陈洪葛的博客

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
\[ \int_{0}^{\pi}{\left(\frac{1}{x}+\frac{1}{\pi-x}-\frac{1}{\sin{x}}\right)dx}=2\ln{\pi}-\ln{4}\]

$\square$
 

计算

\[\int_0^1 \int_0^1 \frac{\ln(2-xy)}{1-xy}dx \ dy \]

(tian_275461)

解

\begin{align*}
\int_0^1 \int_0^1 \frac{\ln(2-xy)}{1-xy}dx \ dy &= \int_0^1 \int_0^1 \frac{\ln \left(1+(1-xy) \right)}{1-xy}dx \ dy
\\  &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}\int_0^1 \int_0^1 (1-xy)^{n-1}dx \ dy \\ &=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}\int_0^1 \left(\frac{1-(1-y)^n}{y} \right)dy \\  &=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2}H_n\\
&=\frac{(-1)^{n}}{n}\int_{0}^{1}{(1-x)^{n-1}\ln{x}dx}\\
&=-\int_{0}^{1}{\sum_{n=1}^{\infty}{\frac{(x-1)^{n-1}}{n}\ln{x}}dx}\\
&=\int_{0}^{1}{\frac{\Li_{1}(x-1)\ln{x}}{1-x}dx}\\
&=-\int_{0}^{1}{\frac{\ln{(1+t)}\ln{(1-t)}}{t}dt}
\end{align*}
为了计算上面的式子，我们先算
\[ \int_{0}^{1}{\frac{\ln^{2}{(1+x)}}{x}dx} \]
为此考虑一般的\begin{align*}
I(t) &= \int_0^t \frac{\ln^2(1+x)}{x}dx \\
&= \ln t\ln^2(1+t)-2\int_0^t \frac{\ln(x)\ln(1+x)}{1+x}dx \\
&= \ln^2{t}\ln(1+t)-2\int_1^{1+t}\frac{\ln(y) }{y} \left( \ln y+\ln \left( 1-\frac{1}{y}\right)\right)dy \quad (y=x+1)\\
&= \ln t\ln^2(1+t)-\frac{2}{3}\ln^3(1+t)-2\int_{1}^{1+t}{\frac{\ln{y}}{y}\ln{\left(1-\frac{1}{y}\right)}dy} \quad (z=\frac{1}{y})\\
&=\ln t\ln^2(1+t)-\frac{2}{3}\ln^3(1+t)+2\int_{\frac{1}{1+x}}^{1}{\frac{\ln{z}\ln{(1-z)}}{z}dz}\\
&=\ln{t}\ln^2(1+t)-\frac{2}{3}\ln^3(1+t)-2\int_{\frac{1}{1+x}}^{1}{\ln{z}d \Li_{2}(z)}\\
&=\ln{t}\ln^2(1+t)-\frac{2}{3}\ln^3(1+t)-2\left(\ln{z}\Li_{2}(z)\bigg|_{\frac{1}{1+x}}^{1}-\int_{\frac{1}{1+x}}^{1}{\frac{\Li_{2}(z)}{z}dz}\right)\\
&=\ln{t}\ln^2(1+t)-\frac{2}{3}\ln^3(1+t)-2\ln{(1+t)}\Li_{2}\left(\frac{1}{1+t}\right)-2\Li_{3}\left( \frac{1}{1+t} \right)+2\Li_{3}(1)
 \end{align*}
故
\[ I(1)=-\frac{2}{3}\ln^3{2}-2\ln{2}\Li_{2}\left(\frac{1}{2}\right)-2\Li_{3}\left(\frac{1}{2}\right)+2\Li_{3}(1)\]
由
\[ \Li_{2}\left(\frac{1}{2}\right)=\frac{1}{12}(\pi^{2}-6\ln^2{2}) \]
\[ \Li_{3}\left(\frac{1}{2}\right)=\frac{1}{24}[4\ln^3{2}-2\pi^{2}\ln{2}+21\zeta(3)] \]
\[ I(1)=\frac{\zeta(3)}{4} \]
对已知结论
\[ \int_{0}^{1}{\frac{\ln^2{(1-x)}}{x}dx}=2\zeta(3) \]
作$x=t^{2}$
\[ \Rightarrow \int_{0}^{1}{\frac{\ln^2{(1-t^2)}}{t}dx}=\zeta(3) \]
\[ \Leftrightarrow \int_{0}^{1}{\frac{\ln^2{(1-t)}}{t}dx}+\int_{0}^{1}{\frac{\ln^2{(1+t)}}{t}dx}+2\int_{0}^{1}{\frac{\ln{(1-t)}\ln{(1+t)}}{t}dx}=\zeta(3) \]
\[ \Rightarrow \int_{0}^{1}{\frac{\ln{(1-t)}\ln{(1+t)}}{t}dx}=-\frac{5}{8}\zeta(3) \]
故有
\[ \int_0^1 \int_0^1 \frac{\ln(2-xy)}{1-xy}dx \ dy=\frac{5}{8}\zeta(3) \]

