6 + \sqrt {35} }}} \right)}}{{1 + x}}dx} = \frac{{{\pi ^2}}}{{12}}\left( {3 - \sqrt {35} } \right) + \ln \frac{{1 + \sqrt 5 }}{2} \cdot \ln \left( {8 + 3\sqrt 7 } \right) + + \ln 2 \cdot \ln \left( {\sqrt 5 + \sqrt 7 } \right)$
4. $\int_0^1 {\frac{{ar\tanh x\ln x}}{{x\left( {1 - x} \right)\left( {1 + x} \right)}}dx}$
5. $\int_0^1 {\frac{{\arctan {x^{3 + \sqrt 8 }}}}{{1 + {x^2}}}dx}$
6. $\int_0^{\frac{\pi }{3}} {x{{\ln }^2}\left( {2\sin \frac{x}{2}} \right)dx}$
7. $\int_0^{\frac{\pi }{2}} {{{\ln }^n}\sin xdx}$
8. $\int_0^\infty {\frac{{\sin x}}{{\cosh x - \cos x}} \cdot \frac{{{x^n}}}{{n!}}dx}$
9. $\int_0^\infty {\frac{{\sin x}}{{\cosh x + \cos x}} \cdot \frac{{{x^n}}}{{n!}}dx}$
10. $\int_{\frac{\pi }{4}}^{\frac{\pi }{2}} {\ln \ln \left( {\tan x} \right)dx}$
11. $\int_0^\infty {\frac{{x - \sin x}}{{\left( {{\pi ^2} + {x^2}} \right){x^3}}}dx} = \frac{1}{{2{\pi ^3}}}\left( {1 + \frac{{{\pi ^2}}}{2} - \pi - \frac{1}{{{e^\pi }}}} \right)$
12. $\int_0^\infty {\frac{1}{{{a^2} + {x^2}}}\frac{{x - \sin x}}{{{x^3}}}dx} = \frac{{{a^2} - 2a + 2 - 2{e^{ - a}}}}{{4{a^3}}},a > 0$
13. $\prod\limits_{k = 4}^\infty {\left[ {1 - {{\left( {\frac{3}{k}} \right)}^3}} \right]} = \frac{8}{{15561}} \cdot \frac{{\cos \left( {\frac{{3\sqrt 3 }}{2}\pi } \right)}}{\pi }$
14. $\sum\limits_{m = 1}^\infty {\sum\limits_{n = {2^{m - 1}}}^{{2^m} - 1} {\frac{m}{{\left( {2n + 1} \right)\left( {2n + 2} \right)}}} } = 1 - \gamma$
15. $\int_0^1 {\frac{{\left( {1 - x} \right)\ln x \cdot {e^{ - x}}}}{{\pi - x}}dx}$
16. $\mathop {\lim }\limits_{n \to \infty } \left[ { - \frac{1}{{2m}} + \ln \left( {\frac{e}{m}} \right) + \sum\limits_{n = 2}^m {\left( {\frac{1}{n} - \frac{{\varsigma \left( {1 - n} \right)}}{{{m^n}}}} \right)} } \right] = \gamma$
17. $\int_0^\infty {\frac{1}{{x{e^x}\left( {{\pi ^2} + {{\ln }^2}x} \right)}}dx}$
18. $\int_0^\infty {\frac{{\ln \left( {1 + {x^2}} \right)}}{{{e^{2\pi x}} - 1}}dx}$
19. $\sum\limits_{n = 1}^\infty {\frac{{\sum\limits_{k = 1}^n {\frac{1}{{{k^4}}}} }}{{{n^2}}}}$
20. $\int_0^1 {\int_0^1 {\int_0^1 {\int_0^1 {\frac{{dwdxdydz}}{{\left( {wxy - 1} \right)\left( {zwx - 1} \right)\left( {yzw - 1} \right)\left( {xyz - 1} \right)}}} } } }$
21. $\int_0^\infty {\frac{{{e^{ - {x^2}}}}}{{{\pi ^2} + {{\left( {\gamma + x} \right)}^2}}}dx}$
22. $\sqrt[3]{{1 + \sqrt[3]{{1 + \sqrt[3]{{1 + \sqrt[3]{{1 + \sqrt[3]{{1 + \cdots }}}}}}}}}} = \frac{2}{{\sqrt 3 }}\cos \left( {\frac{1}{3}\arccos \frac{{3\sqrt 3 }}{2}} \right)$
