陈洪葛的博客

设$a,b,c>0$ 证明
\[ \frac{1}{(3a+2b+c)^2}+\frac{1}{(3b+2c+a)^2}+\frac{1}{(3c+2a+b)^2}\leq \frac{1}{4(ab+bc+ca)} \]
证明:
1、$ c\geq b\geq a$
由 AM-GM
\[ \frac{1}{(3a+2b+c)^2}=\frac{1}{[(a+2b)+(c+2a)]^2}\leq \frac{1}{4(a+2b)(c+2a)} \]
\[ \Leftrightarrow \frac{1}{(a+2b)(c+2a)}+\frac{1}{(b+2c)(a+2b)}+\frac{1}{(c+2a)(b+2c)}\leq \frac{1}{ab+bc+ca} \]
\[ \Leftrightarrow (a+2b)(b+2c)(c+2a)\geq 3(a+b+c)(ab+bc+ca) \]
\[ \Leftrightarrow ab^2+bc^2+ca^2\geq a^2b+b^2c+c^2a \]
\[ \Leftrightarrow (a-b)(b-c)(c-a)\geq 0 \]
上式成立，由于$c\geq b\geq a$
2、$a\geq b\geq c$
由AM-GM
\[  \frac{1}{[(a+b+c)+(2a+b)]^2}\leq \frac{1}{4(a+b+c)(2a+b)} \]
\[\Leftrightarrow \frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\leq \frac{a+b+c}{ab+bc+ca}\]
\[  \Leftrightarrow (2a+b)(2b+c)(2c+a)(a+b+c)\geq (ab+bc+ca)[2(a^2+b^2+c^2)+7(ab+bc+ca)]\]
由于$a\geq b\geq c$，所以
\[ (2a+b)(2b+c)(2c+a)\geq 3(a+b+c)(ab+bc+ca) \]
故只要证明
\[ 3(a+b+c)^2\geq 2(a^2+b^2+c^2)+7(ab+bc+ca) \]
显然成立。

证法2（西西）
\begin{align*}
 (a+2b+3c)^2&=[(a+c)+2(b+c)]^2\\
 &=(a+c)^2+4(b+c)^2+4(a+c)(b+c)\\
  &\geq 3(b+c)^2+6(a+c)(b+c)\\
  &=3(b+c)(2a+b+3c)
\end{align*}
等价证明
\[ \sum_{cyc}{\frac{1}{(b+c)(2a+b+3c)}}\leq \frac{3}{4(ab+bc+ca)}\]
就是
\[ \sum_{cyc}{\left(\frac{1}{2}-\frac{ab+bc+ca}{(b+c)(2a+b+3c)}\right)}\geq \frac{3}{4}\]
\[ \Leftrightarrow\sum_{cyc}{\frac{b+c}{2a+b+3c}}+2\sum_{cyc}{\frac{c^2}{(b+c)(2a+b+3c)}}\geq \frac{3}{2} \]
由Cauchy-Schwarz
\[ \sum_{cyc}{\frac{b+c}{2a+b+3c}}\geq \frac{4(a+b+c)^2}{\sum{(b+c)(2a+b+3c}}=1 \]
\[ 2\sum_{cyc}{\frac{c^2}{(b+c)(2a+b+3c)}}\geq \frac{2(a+b+c)^2}{\sum{(b+c)(2a+b+3c)}}=\frac{1}{2} \]
Done!

证法3（天书）
用替换
\[ \left\{
\begin{array}{ll}
x=3a+2b+c, &  \\
y=2b+2c+a, & \\
z=3c+2a+b, &
\end{array}\right.\]
得到\[ \left\{
\begin{array}{ll}
a=\dfrac{7x-5y+z}{18}, &  \\
b=\dfrac{x+7y-5z}{18}, & \\
c=\dfrac{y+7z-5x}{18}, &
\end{array}\right.\]
不等式等价于
\[ \Leftrightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\leq \frac{27}{14(xy+yz+xz)-11(x^2+y^2+z^2)} \]
\[ \Leftrightarrow 14\sum{xy}\cdot\sum{x^2y^2}\leq 27x^2y^2z^2+11\sum{x^2}\sum{x^2y^2} \]
\[ \Leftrightarrow 14\sum{x^3y^3}+14xyz\sum{x^2(y+z)}\leq 60x^2y^2z^2+11\sum{x^4(y^2+z^2)} \]
\[ \Leftrightarrow \sum{\left(7x^2y^2+2z^2(2x^2-3xy+2y^2)\right)(x-y)^2}\geq 0 \]
Done

