## 两道积分习题

1.设$f:[0,1]\to\mathbb{R}$连续,求极限$$\lim\limits_{n\rightarrow \infty}\int_0^1\int_0^1\cdots\int_0^1 f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)dx_1dx_2\cdots dx_n.$$

\begin{align*}&\int_{[0,1]^n}\left| f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)-f(\frac{1}{2})\right|dx_1dx_2\cdots dx_n\\\leq &\int_{\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|\geq\delta}\left| f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)-f(\frac{1}{2})\right|dx_1dx_2\cdots dx_n\\+&\int_{\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|<\delta}\left| f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)-f(\frac{1}{2})\right|dx_1dx_2\cdots dx_n\\\leq&2M\int_{\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|\geq\delta}dx_1dx_2\cdots dx_n+\varepsilon\\\leq&\frac{2M}{\delta^2}\int_{\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|\geq\delta}\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|^2dx_1dx_2\cdots dx_n+\varepsilon\\\leq&\frac{2M}{\delta^2}\int_{[0,1]^n}\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|^2dx_1dx_2\cdots dx_n+\varepsilon\\=&\frac{M}{6n\delta^2}+\varepsilon.\end{align*}

$$\mathop {\lim }\limits_{n \to \infty } E\left( {f\left( {\frac{{{X_1} + {X_2} + \cdots + {X_n}}}{n}} \right)} \right) = E\left( {\mathop {\lim }\limits_{n \to \infty } f\left( {\frac{{{X_1} + {X_2} + \cdots + {X_n}}}{n}} \right)} \right) = E\left( {f\left( {\mathop {\lim }\limits_{n \to \infty } \frac{{{X_1} + {X_2} + \cdots + {X_n}}}{n}} \right)} \right) = f\left( {\frac{1}{2}} \right).$$

2.求证$$\int_0^1\prod_{n=1}^\infty(1-x^n)dx=\frac{4\pi\sqrt3}{\sqrt{23}}\frac{\sinh\frac{\pi\sqrt{23}}3}{\cosh\frac{\pi\sqrt{23}}2}.$$

## 利用定积分证明组合恒等式

$C_n^1 - \frac{1}{2}C_n^2 + \cdots + \frac{{{{\left( { - 1} \right)}^{n + 1}}}}{n}C_n^n = 1 + \frac{1}{2} + \cdots + \frac{1}{n}.$

\begin{align*}\int_0^1 {\frac{{1 - {{\left( {1 - x} \right)}^n}}}{x}dx}&=\int_0^1 {\left[ {1 - {{\left( {1 - x} \right)}^n}} \right]d\left( {\ln x} \right)}  =  - n\int_0^1 {\ln x{{\left( {1 - x} \right)}^{n - 1}}dx} \\&=  - n\int_0^1 {{y^{n - 1}}\ln \left( {1 - y} \right)dy}  = n\int_0^1 {{y^{n - 1}}\sum\limits_{k = 1}^\infty  {\frac{{{y^k}}}{k}} dy}  = \sum\limits_{k = 1}^\infty  {\frac{n}{k}\int_0^1 {{y^{n + k - 1}}dy} } \\&= \sum\limits_{k = 1}^\infty  {\frac{n}{{k\left( {n + k} \right)}}}  = \sum\limits_{k = 1}^\infty  {\left[ {\frac{1}{k} - \frac{1}{{n + k}}} \right]}  = 1 + \frac{1}{2} +  \cdots  + \frac{1}{n}.\end{align*}

## 利用导数不等式进行数列求和估计

（1）证明$\{x_n\}$为递减数列;

（2）证明$\frac{1}{{p + 1}} \cdot \frac{{{2^n} - 1}}{{{2^{n - 1}}}} < \ln \left( {{x_1}{x_2} \cdots {x_n}} \right) < \frac{1}{p} \cdot \frac{{{2^n} - 1}}{{{2^{n - 1}}}}.$

${x_{n + 1}} = \frac{{{x_n} - 1}}{{\ln {x_n}}} \ge \frac{{{x_n} - 1}}{{{x_n} - 1}} = 1.$