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根据上面的办法，我们可以得到

\[ \int_{0}^{1}{\frac{\ln^{2}(1-t)}{t}dt}=2\zeta(3) \]

\[ \int_{0}^{1}{\frac{\ln^{2}(1+t)}{t}dt}=\frac{1}{4}\zeta(3) \]

\[ \sum_{n=1}^{\infty}{\frac{H_{n}}{n^2}}=2\zeta(3) \]

\[ \int_{0}^{1}\int_{0}^{1}\frac{\ln(1-xy)}{1-xy}dxdy =-\zeta(3) \]

\[ \sum_{n=1}^{\infty}{\frac{H_{n}}{n^3}}=\frac{\pi^4}{72} \]
 

记得这个问题曾经在大约1年前和网友讨论过，那时候真心不会。前几天在AMM上找到了一样的。

设$f(x)$是$[0,1]\rightarrow R$上的连续函数，且记$\displaystyle F(x)=\int_{0}^{x}{f(t)dt}$,并有
\[ \int_{0}^{1}{x^2f(x)dx}=-2\int_{\frac{1}{2}}^{1}{F(t)dt} \]
求证
\[\int_{0}^{1}{f^2(x)dx}\geq 80\left( \int_{0}^{1}{f(x)dx}\right)^2\]
(tian_275461)

 首先由条件，我们看到
 \[  \int_{0}^{1}{x^2F'(x)dx}=x^2F(x)\bigg|_{0}^{1}-2\int_{0}^{1}{xF(x)dx}=F(1)-2\int_{0}^{1}{xF(x)dx}\]
 \[ \Rightarrow\qquad F(1)=2\int_{0}^{1}{xF(x)dx}-2\int_{\frac{1}{2}}^{1}{F(t)dt}=2\left(\int_{0}^{\frac{1}{2}}{xF(x)dx}+\int_{\frac{1}{2}}^{1}{(x-1)F(x)dx}  \right)\]
 而要证的不等式等价于
 \[ \int_{0}^{1}{(F'(x))^{2}dx}\geq 24(F(1))^2=80\left[2\left(\int_{0}^{\frac{1}{2}}{xF(x)dx}+\int_{\frac{1}{2}}^{1}{(x-1)F(x)dx}  \right)\right]^{2} \]
 注意到由Cauchy-Schwarz不等式，我们有
 \begin{align*}
 \int_{0}^{\frac{1}{2}}{x^4} \int_{0}^{\frac{1}{2}}{(F'(x))^{2}dx}&\geq \left(\int_{0}^{\frac{1}{2}}{x^2F'(x)dx}\right)^{2}\\
 &=\left[\frac{1}{4}F\left(\frac{1}{2}\right)-2\int_{0}^{\frac{1}{2}}{xF(x)dx}\right]^{2}\\
 &=A^2
 \end{align*}
 \begin{align*}
 \int_{\frac{1}{2}}^{1}{(x-1)^4}\int_{\frac{1}{2}}^{1}{(F'(x))^{2}dx}&\geq  \left(\int_{\frac{1}{2}}^{1}{(x-1)^2F'(x)dx}\right)^{2}\\
 &= \left[-\frac{1}{4}F\left(\frac{1}{2}\right)-2\int_{\frac{1}{2}}^{1}{(x-1)F(x)dx}\right]^{2}\\
 &=B^2
 \end{align*}
 而由Cauchy-Schwarz不等式，我们有
 \[A^2+B^2\geq \frac{1}{2} (A+B)^2=\frac{1}{2} F^{2}(1) \]
 而
 \[ \int_{0}^{\frac{1}{2}}{x^4dx}=\int_{\frac{1}{2}}^{1}{(1-x)^4dx}=\frac{1}{160}\]
 故
 \[ \int_{0}^{1}{f^2(x)dx}\geq 80\left( \int_{0}^{1}{f(x)dx}\right)^2\]
 

