陈洪葛的博客

证明

\[I= \lim_{n\to\infty}\left(\int_{0}^{\frac{\pi}{2}}\left|\frac{\sin(2n+1)x}{\sin{x}}\right|dx-\frac{2\ln{n}}{\pi}\right)=\frac{6\ln{2}}{\pi}+\frac{2\gamma}{\pi}+\frac{2}{\pi}\cdot\sum_{k=1}^{\infty}\frac{1}{2k+1}\ln\left(1+\frac{1}{k} \right)\]

_________________________________________________________________________________________________________

证明:
我们应该有结论:对定义在$\left[0,\frac{\pi}{2}\right]$任意连续函数$f(x)$,有
\[ \lim_{n\to\infty}\int_{0}^{\frac{\pi}{2}}|\sin nx|f(x)dx=\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}f(x)dx \]
特别的，对于奇数的$n$也成立.这时
\[ \lim_{n\to\infty}\int_{0}^{\frac{\pi}{2}}|\sin(2n+1)x|\left(\frac{1}{\sin{x}}-\frac{1}{x}\right)dx=\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{\sin{x}}-\frac{1}{x} \right)dx \]
注意到
\[ \frac{1}{\sin{x}}=\frac{1}{x}+\sum_{n=1}^{\infty}(-1)^{n}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right) \]
就有
\[\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{\sin{x}}-\frac{1}{x} \right)dx=\sum_{n=1}^{\infty}(-1)^{n}\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{x-n\pi}+\frac{1}{x+n\pi}\right)dx=\frac{2}{\pi}\ln\frac{4}{\pi}\]
而
\[ \int_{0}^{\frac{\pi}{2}}\frac{|\sin(2n+1)x|}{x}dx=\int_{0}^{\frac{\pi}{2}(2n+1)}\frac{|\sin{x}|}{x}dx=\int_{0}^{n\pi}\frac{|\sin{x}|}{x}dx+\int_{n\pi}^{n\pi+\frac{\pi}{2}}\frac{|\sin{x}|}{x}dx
\]
同时
\[ \int_{n\pi}^{n\pi+\frac{\pi}{2}}\frac{|\sin{x}|}{x}dx\sim o\left(\frac{1}{n}\right) \qquad (n\to\infty) \]
所以
\[ \int_{0}^{\frac{\pi}{2}}\frac{|\sin(2n+1)x|}{x}dx=\int_{0}^{n\pi}\frac{|\sin{x}|}{x}dx+o\left(\frac{1}{n}\right) \qquad (n\to\infty) \]
\begin{align*}
\int_{0}^{n\pi}\frac{|\sin{x}|}{x}dx&=\int_{0}^{\pi}\left(\sum_{k=0}^{n-1}\frac{1}{x+k\pi}\right)\sin{x}dx\\
&=\int_{0}^{1}\left(\sum_{k=0}^{n-1}\frac{1}{x+k}\right)\sin{\pi x}dx\\
&=\int_{0}^{1}\sin{\pi x}\left(\psi(x+n)-\psi(x)\right)dx
\end{align*}
这里
\[ \psi(x)=\frac{\Gamma'(x)}{\Gamma(x)} \]
我们知道它有渐进展开式
\[ \psi(x)=\ln{x}-\frac{1}{2x}-\sum_{r=1}^{n}\frac{(-1)^{r-1}B_{r}}{2r}x^{-2r}+O(x^{-2n-2}) \]
而
\[ \zeta(1-2m)=\frac{(-1)^{m}B_{m}}{2m}\qquad (m\geq 1) \]
所以
\[ \psi(x)=\ln{x}-\frac{1}{2x}+\sum_{k=1}^{\infty}\frac{\zeta(1-2k)}{x^{2k}}+O(x^{-2n-2}) \]
另一方面，有
\[\int_0^1 \sin(\pi x)\psi(x+n) dx = \frac{2}{\pi} \ln n + O\left(\frac{1}{n}\right)\qquad (n\to\infty)\]
于是
\[\lim_{n\to\infty}\left(\int_{0}^{\frac{\pi}{2}}\left|\frac{\sin(2n+1)x}{\sin{x}}\right|dx-\frac{2\ln{n}}{\pi}\right)=\frac{2}{\pi}\ln\frac{4}{\pi}-\int_{0}^{1}\sin{\pi x}\psi(x)dx \]
下面来弄后面那个积分,分部积分有
\[ \int_{0}^{1}\sin\pi{x}\cdot \psi(x)dx=\int_{0}^{1}\sin{\pi x}d\ln\Gamma(x)=-\pi\int_{0}^{1}\cos\pi x\ln\Gamma(x)dx \]
然后把$\ln\Gamma(x)$的Fourier展开，就是
\[ \ln\Gamma(x)=\frac{1}{2}\ln\frac{\pi}{\sin{\pi x}}+[\gamma+\ln{2\pi}]\left(\frac{1}{2}-x\right)+\frac{1}{\pi}\sum_{k=2}^{\infty}\frac{\ln{k}}{k}\sin(2\pi kx)\]
接着就是3个式子