问题：
设$a,b,c>0$，$a^2+b^2+c^2=3$证明
\[ \frac{(b+c)^2}{a^a+1}+\frac{(c+a)^2}{b^b+1}+\frac{(a+b)^2}{c^c+1}\leq 6 \]
证明:
注意到
\[ x^x\geq \frac{x^2+1}{2} \]
所以
\[ \sum{\frac{(b+c)^2}{a^a+1}}\leq \frac{2(b+c)^2}{a^2+3}=\sum{\frac{2(b+c)^2}{2a^2+b^2+c^2}}\]
再由Cauchy-Schwarz
\[ \sum{\frac{(b+c)^2}{2a^2+b^2+c^2}}\leq \sum{\left(\frac{b^2}{a^2+b^2}+\frac{c^2}{a^2+c^2}\right)}=3 \]
Done!

计算
\[ I=\int_{0}^{\pi}{\frac{x\cos{x}}{1+\sin^{2}{x}}dx} \]

(tian_275461)
解
\[ I=\int_{0}^{\pi}{xd\arctan{(\sin{x})}}=-\int_{0}^{\pi}{\arctan{(\sin{x})}dx}=-2\int_{0}^{\frac{\pi}{2}}{\arctan{(\sin{x})}dx}\]

注意到
\[ \arctan{\left(\frac{\sin{x}}{+\infty}\right)}-\arctan{\left(\frac{\sin{x}}{1}\right)}=-\int_{1}^{+\infty}{\frac{\sin{x}}{y^2+\sin^2{x}}dy}\]
故
\begin{align*}
I&=-2\int_{0}^{\frac{\pi}{2}}{\int_{1}^{+\infty}{\frac{\sin{x}}{y^2+\sin^2{x}}dy}dx}\\
&=-2\int_{1}^{+\infty}{\int_{0}^{\frac{\pi}{2}}{\frac{\sin{x}}{y^2+\sin^2{x}}dx}dy}\\
&=-2\int_{1}^{+\infty}{\int_{0}^{1}{\frac{1}{y^2+1-t^2}dt}dy}\qquad (t=\cos{x}) \\
&=-\int_{1}^{+\infty}{\frac{1}{\sqrt{y^2+1}}\ln{\left(\frac{\sqrt{y^2+1}+1}{\sqrt{y^2+1}-1}  \right)}dy}\\
&=-\int_{\text{arcsinh}1}^{+\infty}{\ln\left(\frac{\cosh{z}+1}{\cosh{z}-1}\right)dz}\qquad (y=\sinh{z})\\
&=2\int_{\text{arcsinh}1}^{+\infty}{\ln\left(\frac{1-e^{-z}}{1+e^{-z}}\right)dz}\\
&=2\int_{0}^{\sqrt{2}-1}{\frac{\ln{(1-t)}-\ln(1+t)}{t}dt}\qquad (t=e^{-z})\\
&=2\text{Li}_{2}(1-\sqrt{2})-2\text{Li}_{2}(\sqrt{2}-1)
\end{align*}
套用多重对数函数的性质(2)(3)(4)
(2)
\[ \text{Li}_{2}(1-x)+\text{Li}_{2}\left(1-\frac{1}{x}\right)=-\frac{1}{2}\ln^2{x} \]
(3)
\[ \text{Li}_{2}(x)+\text{Li}_{2}(1-x)=\frac{1}{6}\pi^2-\ln{x}\cdot\ln(1-x) \]
(4)
\[ \text{Li}_{2}(-x)-\text{Li}_{2}(1-x)+\frac{1}{2}\text{Li}_{2}(1-x^2)=-\frac{1}{12}\pi^2-\ln{x}\cdot\ln{(x+1)}\]
得到
\begin{align*}
   \text{Li}_{2}(2-\sqrt{2})+ \text{Li}_{2}(-\sqrt{2})&=-\frac{1}{2}\ln^2(\sqrt{2}-1) \qquad (\text{用$x=\sqrt{2}-1$套第2个})\\
    \text{Li}_{2}(2-\sqrt{2})+ \text{Li}_{2}(\sqrt{2}-1)&=\frac{\pi^2}{6}-\ln{(\sqrt{2}-1)}\cdot\ln{(2-\sqrt{2})}\qquad (\text{用$x=\sqrt{2}-1$套第3个})\\
     \text{Li}_{2}(-\sqrt{2})- \text{Li}_{2}(1-\sqrt{2})+\frac{1}{2} \text{Li}_{2}(1-(-\sqrt{2})^2)&=-\frac{\pi^2}{12}-\ln{\sqrt{2}}\cdot\ln{(1+\sqrt{2})}\qquad (\text{用$x=-\sqrt{2}$套第4 个})
\end{align*}
用第2个式子减去第1个式子加上第3个式子，得到
\[  \text{Li}_{2}(\sqrt{2}-1)- \text{Li}_{2}(1-\sqrt{2})=\frac{\pi^2}{8}-\frac{1}{2}\ln^2(\sqrt{2}+1)\]
故
\[ I=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4} \]