$\frac{{{x_{n + 1}}}}{{{x_n}}} = \frac{{{x_n} - 1}}{{{x_n}\ln {x_n}}} < \frac{{{x_n} - 1}}{{{x_n} \cdot \frac{{{x_n} - 1}}{{{x_n}}}}} = 1.$

（2） 先证明左边不等式.首先有

$\frac{{\ln {x_{n + 1}}}}{{\ln {x_n}}} = \frac{{\ln {x_{n + 1}}}}{{\ln {x_n}}} = \frac{{\ln \left( {\frac{{{x_n} - 1}}{{\ln {x_n}}}} \right)}}{{\ln {x_n}}} > \frac{1}{2},$

$\sqrt {{x_n}} - \frac{1}{{\sqrt {{x_n}} }} > \ln {x_n}.$

$\ln {x_n} > \frac{1}{2}\ln {x_{n - 1}} > \cdots > \frac{1}{{{2^{n - 1}}}}\ln {x_1} > \frac{1}{{p + 1}} \cdot \frac{1}{{{2^{n - 1}}}}.$

$\frac{1}{{p + 1}} \cdot \frac{{{2^n} - 1}}{{{2^{n - 1}}}} < \ln \left( {{x_1}{x_2} \cdots {x_n}} \right).$

$\ln x > \frac{{2\left( {x - 1} \right)}}{{x + 1}},\quad x > 1$

$\ln \left( {{x_1}{x_2} \cdots {x_n}} \right) < \frac{1}{p} \cdot \frac{{{2^n} - 1}}{{{2^{n - 1}}}}.$

## FoxTrot Series

$F = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}{n^2}}}{{{n^3} + 1}}} = \frac{1}{3}\left[ {1 - \ln 2 + \pi \mathrm{sech}\left( {\frac{{\sqrt 3 }}{2}\pi } \right)} \right].$

## Ahmed’s integrals 和 Coxeter’s integrals

1. 主要结果

$$A(p, q, r) := \int_{0}^{1} \frac{\arctan q \sqrt{p^{2}x^{2} + 1} }{q \sqrt{p^{2}x^{2} + 1}} \frac{p q r}{(r^{2}+1)p^{2}x^{2} + 1} \, dx.$$

\begin{align*} \rho &= \frac{1-pqr}{1+pqr}, & \alpha &= 2\arctan\left( \frac{qr}{\sqrt{r^{2}+1}} \right), \\ \beta &= 2\arctan\left( \frac{rp}{\sqrt{p^{2}+1}} \right), & \gamma &= 2\arctan\left( r\sqrt{q^{2}+1} \right). \end{align*}

\begin{align*} A(p, q, r) &= \frac{1}{8} \left( \alpha(2\pi-\alpha) + \beta(2\pi-\beta) – \gamma(2\pi-\gamma) \right) \\ &\qquad – \frac{1}{2}\Re \left( \operatorname{Li}_{2}(\rho) – \operatorname{Li}_{2}(\rho e^{i\alpha}) – \operatorname{Li}_{2}(\rho e^{i\beta}) + \operatorname{Li}_{2}(\rho e^{i\gamma}) \right). \end{align*}

$$A(p, q, r) = A(\rho \mid \alpha, \beta, \gamma)$$

2. 推论

$$A(p, q, r) = \frac{1}{8} \left( \alpha(2\pi-\alpha) + \beta(2\pi-\beta) – \gamma(2\pi-\gamma) \right).$$

\begin{align*} \int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{dx}{x^{2}+1} &= A\left( \frac{1}{\sqrt{2}}, \sqrt{2}, 1 \right) \\ &= A\left( 0 \, \middle| \, \frac{\pi}{2}, \frac{\pi}{3}, \frac{2\pi}{3} \right) \\ &= \frac{5\pi^{2}}{96}. \end{align*}

$$A(p, q, r) = \frac{\pi}{4} \left( \alpha + \beta – \gamma \right).$$

$$\Re\operatorname{Li}_{2}(-e^{i\theta}) = \operatorname{Li}_{2}(-1) + \frac{\theta^{2}}{4} \quad |\theta| \leq \pi,$$

\begin{align*} A(-1 \mid \alpha, \beta, \gamma) &= \frac{\pi}{4} \left( \alpha + \beta – \gamma \right). \end{align*}