\[\int_0^1 \cos\pi x\ln\frac{\pi}{\sin(\pi x)} dx = 0  \tag{1}\]
\[ \int_{0}^{1}\cos\pi x\left(\frac{1}{2}-x\right)dx=\frac{2}{\pi^2} \tag{2}\]
 \[ \int_{0}^{1}\cos\pi x\sin2\pi kx dx=\frac{4k}{(4k^2-1)\pi} \tag{3} \]

代进去计算得到
\begin{align*}
  I&=\lim_{n\to\infty}\left(\int_{0}^{\frac{\pi}{2}}\left|\frac{\sin(2n+1)x}{\sin{x}}\right|dx-\frac{2\ln{n}}{\pi}\right)\\
&=\frac{2}{\pi}\ln\frac{4}{\pi}+\pi\left[\frac{2}{\pi^2}(\gamma+\ln2\pi)+\frac{4}{\pi^2}\sum_{k=2}^{\infty}\frac{\ln{k}}{4k^2-1} \right]\\
&=\frac{2}{\pi}\left[\ln{8}+\gamma+\sum_{k=2}^{\infty}\ln{k}\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\right]\\
&=\frac{6\ln{2}}{\pi}+\frac{2\gamma}{\pi}+\frac{2}{\pi}\cdot\sum_{k=1}^{\infty}\frac{1}{2k+1}\ln\left(1+\frac{1}{k} \right)
\end{align*}

问题：设函数$f$在区间$[a,b]$上处处大于$0$,且对于$L>0$满足Lipschitz条件
\[ |f(x_1)-f(x_2)|\leq L|x_{1}-x_{2}| \]
又已知对于$a\leq c\leq d\leq b$有
\[ \int_{c}^{d}\frac{dx}{f(x)}=\alpha,\qquad \int_{a}^{b}\frac{dx}{f(x)}=\beta \]
证明下列积分不等式:
\[ \int_{a}^{b}f(x)dx\leq \frac{e^{2L\beta}-1}{2L\alpha}\int_{c}^{d}f(x)dx \]

