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

解法一.设$|f|$最大值为$M$.对任何$\varepsilon>0$,存在$\delta>0$,使得当$|x-1/2|<\delta$时,有$$\left|f(x)-f(\frac{1}{2})\right|<\varepsilon.$$

\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*}

因此$$\limsup_{n\rightarrow\infty}\int_{[0,1]^n}\left| f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)-f(\frac{1}{2})\right|dx_1dx_2\cdots dx_n\leq \varepsilon.$$

令$\varepsilon\rightarrow0$即可.

解法二.由科尔莫格罗夫强大数定律得$$\frac{{{X_1} + {X_2} + \cdots + {X_n}}}{n}\mathop \to \limits^{a.s.} E\left( {{X_i}} \right) = \frac{1}{2}\left( {n \to + \infty } \right).$$

又因为$f(x)$连续有界,由控制收敛定理可知

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

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

解.注意到Pentagonal number theorem,我们知$$\int_{0}^{1}\prod_{n\geq1}\left(1-x^{n}\right)dx=\sum_{k\in\mathbb{Z}}\left(-1\right)^{k}\int_{0}^{1}x^{k\left(3k-1\right)/2}dx=\sum_{k\in\mathbb{Z}}\left(-1\right)^{k}\frac{2}{3k^{2}-k+2}.$$再利用求和公式可知$$\sum_{n\in\mathbb{Z}}\left(-1\right)^{n}f\left(n\right)=-\sum\left\{ \pi\csc\left(\pi z\right)f(z) \textrm{ 在 } f\left(z\right)\textrm{ 的极点上的留数}\right\}.$$而极点为$z=\frac{1}{6}\left(1\pm i\sqrt{23}\right)$,由此求得.

利用定积分证明恒等式：

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

证.首先注意到\[{\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k + 1}}}}{k}C_n^k{x^k}} } \right)^\prime } = \sum\limits_{k = 1}^n {C_n^k{{\left( { - x} \right)}^{k - 1}}}  = \frac{1}{{ - x}}\sum\limits_{k = 1}^n {C_n^k{{\left( { - x} \right)}^k}}  = \frac{1}{{ - x}}\left[ {{{\left( {1 - x} \right)}^n} - 1} \right].\]

因此\[\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k + 1}}}}{k}C_n^k}  = \int_0^1 {\frac{{{{\left( {1 - x} \right)}^n} - 1}}{{ - x}}dx}  = \int_0^1 {\frac{{1 - {{\left( {1 - x} \right)}^n}}}{x}dx} .\]

因此

当我能解数学题时，表明我从阴影中走出来了。

设$x_1=\frac{p+1}{p},p>1,x_{n+1}=\frac{x_n-1}{\ln x_n}$.

（1）证明$\{x_n\}$为递减数列;

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

证.（1）首先利用$\ln(1+x)\leq (x>-1)$可知

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

又$x_1=\frac{p+1}{p}>1$,显然有$x_n>1$.再利用$\ln (1+x)\geq \frac x{x+1}$,我们有

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

因此$\{x_n\}$为递减数列.

（2） 先证明左边不等式.首先有

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

即

\[\sqrt {{x_n}} - \frac{1}{{\sqrt {{x_n}} }} > \ln {x_n}.\]

这只要在不等式$\frac{1}{2}\left( {x - \frac{1}{x}} \right) > \ln x,x > 1$中令$x=\sqrt{x_n}>1$便可得到.又利用$\frac x{x+1}\leq \ln (x+1)\leq x$可知\[\frac{1}{{p + 1}} < \ln {x_1} = \ln \left( {1 + \frac{1}{p}} \right) < \frac{1}{p}.\]

于是

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

将上述不等式前$n$项累加便可得到

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

接着证明右边不等式.由不等式

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

可知\[{x_{n + 1}} = \frac{{{x_n} - 1}}{{\ln {x_n}}} < \frac{{{x_n} + 1}}{2}.\]

因此\[{x_{n + 1}} - 1 < \frac{{{x_n} - 1}}{2} \Rightarrow {x_n} \le 1 + \frac{1}{{p{2^{n - 1}}}} \Rightarrow \ln {x_n} < \frac{1}{{p{2^{n - 1}}}},\]

从而有

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

在这里,介绍了FoxTrot Series.

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

另外,在这里有一些极限讨论

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

百年云烟只过眼，不为繁华易素心

一道略火的题，由雷神解答：

证明\[\int_0^{2\left( {k + 1} \right)} {\frac{{{x^k}}}{{k!}}{e^{ - x}}dx} > \frac{k}{{1 + k}},\quad k = 0,1,2, \cdots .\]

证.注意到\[\int {\frac{{{x^k}}}{{k!}}{e^{ - x}}dx} = - {e^{ - x}}\sum\limits_{i = 0}^k {\frac{{{x^i}}}{i}} + C.\]

因此

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

等价于证明\[1 - {e^{ - 2\left( {k + 1} \right)}}\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{i}} > \frac{k}{{1 + k}}.\]

即证

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

下面证明\[\sum\limits_{i = 0}^{2k + 1} {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} > \left( {k + 1} \right)\sum\limits_{i = 0}^k {\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}}} .\]

等价于证明

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

只需证明\[\frac{{{{\left( {2k + 2} \right)}^{i + k + 1}}}}{{\left( {i + k + 1} \right)!}} > k\frac{{{{\left( {2k + 2} \right)}^i}}}{{i!}} \Leftrightarrow {\left( {2k + 2} \right)^{k + 1}} > k\frac{{\left( {i + k + 1} \right)!}}{{i!}}.\]

而

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

证明\[\sum\limits_{n = 1}^\infty {\arctan \frac{{10n}}{{\left( {3{n^2} + 2} \right)\left( {9{n^2} - 1} \right)}}} = \ln 3 - \frac{\pi }{4}.\]

解.(r9m)首先有

\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*}

分母里的无穷乘积$\displaystyle \prod\limits_{n=1}^{\infty}\left(1+\frac{2}{3n^2}\right)$和$\displaystyle \prod\limits_{n=1}^{\infty}\left(1-\frac{1}{9n^2}\right)$ 可以被忽略,当它们收敛于实数时.

因此

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

对分子进行分解

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

我们得

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

在$\displaystyle z = i,\frac{i}{3},\frac{i+1}{3},\frac{i-1}{3}$ 处运用$\displaystyle \frac{1}{\Gamma(z)} = ze^{\gamma z}\prod\limits_{n=1}^{\infty}\left(1+\frac{z}{n}\right)e^{-z/n}$ .

我们可以改写为

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

另一方面运用Gauss-Legendre Triplication Formula

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

令$z = \dfrac{i-1}{3}$我们有

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

因此$$S = \arg \frac{-3^{i}\Gamma(i)}{\Gamma(i-1)} = \arg (3^{i}(1-i)) = \log 3 - \frac{\pi}{4}.$$

解法二.(robjohn)运用$\arctan(x)=\arg(1+ix)$,分解可知

\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*}

因此有

\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*}

裂项可知

\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*}

网友WB问我一道积分题\[\int_0^\infty  {\int_0^\infty  {{e^{ - a\sqrt {{x^2} + {y^2}} }}\cos \alpha x\cos \beta ydxdy} } .\]

经极坐标代换后得到

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

然后利用积化和差公式感觉可以继续往下算.