3. 在Coxeter’s integrals上的应用

\begin{align*} &\int_{0}^{\beta} \arctan \sqrt{\frac{\cos\theta + 1}{a\cos\theta + b}} \, d\theta \\ &= 2 A \left( \sqrt{\frac{a-b}{2} \cdot \frac{1-\cos\beta}{a\cos\beta+ b}}, \sqrt{\frac{2}{a+b}}, \sqrt{\frac{a+b}{a-b}} \right) \\ &= 2 A \left( \frac{1-k}{1+k} \, \middle| \, \alpha, \beta, \gamma\right) \end{align*}

$$k = \left( \frac{1-\cos\beta}{a\cos\beta+ b} \right)^{1/2}, \quad \alpha = \arccos\left(\frac{a-1}{a+1}\right), \quad \gamma = \arccos\left(-\frac{1+b}{1+a}\right).$$定义的.

\begin{align*} \int_{0}^{\frac{\pi}{2}} \arccos \left( \frac{\cos \theta}{1+2\cos \theta} \right) \; d\theta &= 2 \int_{0}^{\frac{\pi}{2}} \arctan \sqrt{\frac{\cos \theta + 1}{3\cos \theta + 1}} \, d\theta \\ &= 4 A \left( 0 \, \middle| \, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3} \right) \\ &= \frac{5\pi^{2}}{24}. \end{align*}

\begin{align*} \int_{0}^{\frac{\pi}{3}} \arccos \left( \frac{\cos \theta}{1+2\cos \theta} \right) \; d\theta &= 2 \int_{0}^{\frac{\pi}{3}} \arctan \sqrt{\frac{\cos \theta + 1}{3\cos \theta + 1}} \, d\theta \\ &= 4 A \left( \frac{\sqrt{5}-1}{\sqrt{5}+1} \, \middle| \, \frac{\pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3} \right). \end{align*}

$$\operatorname{Li}_{2}(\rho) – \operatorname{Li}_{2}(\rho^{2}) – \operatorname{Li}_{2}(\rho^{3}) + \frac{1}{3}\operatorname{Li}_{2}(\rho^{6}), \quad \rho = \frac{\sqrt{5}-1}{\sqrt{5}+1}.$$

$$\int_{0}^{\frac{\pi}{3}} \arccos \left( \frac{\cos \theta}{1+2\cos \theta} \right) \; d\theta = \frac{2\pi^{2}}{15}.$$

\begin{align*}\int_0^{\frac{\pi }{3}} {\arccos \frac{{1 - \cos x}}{{2\cos x}}dx}  &= 2\int_0^{\frac{\pi }{3}} {\arctan \sqrt {\frac{{ - 1 + 3\cos x}}{{1 + \cos x}}} dx}  = 2\int_0^{\frac{\pi }{3}} {\left( {\frac{\pi }{2} - \arctan \sqrt {\frac{{\cos x + 1}}{{3\cos x - 1}}} } \right)dx} \\&= \frac{{{\pi ^2}}}{3} - 2\int_0^{\frac{\pi }{3}} {\arctan \sqrt {\frac{{\cos x + 1}}{{3\cos x - 1}}} dx}  = \frac{{{\pi ^2}}}{3} - 4A\left( {1\left| {\frac{\pi }{3},\frac{\pi }{3},\frac{\pi }{2}} \right.} \right).\end{align*}

\begin{align*}A\left( {0\left| {\frac{\pi }{3},\frac{\pi }{3},\frac{\pi }{2}} \right.} \right) = &\frac{1}{8}\left[ {\frac{\pi }{3}\left( {2\pi  - \frac{\pi }{3}} \right) + \frac{\pi }{3}\left( {2\pi  - \frac{\pi }{3}} \right) - \frac{\pi }{2}\left( {2\pi  - \frac{\pi }{2}} \right)} \right]\\=& \frac{{13{\pi ^2}}}{{288}} .\end{align*}