____________________________________________________________________________________________________________________

证明
记
\[ f(x_{0})=\min_{x\in[a,b]}f(x)=m \]
则
\[ m\leq f(x)\leq m+L|x-x_{0}| \]
就有
\begin{align*}
&2L\int_{c}^{d}\frac{dx}{f(x)}\cdot\int_{a}^{b}f(x)dx\\
&\leq 2L\int_{c}^{d}\frac{dx}{f(x)}\cdot\int_{a}^{b}[m+L|x-x_{0}|]dx\\
&= 2L\int_{c}^{d}\frac{dx}{f(x)}\cdot\left[\int_{a}^{x_{0}}(m+L(x_{0}-x)dx+\int_{x_{0}}^{b}(m+L(x-x_{0}))dx\right]\\
&= 2L\int_{c}^{d}\frac{dx}{f(x)}\left[m(b-a)+\frac{L}{2}[(x_{0}-a)^2+(b-x_{0})^{2}]\right]\\
&\leq m^{2}\int_{c}^{d}\frac{dx}{f(x)}\left[\frac{2L(b-a)}{m}+\frac{L^2(x_{0}-a)^2+L^2(b-x_{0})^{2}}{m^2}\right]\\
&\leq m(d-c)\left[\frac{2L(b-a)}{m}+\frac{L^2(x_{0}-a)^2+L^2(b-x_{0})^{2}}{m^2}\right]\\
&\leq \int_{c}^{d}f(x)dx\left[\frac{2L(b-a)}{m}+\frac{L^2(x_{0}-a)^2+L^2(b-x_{0})^{2}}{m^2}\right]\\
&=\int_{c}^{d}f(x)dx\left[\frac{2L(b-x_{0})}{m}+\frac{2L(x_{0}-a)}{m}+\frac{L^2(x_{0}-a)^2+L^2(b-x_{0})^{2}}{m^2}\right]\\
&=\int_{c}^{d}f(x)dx\left[\left(\frac{L(x_{0}-a)}{m}+1\right)^{2}+\left(\frac{L(b-x_{0})}{m}+1\right)^{2}-2 \right]\\
&\leq \int_{c}^{d}f(x)dx\left[\left(\frac{L(x_{0}-a)}{m}+1\right)^{2}\left(\frac{L(b-x_{0})}{m}+1\right)^{2}-1 \right]
\end{align*}
兹证明
\[ \left[\left(\frac{L(x_{0}-a)}{m}+1\right)^{2}\left(\frac{L(b-x_{0})}{m}+1\right)^{2}-1 \right]\leq e^{2L\beta}-1 \]
就是
\[ \ln\left(\frac{L(x_{0}-a)}{m}+1\right)+\ln\left(\frac{L(b-x_{0})}{m}+1\right)\leq L\beta \]
这时，只要注意到
\begin{align*}
L\beta&=L\int_{a}^{b}\frac{1}{f(x)}dx\\
&\geq L\int_{a}^{b}\frac{1}{m+L|x-x_{0}|}dx\\
&=L\int_{a}^{x_{0}}\frac{1}{m+L(x_{0}-x)}dx+L\int_{x_{0}}^{b}\frac{1}{m+L(x-x_{0})}dx\\
&=\ln\left(\frac{L(x_{0}-a)}{m}+1\right)+\ln\left(\frac{L(b-x_{0})}{m}+1\right)
\end{align*}
就好.

计算无穷级数
\[ \sum_{k=1}^{\infty}\frac{1}{k^2}\cos\left(\frac{9}{k\pi+\sqrt{k^2\pi^2-9}}\right)\]
西神说一般的我们有
\[ \sum_{k=1}^{\infty}\frac{\cos(k\pi-\sqrt{k^2\pi^2-a^2})}{k^2}=\frac{\pi^2}{12}\left(\cosh(a)+\frac{3}{a}\sinh(a)\right)\]
令$a=3$就有
\[ \sum_{k=1}^{\infty}\frac{1}{k^2}\cos\left(\frac{9}{k\pi+\sqrt{k^2\pi^2-9}}\right)=-\frac{\pi^2}{12e^{3}}\]
继续可以推广
\[ \sum_{n=0}^{\infty}\frac{n^{2}\pi^{2}+\phi^{2}}{(n^{2}\pi^{2}-\phi^{2})^{2}}(-1)^{n}\cos\sqrt{n^{2}\pi^{2}+a^{2}-\phi^{2}}=\frac{\cos\sqrt{a^2-\phi^{2}}}{2\phi^{2}}+\frac{a\cos{a}\cot\phi+\phi\sin{a}}{2a\sin\phi}\]

求
\[ \iiint_{D}\ln{x}\ln{y}\ln{z}\cos(x^2+y^2+z^2)dxdydz\]
其中
\[ D:[0,+\infty)\times[0,+\infty)\times[0,+\infty)\]

设$x_{1},x_{2},\cdots,x_{n}(n\geq 3)$是任意实数，且满足$x_{1}x_{2}\cdots x_{n}=1$,求证
\[ \sum_{k=1}^{n}\frac{x_{k}^{2}}{x_{k}^{2}-2x_{k}\cos\frac{2\pi}{n}+1}\geq 1 \]