____________________________________________________________________________

以上是tian_275461的解法，在stackexchange上看到sos440表示

\[\int_{0}^{\pi} \frac{x \cos x}{1+\sin^2 x} \, dx = 4 \chi_{2}(1-\sqrt{2})\]

where

\[\chi_{2}(z) = \sum_{n=0}^{\infty} \frac{z^{2n+1}}{(2n+1)^2}\]

is the Legendre chi function of order 2. By exploiting some identities involving dilogarithm, we find that

\[\chi_{2}(1-\sqrt{2}) = \frac{1}{4} \log^2 (1+\sqrt{2}) - \frac{3}{8}\zeta(2).\]

This gives the answer

\[\int_{0}^{\pi} \frac{x \cos x}{1+\sin^2 x} \, dx = \log^2(1+\sqrt{2}) - \frac{\pi^2}{4}.\]

______________________________________________________________________________

reference：

http://math.stackexchange.com/questions/323603/integrate-displaystyle-int-0-pi-fracx-cosx1-sin2xdx

http://sos440.tistory.com/83

       多重对数函数（polylogarithm）也称：Jonquière's function 是数学中一种特殊的幂级数，定义为：
\[ \text{Li}_{s}(z)=\sum_{k=1}^{\infty}{\frac{z^{k}}{k^{s}}} \]
一般来说，多重对数函数不像对数函数那样是一个初等函数。上述定义中，自变量$|z| < 1$，$s$对所有复数值有效。通过解析延拓，可以将$z$的定义域扩展到更大的范围。$s=1$时的多重对数函数可以用自然对数表示$\text{Li}_{1}(z)=-\ln{(1-z)}$，$s = 2$和$3$的多重对数函数分别称为 Dilogarithm 及 Trilogarithm，其名称的由来是多重对数函数表示为以下的递推积分式
\[ \text{Li}_{s+1}(z)=\int_{0}^{z}{\frac{\text{Li}_{s}(t)}{t}dt} \]
其中 Dilogarithm 可以表示成
\[ \text{Li}_{2}(z)=\sum_{k=1}^{\infty}{\frac{z^{k}}{k^2}}=-\int_{0}^{z}{\frac{\ln(1-t)}{t}dt}\]

http://zh.wikipedia.org/wiki/%E5%A4%9A%E9%87%8D%E5%AF%B9%E6%95%B0%E5%87%BD%E6%95%B0