$\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx} = \int_0^{\frac{\pi }{2}} {\arctan \sqrt {\frac{{\cos x + 1}}{{\cos x}}} dx} = 2A\left( { - 1\left| {\frac{\pi }{2},\frac{\pi }{2},\frac{{2\pi }}{3}} \right.} \right)$

\begin{align*}A\left( { - 1\left| {\frac{\pi }{2},\frac{\pi }{2},\frac{{2\pi }}{3}} \right.} \right) = &\frac{1}{8}\left[ {\frac{\pi }{2}\left( {2\pi  - \frac{\pi }{2}} \right) + \frac{\pi }{2}\left( {2\pi  - \frac{\pi }{2}} \right) - \frac{{2\pi }}{3}\left( {2\pi  - \frac{{2\pi }}{3}} \right)} \right]\\&- \frac{1}{2}{\mathop{\rm Re}\nolimits} \left[ { \operatorname{Li}_2\left( { - 1} \right) - \operatorname{Li}_2\left( { - {e^{i\frac{\pi }{2}}}} \right) - \operatorname{Li}_2\left( { - {e^{i\frac{\pi }{2}}}} \right) + \operatorname{Li}_2\left( { - {e^{i\frac{{2\pi }}{3}}}} \right)} \right]\\= &\frac{{11{\pi ^2}}}{{144}} - \frac{1}{2}\left[ { - \frac{{{\pi ^2}}}{{12}} + \frac{{{\pi ^2}}}{{48}} + \frac{{{\pi ^2}}}{{48}} + \frac{{{\pi ^2}}}{{36}}} \right] = \frac{{{\pi ^2}}}{{12}}.\end{align*}

\begin{align*}&{\mathop{\rm Re}\nolimits} \left\{ {\operatorname{Li}_2\left( {{e^{i\frac{\pi }{3}}}} \right)} \right\} = {\mathop{\rm Re}\nolimits} \left\{ {\operatorname{Li}_2\left( {{e^{ - i\frac{{2\pi }}{3}}}} \right)} \right\} = \frac{1}{2}\left\{ { \operatorname{Li}_2\left( {{e^{i\frac{\pi }{3}}}} \right) + \operatorname{Li}_2\left( {1 - {e^{i\frac{\pi }{3}}}} \right)} \right\}\\= &\frac{1}{2}\left\{ {\frac{1}{6}{\pi ^2} - \ln {e^{i\frac{\pi }{3}}}\ln {e^{ - i\frac{\pi }{3}}}} \right\} = \frac{1}{2}\left\{ {\frac{1}{6}{\pi ^2} - \left( {i\frac{\pi }{3}} \right)\left( { - i\frac{\pi }{3}} \right)} \right\} = \frac{{{\pi ^2}}}{{36}}.\end{align*}

$\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx} = \frac{{{\pi ^2}}}{6}.$

## 级数方法解答的一道积分不等式题

$\int_0^{2\left( {k + 1} \right)} {\frac{{{x^k}}}{{k!}}{e^{ - x}}dx} = - \left. {{e^{ - x}}\sum\limits_{i = 0}^k {\frac{{{x^i}}}{i}} } \right|_0^{2\left( {k + 1} \right)} = 1 - {e^{ - 2\left( {k + 1} \right)}}\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{i}} .$

${e^{2k + 2}} > \left( {k + 1} \right)\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} \Leftrightarrow \sum\limits_{i = 0}^\infty {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} > \left( {k + 1} \right)\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} .$

$\sum\limits_{i = k + 1}^{2k + 1} {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} = \sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^{i + k + 1}}}}{{\left( {i + k + 1} \right)!}}} > k\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} .$

${\left( {2k + 2} \right)^{k + 1}} > k\frac{{\left( {2k + 1} \right)!}}{{k!}} = k\left( {k + 1} \right) \cdots \left( {2k + 1} \right) > k\frac{{\left( {i + k + 1} \right)!}}{{i!}}$是显然的.

## Sion's minimax theorem

(1)试比较$\inf\limits_{y \in I}\sup_{x \in I}f(x,y)$ 与$\sup\limits_{x \in I}\inf_{y\in I}f(x,y)$的大小并证明之;
(2)给出并证明使等式$\inf\limits_{y\in I}\sup\limits_{x \in I}f(x,y) = \sup\limits_{x \in I}\inf\limits_{y \in I}f(x,y)$成立的充分必要条件.