设$x_{1},x_{2},\cdots,x_{n}(n\geq 3)$是正实数，且满足
\[ \sum_{1\leq i,j\leq n}|1-x_{i}x_{j}|=\sum_{1\leq i,j\leq n}|x_{i}-x_{j}| \]
求证：对任意的实数$a_{1},a_{2},\cdots,a_{n}$,存在实数$t$,使得
\[ \sum_{i=1}^{n}|\sin(t-a_{i})|\leq \cot\left(\frac{\pi}{2\sum\limits_{i=1}^{n}x_{i}}\right)\]

Prove that for any finite sequence $\{a_{k}\}_{k=1}^{n}$ of real numbers there is an index $m\in\{0,\cdots,n\}$ such that
\[ \left|\sum_{1\leq k\leq m}a_{k}-\sum_{m<k\leq n}a_{k}\right|\leq \max_{1\leq k\leq n}|a_{k}| \]

Proof:

设
\[ S_{m}=\sum_{1\leq k\leq m}a_{k}-\sum_{m<k\leq n}a_{k}\]
有
\[ S_{0}=-S_{n} \]
而数列$\{S_{m}\}$将会改变符号，于是可以选择一个$p$,使得$S_{p}$和$S_{p-1}$不同号，必然存在一个$p$使得
$|S_{p}|$或者$|S_{p-1}|$不超过$\displaystyle\max_{1\leq k\leq n}|a_{k}|$，要是不会这样，就是说对所有能让$S_{p}$和$S_{p-1}$ 不同号的那些$p$,都有
\[ |S_{p}|>\max_{1\leq k\leq n}|a_{k}|,|S_{p-1}|>\max_{1\leq k\leq n}|a_{k}| \]
那么
\[ 2\max_{1\leq k\leq n}|a_{k}|<|S_{p}|+|S_{p-1}|=|S_{p}-S_{p-1}|=2|a_{p}|\leq 2\max_{1\leq k\leq n}|a_{k}|\]
便得到了矛盾，所以结论成立。

求极限
\[ \lim_{n\to\infty}\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin{nx}}{\sin{x}}\right)^{4}dx \]
解:
首先注意到极限
\[ \lim_{x\to 0}x^2\left(\frac{1}{\sin^{4}{x}}-\frac{1}{x^{4}}\right)=\frac{2}{3} \]
则有
\[ \frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin{nx}}{\sin{x}}\right)^{4}dx=\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}(\sin{nx})^{4}\cdot\frac{1}{x}\cdot x^2\left(\frac{1}{\sin^{4}{x}}-\frac{1}{x^{4}}\right)dx+\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}\frac{(\sin{nx})^{4}}{x^{3}}dx=A_{1}+A_{2} \]
由于$F(x)=x^2\left(\frac{1}{\sin^{4}{x}}-\frac{1}{x^{4}}\right)$可以通过补充定义让它连续，则在$\left[0,\frac{\pi}{2}\right]$上$F(x)$有最大值$M$,而$(\sin{nx})^{4}\leq |\sin{nx}|\leq nx $,则
\[ A_{1}\leq \frac{M}{n}\to 0\qquad (n\to+\infty) \]
而
\begin{align*}
 \lim_{n\to\infty}\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}\frac{\sin^{4}nx}{x^{3}}dx&=\lim_{n\to+\infty}\int_{0}^{\frac{n\pi}{2}}\frac{\sin^{4}{x}}{x^{3}}dx\\
&=\int_{0}^{+\infty}\frac{\sin^{4}{x}}{x^{3}}dx\\
&=\int_{0}^{+\infty}\frac{2\sin^{3}{x}\cos{x}}{x^{2}}dx\\
&=\int_{0}^{+\infty}\frac{6\sin^{2}x\cos^2{x}-2\sin^{4}{x}}{x}dx\\
&=\int_{0}^{+\infty}\frac{\cos{2x}-\cos{4x}}{x}dx\\
&=\ln{2}
\end{align*}
因此
\[ \lim_{n\to\infty}\frac{1}{n^2}\int_{0}^{\frac{\pi}{2}}x\left(\frac{\sin{nx}}{\sin{x}}\right)^{4}dx=\ln{2} \]