Eufisky - The lost book

美国数学月刊无穷乘积与西西新年祝福题

若$a_i$是超越方程\[\left( {\cos x} \right)\left( {\cosh x} \right) + 1 = 0\]的正实数根从小到大排成的数列, 求证:

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


首先可以肯定,这个方程里$a_i$肯定是解不出的.其次我们有\[{\left( {\sinh {a_i}} \right)^2} = {\left( {\cosh {a_i}} \right)^2} - 1 = \frac{1}{{{{\cos }^2}{a_i}}} - 1 = {\tan ^2}{a_i} \Rightarrow \sinh {a_i} = \left| {\tan {a_i}} \right|.\]
当$\sinh {a_i} = \tan {a_i}$时,我们有\[{\left( {\frac{{\sin {a_i} - \sinh {a_i}}}{{\cos {a_i} + \cosh {a_i}}}} \right)^2} = {\left( {\frac{{\sin {a_i} - \tan {a_i}}}{{\cos {a_i} - \frac{1}{{\cos {a_i}}}}}} \right)^2} = {\tan ^2}\frac{{{a_i}}}{2}.\]
当$\sinh {a_i} = -\tan {a_i}$时,我们有\[{\left( {\frac{{\sin {a_i} - \sinh {a_i}}}{{\cos {a_i} + \cosh {a_i}}}} \right)^2} = {\left( {\frac{{\sin {a_i} + \tan {a_i}}}{{\cos {a_i} - \frac{1}{{\cos {a_i}}}}}} \right)^2} = {\cot ^2}\frac{{{a_i}}}{2}.\]事实上,两种情况都会出现.接下来大家一起来思考下哈!

以下几个也是不同寻常的题,正是因为莫名其妙、不明觉厉才想一探究竟,希望大家一起来玩!


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

 

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 ?
陶哲轩解答:Yes.  If $f$ were not constant, then (since ${\bf R}^+$ is connected) it could not be locally constant, thus there exists $x_0 \in {\bf R}^+$ such that $f$ is not constant in any neighbourhood of $x_0$.  By rescaling (replacing $f(x)$ with $f(x_0 x)$) we may assume without loss of generality that $x_0=1$.
 
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.
来源:这里
(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.

也就是求\[\sum\limits_{n = 0}^\infty  {\frac{{{x^n}}}{{\prod\limits_{k = 0}^n {\left( {a + kd} \right)} }}} .\]

解.首先有$$\prod_{k=0}^n{\frac{1}{a+kd}}=\frac{\Gamma \left( \frac{a}{d} \right)}{d^{n+1}\Gamma \left( \frac{a}{d}+n+1 \right)},$$

又因为$$\gamma \left( s,x \right) =\sum_{k=0}^{\infty}{\frac{x^se^{-x}x^k}{s\left( s+1 \right) ...\left( s+k \right)}}=x^s\,\Gamma \left( s \right) \,e^{-x}\sum_{k=0}^{\infty}{\frac{x^k}{\Gamma \left( s+k+1 \right)}},$$我们有

\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*}
其中$\displaystyle\Gamma(s,x) = \int_x^{\infty} t^{s-1}\,e^{-t}\,{\rm d}t$为the upper incomplete gamma function,而$\displaystyle\gamma(s,x) = \int_0^x t^{s-1}\,e^{-t}\,{\rm d}t$为the lower incomplete gamma function.参考这里.

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


令$a\in (0,\pi)$,设$n$为正整数.证明$$\int_0^{\pi}{\frac{\cos \left( nx \right) -\cos \left( na \right)}{\cos x-\cos a}dx}=\pi \frac{\sin \left( na \right)}{\sin a}.$$

求$$\int_1^{\frac{\sqrt{5}+1}{2}}{\left( \frac{\arctan x}{\arctan x-x} \right) ^2dx},$$
$$\int_0^1{\frac{\arctan x}{x\sqrt{1-x^2}}dx}.$$

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

西西在大学群里的一道题,也是2014年第六届非数竞赛预赛最后一题的推广:


设${A_n} = \frac{n}{{{n^2} + 1}} + \frac{n}{{{n^2} + {2^2}}} + \cdots + \frac{n}{{{n^2} + {n^2}}}$,求极限

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


这里提供个一般的方法.


Euler-Maclaurin求和公式

设函数$f\in C^{(2m+2)}[a,b],h=\frac{b-a}{n},x_i=a+ih,i=0,1,\cdots,n$,则

\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*}

其中$\xi\in [a,b]$, $B_{2k}(k=1,2,\cdots,m+1)$是Bernoulli数且$B_2=\frac16,B_4=-\frac{1}{30},B_6=\frac{1}{42}$.


enlightened解:取$a=0,b=1,f(x)=\frac{1}{1+x^2}$,则$h=\frac1n,x_i=\frac{i}{n},A_n=\frac{1}{n}\sum\limits_{i = 1}^n {f\left( {{x_i}} \right)}$,则

\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*}

其中$\xi\in [0,1]$,也即

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

注意到${f^{\left( 8 \right)}}\left( \xi \right)$有界,因此$n\to\infty$时,所求极限为$\frac{1}{{2016}}$.

西西爆难积分题求解

在这里,主要展示西西12年7月在百度贴吧数学吧中贴出的30个积分题的求解,本文中主要参考的是自己以前的摘录,不知何故,与西哥的版本有些出入,但基本包含了西哥的所有问题,来源于网友的解答均会注明出处。

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. \[\int_0^{ + \infty } {\frac{{dx}}{{{x^2} + {{\left( {{n^2}{x^2} - 1} \right)}^2}}}}\]

参阅:[1]http://tieba.baidu.com/p/2114477017

[2]http://tieba.baidu.com/p/3148596990

[3]http://tieba.baidu.com/p/2121721064

[4]http://tieba.baidu.com/p/2908551228

[5]http://tieba.baidu.com/p/1700279486?qq-pf-to=pcqq.c2c