对$\forall m\neq 0$,若一个连续函数$f:\mathbb{R}\rightarrow\mathbb{R}$满足函数方程
\[ f\left(2x-\frac{f(x)}{m}\right)=mx\]
则有$f(x)=m(x-c)$.
证明:
我们设$g(x)=2x-\frac{f(x)}{m}$,显然$g(x)$是连续函数且有
\[ g(g(x))=2g(x)-x \qquad \forall x\in\mathbb{R}\]
若$g(x_1)=g(x_2)$则有$g(g(x_{1}))=g(g(x_{2}))$,我们得到
\[ x_{1}=x_{2}\]
故$g(x)$是一个单射，而我们知道，若$g(x)$是一个连续的单射，则$g(x)$严格单调。（关于这一点可以用反证法证明）,因此，$g(x)$有2种情况,严格递增或者严格递减。下证明$g(x)$只能严格递增。
(反证)设$g(x)$严格递减，则对于$x_{1}<x_{2}$,我们有$g(x_{1})>g(x_{2})$,接着又有$g(g(x_{1}))<g(g(x_{2}))$.而这等价于
\[ 2g(x_{1})-x_{1}<2g(x_{2})-x_{2}\]
\[ \Leftrightarrow 2[g(x_{1})-g(x_{2})]<x_{1}-x_{2}\]
上面不可能成立，因为左边大于0而右边小于0。故$g(x)$只能严格递增。
改写$g(g(x))=2g(x)-x$为
\[ g(g(x))-g(x)=g(x)-x \]
递推后得到
\[ g^{n}(x)=ng(x)-(n-1)x  \qquad (n\geq 1) \]
这里$g^(n)(x)$表示$n$次符合。
则有
\[ g^{n}(x)-g^{n}(0)=n[g(x)-x-g(0)]+x \]
\[ \Leftrightarrow \frac{ g^{n}(x)-g^{n}(0)}{n}=g(x)-x-g(0)+\frac{x}{n}\]
而$g(x)$严格递增，$g^{n}(x)$也严格递增，故对上式令$n\rightarrow\infty$,由$g(x)$的单调性，我们得到

$$g(x)\leq x+g(0),\qquad x<0$$
$$g(x)\geq x+g(0),\qquad x>0$$

这样，我们得到$g(x)$的值域也是$\mathbb{R}$，故$g(x)$是一个一一映射。且$g^{-1}$存在。现在，用$x=g^{-1}(g^{-1}(y))$带入原来的方程，则有
\[ g^{-1}(g^{-1}(y))=2g^{-1}(y)-y\]
$g^{-1}(y)$同样满足这个方程，则用相同的手段，我们得到

$$g^{-1}(y)\leq y+g(0),\qquad y<0$$
$$g^{-1}(y)\geq y+g(0),\qquad y>0$$

现在，用$x=g^{-1}(y)$带入
\[ g(g(x))-g(x)=g(x)-x \]
得到
\[ g(y)-y=y-g^{-1}(y)\]
令$y=0$得到$g^{-1}(0)=-g(0)$
假设$g(0)\geq 0$,则对$x>0$有$g(x)\geq x+g(0)>0$,则对$y=g(x)>0$有$x>g(x)+g^{-1}(0)=g(x)-g(0)$.故得到
\[ g(x)=x+g(0) \qquad (x>0) \]
同理可得
\[ g(x)=x+g(0) \qquad (x<0) \]
这样我们得到$f(x)=m(x-g(0)）$对$x\in \mathbb{R}$成立。