http://mathworld.wolfram.com/Dilogarithm.html

http://mathworld.wolfram.com/Polylogarithm.html

有以下4条常用性质值得注意
(1)
\[ \text{Li}_{2}(x)+\text{Li}_{2}(-x)=\frac{1}{2}\text{Li}_{2}(x^2) \]
(2)
\[ \text{Li}_{2}(1-x)+\text{Li}_{2}\left(1-\frac{1}{x}\right)=-\frac{1}{2}\ln^2{x} \]
(3)
\[ \text{Li}_{2}(x)+\text{Li}_{2}(1-x)=\frac{1}{6}\pi^2-\ln{x}\cdot\ln(1-x) \]
(4)
\[ \text{Li}_{2}(-x)-\text{Li}_{2}(1-x)+\frac{1}{2}\text{Li}_{2}(1-x^2)=-\frac{1}{12}\pi^2-\ln{x}\cdot\ln{(x+1)}\]
证明
(1)
\begin{align*}
   \text{Li}_{2}(x)+\text{Li}_{2}(-x)&=\sum_{n=1}^{\infty}{\frac{x^n+(-x)^n}{n^2}}\\
   &=\sum_{k=1}^{\infty}{2\cdot\frac{x^{2k}}{(2k)^2}}\\
   &=\sum_{k=1}^{\infty}{\frac{x^{2k}}{2k^2}}=\frac{1}{2}\text{Li}_{2}(x^2)
\end{align*}
(2)
记
\[ \int_{0}^{1-x}{\frac{-\ln{(1-t)}}{t}dt}+\int_{0}^{1-\frac{1}{x}}{\frac{-\ln{(1-t)}}{t}dt}=I_{1}+I_{2} \]
对$I_{2}$作替换$\displaystyle t=1-\frac{1}{1-y}\Rightarrow dt=-\frac{1}{(1-y)^2}dy$
\[ I_{2}=\int_{0}^{1-x}{\frac{\ln(1-y)}{y(1-y)}dy}=\int_{0}^{1-x}{\frac{\ln(1-y)}{y}dy}+\int_{0}^{1-x}{\frac{\ln(1-y)}{1-y}dy} \]
所以
\[ I_{1}+I_{2}= \int_{0}^{1-x}{\frac{\ln(1-y)}{1-y}dy}=-\frac{1}{2}\ln^{2}(1-y)\bigg|_{0}^{1-x}=-\frac{1}{2}\ln^2{x} \]
(3)
记
\[ \int_{0}^{x}{\frac{-\ln(1-t)}{t}dt}+\int_{0}^{1-x}{\frac{-\ln(1-t)}{t}dt}=I_{1}+I_{2} \]
对$I_{1}$作替换$y=1-t $
\begin{align*}
I_{1}+I_{2}&=\int_{1-x}^{1}{\frac{-\ln{y}}{1-y}dy}+\int_{0}^{1-x}{\frac{-\ln(1-t)}{t}dt}\\
&=-\ln{x}\cdot\ln{(1-x)}+\int_{0}^{1}{\frac{-\ln(1-t)}{t}dt}\\
&=-\ln{x}\cdot\ln{(1-x)}+\sum_{n=1}^{\infty}{\frac{1}{n^2}}\\
&=\frac{\pi^2}{6}-\ln{x}\cdot\ln(1-x)
\end{align*}
(4)
记\[ A=\text{Li}_{2}(-x)-\text{Li}_{2}(1-x)+\frac{1}{2}\text{Li}_{2}(1-x^2)\]
\begin{align*}
A+\text{Li}_{2}(x)+\text{Li}_{2}(1-x)&=\frac{1}{2}\left(\text{Li}_{2}(x^2)+\text{Li}_{2}(1-x^2)\right)\\ 
&=\frac{1}{2}\cdot\left(\frac{1}{6}\pi^2-\ln{x^2}\cdot\ln(1-x^2)\right)\\
&=\frac{1}{12}\pi^2-\ln{x}\cdot\ln(1-x^2)
\end{align*}
马上得到
\[ A=-\frac{1}{12}\pi^2-\ln{x}\cdot\ln(1+x) \]

——————————————————————————————————————————————————————

相关的特殊值\begin{align*}
\text{Li}_{2}(0)&=0\\
\text{Li}_{2}(1)&=\frac{1}{6}\pi^2\\
\text{Li}_{2}(-1)&=-\frac{1}{12}\pi^2\\
\text{Li}_{2}\left(\frac{1}{2}\right)&=\frac{1}{12}\pi^2-\frac{1}{2}\ln^2{2}\\
\text{Li}_{2}(-\phi)&=-\frac{1}{10}\pi^2-\ln^2{\phi}\\
\text{Li}_{2}\left(-\frac{1}{\phi}\right)&=-\frac{1}{15}\pi^2+\frac{1}{2}\ln^2{\phi}\\
\text{Li}_{2}\left(-\frac{1}{\phi^{2}}\right)&=\frac{1}{15}\pi^2-\ln^2{\phi}\\
\text{Li}_{2}\left(\frac{1}{\phi}\right)&=\frac{1}{10}\pi^2-\ln^2{\phi}\\
\end{align*}
这里$\displaystyle \phi=\frac{\sqrt{5}+1}{2}$是Golden ratio。