## 某家公司的笔试题

1.设实数列$\{a_n\}$满足$a_{n+p}\leq a_{n}+a_{p}$对于任意的正整数$p,n$，证明：$$\lim\limits_{n\to +\infty}\dfrac{a_n}{n}=\inf\limits_n \dfrac{a_n}{n}.$$
2.设实函数$f(x)$在$(0,1)$内一阶可导且满足$f(1)=1,f(0)=0$，设
\begin{equation*}\int_0^1|f'(x)-f(x)|\geq {u}.\end{equation*}求$u$的最大值。
3.给定一个圆求在这个圆里面随机选择四个点围成一个凸集的概率。

## 与$\sum \arctan$有关的一些问题

\begin{align*}S&=\sum\limits_{n=1}^{\infty} \arctan \frac{10n}{(3n^2+2)(9n^2-1)} \\&= \sum\limits_{n=1}^{\infty} \arg \left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(1+\frac{10in}{(3n^2+2)(9n^2-1)}\right)\\&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4\left(1+\frac{2}{3n^2}\right)\left(1-\frac{1}{9n^2}\right)}\right).\end{align*}

\begin{align*}S&= \arg \prod\limits_{n=1}^{\infty}\left(\frac{(3n^2+2)(9n^2-1)+10in}{27n^4}\right).\end{align*}

$$(3n^2+2)(9n^2-1)+10in = (n-i)(3n+i)(3n+i+1)(3n+i-1).$$

$$S = \arg \prod\limits_{n=1}^{\infty}\frac{\left(1+\frac{i}{3n}\right)\left(1+\frac{i+1}{3n}\right)\left(1+\frac{i-1}{3n}\right)}{\left(1+\frac{i}{n}\right)}.$$

$$S = \arg \frac{-\Gamma(i)}{\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right)\Gamma\left(\frac{i-1}{3}\right)}.$$

$$\Gamma(3z) = \frac{1}{2\pi}3^{3z - \frac{1}{2}}\Gamma\left(z\right)\Gamma\left(z+\frac{1}{3}\right)\Gamma\left(z+\frac{2}{3}\right).$$

$$\Gamma\left(\frac{i-1}{3}\right)\Gamma\left(\frac{i}{3}\right)\Gamma\left(\frac{i+1}{3}\right) = 2\pi 3^{-i+\frac{3}{2}}\Gamma(i-1)$$

\begin{align*}&1+\frac{10in}{\left(3n^2+2\right)\left( 9n^2-1\right)}\\=&\frac{\left(1-\frac in\right)\left(1+\frac i{3n-1}\right)\left(1+\frac i{3n+1}\right)\left(1+\frac i{3n}\right)}{1+\frac2{3n^2}}.\end{align*}

\begin{align*}&\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right)\\=&\arctan\left(\frac1{3n-1}\right)+\arctan\left(\frac1{3n}\right)+\arctan\left(\frac1{3n+1}\right)-\arctan\left(\frac1n\right).\end{align*}

\begin{align*}&\sum_{n=1}^\infty\arctan\left(\frac{10n}{\left(3n^2+2\right)\left( 9n^2-1\right)}\right)\\=&\lim_{m\to\infty}\sum_{n=1}^m\left[\arctan\left(\frac1{3n-1}\right)+\arctan\left(\frac1{3n}\right)+\arctan\left(\frac1{3n+1}\right)-\arctan\left(\frac1n\right)\right]\\=&-\arctan(1)+\lim_{m\to\infty}\sum_{n=m+1}^{3m+1}\arctan\left(\frac1n\right)\\=&-\arctan(1)+\lim_{m\to\infty}\sum_{n=m+1}^{3m+1}\left[\frac1n+O\left(\frac1{n^3}\right)\right]\\=&\log(3)-\frac\pi4.\end{align*}

## 又一道二元反常积分题

$\int_0^\infty {dr} \int_0^{\frac{\pi }{2}} {r{e^{ - ar}}\cos \left( {\alpha r\cos \theta } \right)\cos \left( {\beta r\sin \theta } \right)d\theta } .$