设$x_{1},x_{2},\cdots,x_{n}$是非负实数，求证:
\[ \prod_{k=1}^{n}(1+x_{k})\leq 1+\sum_{k=1}^{n}\frac{1}{k!}\left(1-\frac{k}{2n}\right)^{k-1}\left(\sum_{k=1}^{n}x_{k} \right)^{k}\]
证明：为了方便，我们记
\[ S=\sum_{k=1}^{n}x_{k}\]
由AM-GM不等式，得到
\[ \prod_{k=1}^{n}(1+x_{k})\leq \left(1+\frac{S}{n}\right)^{n}\]
这时，只要证明
\[ \left(1+\frac{S}{n}\right)^{n}\leq 1+\sum_{k=1}^{n}\frac{1}{k!}\left(1-\frac{k}{2n}\right)^{k-1}\left(\sum_{k=1}^{n}x_{k} \right)^{k}\]
注意到
\[ \left(1+\frac{S}{n}\right)^{n}=1+C_{n}^{1}\cdot\frac{S}{n}+\cdots+C_{n}^{k}\left(\frac{S}{n}\right)^{k}+\cdots+C_{n}^{n}\left(\frac{S}{n}\right)^{n}\]
兹证明
\[ C_{n}^{k}\left(\frac{S}{n}\right)^{k}\leq \frac{1}{k!}\left(1-\frac{k}{2n}\right)^{k-1}\left(\sum_{k=1}^{n}x_{k} \right)^{k}\]
稍微化简下，就是
\[ [2n-2(k-1)][2n-2(k-2)]\cdots (2n-2)\leq (2n-k)^{k-1} \]
注意到首尾项的细节，就有AM-GM
\[ [2n-2(k-1)](2n-2)\leq \left(\frac{2n-2(k-1)+2n-2}{2}\right)^{2} \]
把它们配对，马上得到不等式。
问题：计算
\[ I=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sin^{2}(\tan{x})}dx \]
首先，我们用$t=\tan{x}$,得到
\[ I=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sin^{2}(\tan{x})}dx=\int_{0}^{+\infty}\frac{1}{(1+t^2)(1+\sin^2{x})}dx\]
而通过倍角公式
\[ 1+\sin^{2}{x}=1+\frac{1}{2}(1-\cos{2x})=\frac{1}{2}(3-\cos{2x}) \]
于是
\[ I=2\int_{0}^{+\infty}\frac{1}{(1+t^2)(3-\cos{2x})}dx\]
这时，回忆起一个恒等式
\[ \frac{1-a\cos{x}}{1-2a\cos{x}+a^2}=1+\sum_{k=0}^{\infty}a^{k}\cos{kx}\qquad (|a|<1) \]
要证明它并不难，只要用$z=a(\cos{x}+i\sin{x})$带入熟悉的展开式
\[ \frac{1}{1-z}=1+z+z^2+\cdots+z^{n}+\cdots \]
并考察实部就好。对于这个式子，又可以变形成
\[ \frac{1-a^2}{1-2a\cos{x}+a^2}=1+2\sum_{k=1}^{\infty}a^{k}\cos{kx} \]
这时，我们有
\begin{align*}
\int_{0}^{+\infty}\frac{dx}{(1+x^2)(1-2a\cos{2x}+a^2)}&=\frac{1}{1-a^2}\left(\int_{0}^{+\infty}\frac{1}{1+x^2}dx+2\sum_{k=1}^{\infty}\int_{0}^{+\infty}\frac{\cos{2kx}}{1+x^2}dx \right)\\
&=\frac{1}{1-a^2}\left(\frac{\pi}{2}+2\sum_{k=1}^{\infty}a^{k}\int_{0}^{+\infty}\frac{\cos{2kx}}{1+x^2}dx \right)
\end{align*}
显然，我们得计算积分
\[ \int_{0}^{+\infty}\frac{\cos{2kx}}{1+x^2}dx \]
下面设$\displaystyle f(a)=\int_{0}^{+\infty}\frac{\cos{ax}}{1+x^2}dx$,$f(a)$的Laplace变换就是
\begin{align*}
\mathcal{L}(f(a)) & = \int_{0}^{\infty}\int_{0}^{\infty}\frac{\cos(ax)}{1+x^2}e^{-as}\;{da}\;{dx} \\
&= \int_{0}^{\infty}\frac{s}{(1+x^2)(s^2+x^2)}\;{dx} \\
& = \frac{\pi}{2(s+1)}
\end{align*}
因此
\[ f(a) =\mathcal{L}^{-1}\left(\frac{\pi}{2(s+1)}\right) =\frac{\pi}{2}e^{-|a|}\]
我们有
\[ \int_{0}^{+\infty}\frac{\cos{2kx}}{1+x^2}dx=f(2k)=\frac{\pi}{2}e^{-2k} \]
于是
\begin{align*}
\int_{0}^{+\infty}\frac{dx}{(1+x^2)(1-2a\cos{2x}+a^2)}
&=\frac{1}{1-a^2}\left(\frac{\pi}{2}+2\sum_{k=1}^{\infty}a^{k}\int_{0}^{+\infty}\frac{\cos{2kx}}{1+x^2}dx\right)\\
&=\frac{\pi}{2(1-a^2)}+\frac{\pi}{1-a^2}\sum_{k=1}^{\infty}\left(\frac{a}{e^2}\right)^{k}\\
&=\frac{\pi}{2}\cdot\frac{1}{1-a^2}\cdot\frac{e^2+a}{e^2-a}
\end{align*}
这时，只要令$a=3-2\sqrt{2}$,就能方便的计算出
\[ I=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\sin^{2}(\tan{x})}dx=\int_{0}^{+\infty}\frac{1}{(1+t^2)(1+\sin^2{x})}dx=\frac{\sqrt{2}\pi}{4}\cdot\frac{e^2+3-2\sqrt{2}}{e^2-3+2\sqrt{2}}\]
证明
\[I= \lim_{n\to\infty}\left(\int_{0}^{\frac{\pi}{2}}\left|\frac{\sin(2n+1)x}{\sin{x}}\right|dx-\frac{2\ln{n}}{\pi}\right)=\frac{6\ln{2}}{\pi}+\frac{2\gamma}{\pi}+\frac{2}{\pi}\cdot\sum_{k=1}^{\infty}\frac{1}{2k+1}\ln\left(1+\frac{1}{k} \right)\]
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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:
\[ \int_{a}^{b}f(x)dx\leq \frac{e^{2L\beta}-1}{2L\alpha}\int_{c}^{d}f(x)dx \]
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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
\[ \sum_{i=1}^{n}|\sin(t-a_{i})|\leq \cot\left(\frac{\pi}{2\sum\limits_{i=1}^{n}x_{i}}\right)\]