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附上搜集到的一本好书《Polylogarithms and Associated Functions》 by Lewin

http://pan.baidu.com/share/link?shareid=2376620720&uk=3456571357
 

证明
\[\lim_{n\rightarrow\infty}{\frac{\ln{n}}{n}\left(\frac{\sum_{k=1}^{n=1}{\csc{\left(\frac{k\pi}{n}\right)}}}{\ln{n}}-\frac{2}{\pi}n\right)}=\frac{2\gamma}{\pi}-\frac{2\ln{\pi}-\ln{4}}{\pi}\]

(tian_275461)

证明
记
\begin{align*}
 I&= \frac{\ln{n}}{n}\left(\frac{\sum_{k=1}^{n-1}{\csc{\left(\frac{k\pi}{n}\right)}}}{\ln{n}}-\frac{2}{\pi}n\right)\\
 &=\frac{\pi}{n}\sum_{k=1}^{n-1}\csc{\left(\frac{k\pi}{n}\right)}-2\ln{n}
\end{align*}
只要证
\[ \lim_{n\rightarrow\infty}{I}=2\gamma-2\ln{\pi}+\ln{4} \]
也就是
\[ \lim_{n\rightarrow\infty}{(2\gamma-I)}=2\ln{\pi}-\ln{4} \]
记$ S=2\gamma-I$,我们有
\[ \gamma=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}-\ln{n}+c_{n} \quad \text{其中} c_{n}\rightarrow 0\quad  (n\rightarrow\infty) \]
\begin{align*}
 S&=2\sum_{k=1}^{n-1}-\frac{\pi}{n}\sum_{k=1}^{n-1}{\csc{\left(\frac{k\pi}{n}\right)}}+2c_{n}\\
 &=\sum_{k=1}^{n-1}{\left(\frac{1}{k}+\frac{1}{n-k}\right)}-\frac{\pi}{n}\sum_{k=1}^{n-1}{\csc{\left(\frac{k\pi}{n}\right)}}+2c_{n}\\
 &=\frac{\pi}{n}\sum_{k=1}^{n-1}{\left(\dfrac{1}{\frac{k\pi}{n}}+\dfrac{1}{\pi-\frac{k\pi}{n}}\right)}-\frac{\pi}{n}\sum_{k=1}^{n-1}{\csc{\left(\frac{k\pi}{n}\right)}}+2c_{n}
\end{align*}
所以有
\[  \lim_{n\rightarrow\infty}{S}=\int_{0}^{\pi}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx} \]
故只要证
\[ \int_{0}^{\pi}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}=2\ln{\pi}-\ln{4}\]
而
\[ \int_{0}^{\pi}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}=\int_{0}^{\frac{\pi}{2}}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}+\int_{\frac{\pi}{2}}^{\pi}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}\]
 第二部分用替换$ y=\pi-x $
\[ \Rightarrow \lim_{n\rightarrow\infty}{S}=2\int_{0}^{\frac{\pi}{2}}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}\]
注意到
\[ \frac{1}{\sin{x}}=\frac{1}{x}+\sum_{n=1}^{\infty}{(-1)^{n}\left(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\right)} \]
\begin{align*}
 \int_{0}^{\frac{\pi}{2}}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}&=\sum_{n=1}^{\infty}{(-1)^{n+1}\int_{0}^{\frac{\pi}{2}}{\left(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}   \right)dx}}\\
 &=\sum_{n=1}^{\infty}{(-1)^{n+1}\cdot\ln{\left(1-\frac{1}{4n^2}\right)}}
\end{align*}
对$n$分奇偶性讨论
(1) $n=2m-1\ (m=1,2,\cdots) $ 时
\[ (-1)^{n+1}\cdot\ln{\left(1-\frac{1}{4n^2}\right)}=\ln{\left(\frac{(4m-1)(4m-3)}{(4m-2)^2}\right)}\]
(2) $n=2m\ (m=1,2,\cdots) $ 时
\[ (-1)^{n+1}\cdot\ln{\left(1-\frac{1}{4n^2}\right)}=\ln{\left(\frac{(4m)^2}{(4m+1)(4m-1)}\right)}\]
而
\[ \sum_{m=1}^{k}{(-1)^{n+1}\cdot\ln{\left(1-\frac{1}{4n^2}\right)}}=\ln{\left[\frac{1}{4k+1}\left(\frac{(2k)!!}{(2k-1)!!}\right)^{2}\right]}\rightarrow \ln{\frac{\pi}{4}} \quad \text{(用Wallis公式)} \]
马上得到
\[ \int_{0}^{\pi}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}=2\ln{\pi}-\ln{4}\]

$\square$