23. $\int_0^{2ar\cosh \pi } {\frac{{dx}}{{1 + \frac{{{{\sinh }^2}x}}{{{\pi ^4}}}}}}$
24. $\int_0^\infty {\frac{{x{e^x}}}{{\sqrt {4{e^x} - 3} \left( {1 + 2{e^x} - \sqrt {4{e^x} - 3} } \right)}}dx}$
25. $\int_0^{\frac{\pi }{2}} {\frac{{{{\ln }^2}\left( {2\cos x} \right)}}{{{{\ln }^2}\left( {2\cos x} \right) + {x^2}}}dx}$
26. $\mathop {\lim }\limits_{n \to \infty } \frac{{n!}}{{{n^n}}}\left( {\sum\limits_{k = 0}^n {\frac{{{n^k}}}{{k!}}} - \sum\limits_{k = n + 1}^\infty {\frac{{{n^k}}}{{k!}}} } \right)$
27. $\sum\limits_{k = 1}^\infty {\frac{1}{{{k^2}}}\cos \left( {\frac{9}{{k\pi + \sqrt {{k^2}{\pi ^2} - 9} }}} \right)}$
28. $\sum\limits_{n = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{{{{\left( {n\pi + \phi } \right)}^2}}}\cos \left( {\sqrt {{n^2}{\pi ^2} + {a^2} - {\phi ^2}} } \right)} = \frac{{a\cos a\cot \phi + \phi \sin a}}{{a\sin \phi }}$
29. $\sum\limits_{n = 0}^\infty {\frac{{{n^2}{\pi ^2} + {\phi ^2}}}{{{{\left( {{n^2}{\pi ^2} - {\phi ^2}} \right)}^2}}}{{\left( { - 1} \right)}^n}\cos \left( {\sqrt {{n^2}{\pi ^2} + {a^2} - {\phi ^2}} } \right)} = \frac{{\cos \sqrt {{a^2} - {\phi ^2}} }}{{2{\phi ^2}}} + \frac{{a\cos a\cot \phi + \phi \sin a}}{{2a\sin \phi }}$
30. $\sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}\left( {n + \frac{1}{2}} \right)}}{{{{\left( {n + \frac{1}{3}} \right)}^2}{{\left( {n + \frac{2}{3}} \right)}^2}}}} \cos \left[ {\pi \sqrt {\left( {n + \frac{1}{6}} \right)\left( {n + \frac{5}{6}} \right)} } \right] = {\pi ^2}{e^{\frac{{\pi \sqrt 3 }}{6}}}$
31. $\int_{ - 1}^1 {\frac{{\arctan x}}{{1 + x}}\ln \left( {\frac{{1 + {x^2}}}{2}} \right)dx}$
32. $\int_0^1 {\sin \left( {\pi x} \right){x^x}{{\left( {1 - x} \right)}^{1 - x}}dx}$
33. $\int_0^{\frac{\pi }{2}} {{x^2}{{\ln }^2}\left( {2\cos x} \right)dx}$
34. $\int_0^{\frac{\pi }{2}} {\frac{{{x^2}}}{{{x^2} + {{\ln }^2}\left( {2\cos x} \right)}}dx}$
35. $\int_0^{\frac{\pi }{2}} {\frac{{{x^2}{{\ln }^2}\left( {2\cos x} \right)}}{{{x^2} + {{\ln }^2}\left( {2\cos x} \right)}}dx}$
36. $\mathop {\lim }\limits_{n \to \infty } \frac{n}{{\ln n}}\left[ {\frac{1}{\pi }{{\left( {\sum\limits_{k = 1}^n {\sin \frac{\pi }{{\sqrt {{n^2} + k} }}} } \right)}^n} - \frac{1}{{\sqrt[4]{e}}}} \right]$
37. $\int_0^1 {{x^{ - x}}{{\left( {1 - x} \right)}^{ - 1 + x}}\sin \left( {\pi x} \right)dx = \frac{\pi }{e}} = 1.15573$
38. $\int_0^{ + \infty } {\frac{{dx}}{{1 + x\left| {\sin x} \right|}}}$
39. $\int_0^{ + \infty } {\frac{{dx}}{{{x^2} + {{\left( {{n^2}{x^2} - 1} \right)}^2}}}}$