陈洪葛的博客

Prove that

\begin{equation}
\int_{-\infty}^{\infty} \frac{\ln(1+e^{ax})}{1+e^{bx}}\, dx = \frac{\pi^{2}}{12} \frac{a^{2} + b^{2}}{a b^{2}}.
\end{equation}

(sos440)

Proof:
Let $I$ denote the integral  . With the substitution $t=e^{bx}$, we have
\begin{align*}
I&= \int_{-\infty}^{\infty} \frac{\ln(1+e^{ax})}{1+e^{bx}}\, dx
 = \frac{1}{b} \int_{0}^{\infty} \frac{\ln(1+t^{\frac{a}{b}})}{t(1+t)}\, dt.
\end{align*}
Then with the substitution $t\mapsto\frac{1}{t}$, it follows that
\begin{align*}
I&= \frac{1}{b} \int_{0}^{\infty} \frac{\ln(1+t^{\frac{a}{b}}) - \frac{a}{b} \ln t}{1+t}\, dt.
\end{align*}
Summing two identities above, we have

\begin{align*}
2I&= \frac{1}{b} \int_{0}^{\infty} \left( \frac{\ln(1 + t^{\frac{a}{b}})}{t} - \frac{a}{b} \frac{\ln t}{1 + t} \right) \, dt \\
&= \lim_{R\to\infty} \frac{1}{b} \int_{0}^{R} \left( \frac{\ln(1 + t^{\frac{a}{b}})}{t} - \frac{a}{b} \frac{\ln t}{1 + t} \right) \, dt.
\end{align*}
Now we focus on the integral inside the limit. Simplifying, we have
\begin{align*}
& \frac{1}{b} \int_{0}^{R} \left( \frac{\ln(1 + t^{\frac{a}{b}})}{t} - \frac{a}{b} \frac{\ln t}{1 + t} \right) \, dt \\
&= \frac{1}{a} \int_{0}^{R^{\frac{a}{b}}} \frac{\ln(1 + t)}{t} \, dt - \frac{a}{b^{2}} \int_{0}^{R} \frac{\ln t}{1 + t} \, dt \\
&= \frac{1}{a} \int_{0}^{R^{\frac{a}{b}}} \frac{\ln(1 + t)}{t} \, dt - \frac{a}{b^{2}} \ln R \ln (R+1) + \frac{a}{b^{2}} \int_{0}^{R} \frac{\ln (1+t)}{t} \, dt \\
&= -\frac{1}{a} \mathrm{Li}_{2}(-R^{\frac{a}{b}}) - \frac{a}{b^{2}} \mathrm{Li}_{2} (-R) - \frac{a}{b^{2}} \ln R \ln (R+1).
\end{align*}
But we know that the dilogarithm satisfies
\begin{align*}
-\mathrm{Li}_{2}(-x)
&= \mathrm{Li}_{2}\left( \frac{x}{1+x} \right) + \frac{1}{2}\ln^{2}(x + 1)
= \zeta(2) + \frac{1}{2}\ln^2 x + o(1)
\end{align*}
as $x\to\infty$ . Plugging this identity above, it follows that
\begin{align*}
\frac{1}{b} \int_{0}^{R} \left( \frac{\ln(1 + t^{\frac{a}{b}})}{t} - \frac{a}{b} \frac{\ln t}{1 + t} \right) \, dt
&= \frac{a^2 + b^2}{a b^2} \zeta(2) + o(1)
\end{align*}
as $R\to\infty$, and therefore the proof is complete.
 

今天介绍一个用多项式一致逼近函数的定理。

定理：令$f$为定义于区间$[0,1]$上的连续函数，对任意$\varepsilon>0$存在多项式函数$p$使得对所有$x\in[0,1]$不等式
\[ |f(x)-p(x)|<\varepsilon \]
成立。
本定理证明依赖于以下几件事：
(a)定义于闭区间上的连续函数是一致连续的：对任意$\varepsilon>0$，存在$\delta>0$,使得对任意的$x,y$满足$|x-y|<\delta$,就有$|f(x)-f(y)|<\varepsilon $
(b)定义于闭区间上的连续函数是有界的：存在$M>0$使得$|f(x)|<M$。
(c)恒等式
\[ \sum_{k=0}^{n}\left(\frac{k}{n}-x\right)^{2}C_{n}^{k}x^k(1-x)^{n-k}=\frac{x(1-x)}{n} \]

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为了证明(c),令$f(x)=(x+t)^n$,由牛顿二项式定理得
\[ f(x)=\sum_{k=0}^{n}C_{n}^{k}x^{k}t^{n-k} \]
两边对$x$求导，得到
\[ n(x+t)^{n-1}=f'(x)=\sum_{k=0}^{n}C_{n}^{k}kx^{k-1}t^{n-k} \]
故
\[ x(x+t)^{n-1}=\sum_{k=0}^{n}\frac{k}{n}C_{n}^{k}x^{k}t^{n-k} \]
令$g(x)=x(x+t)^{n-1}$.则
\[ g'(x)=(x+t)^{n-1}+(n-1)x(x+t)^{n-2}=\sum_{k=0}^{n}\frac{k}{n}\cdot C_{n}^{k}kx^{k-1}t^{n-k} \]
\[ \Rightarrow\qquad \frac{x}{n}(x+t)^{n-1}+\frac{x^2}{n}(n-1)(x+t)^{n-2}=\sum_{k=0}^{n}\frac{k^2}{n^2}C_{n}^{k}x^{k}t^{n-k} \]
将$t=1-x$代入得到
\begin{equation}
 1=\sum_{k=0}^{n}C_{n}^{k}x^{k}(1-x)^{n-k}
\end{equation}
\begin{equation}
 x=\sum_{k=0}^{n}\frac{k}{n}C_{n}^{k}x^{k}(1-x)^{n-k}
\end{equation}
\begin{equation}
\frac{x}{n}+\frac{x^2}{n}(n-1)=\sum_{k=0}^{n}\frac{k^2}{n^2}C_{n}^{k}x^{k}(1-x)^{n-k} 
\end{equation}
于是
\begin{align*}
  &\sum_{k=0}^{n}\left(\frac{k}{n}-x\right)^{2}C_{n}^{k}x^k(1-x)^{n-k}\\
  &=\sum_{k=0}^{n}\frac{k^2}{n^2}C_{n}^{k}x^{k}(1-x)^{n-k}-2x\sum_{n=0}^{n}\frac{k}{n}C_{n}^{k}x^{k}(1-x)^{n-k}+x^2\sum_{k=0}^{n}C_{n}^{k}x^{k}(1-x)^{n-k}\\
  &=\frac{x}{n}+\frac{x^2}{n}(n-1)-2x^2+x^2\\
  &=\frac{x(1-x)}{n}
\end{align*}
\[\Leftrightarrow\qquad \sum_{k=0}^{n}\left(k-nx\right)^{2}C_{n}^{k}x^k(1-x)^{n-k}=nx(1-x) \]

——————————————————————————————————————————————————————————————
定理证明
令$\varepsilon>0$,存在$\delta>0$,使得当$x,y\in[0,1]$,$|x-y|<\delta$,不等式$|f(x)-f(y)|<\frac{\varepsilon}{2}$成立。同时可以选取$M>0$,使得$|f(x)|<M$.考虑$f(x)$的Bernstein 多项式
\[ \boxed{B_{n}(x)=\sum_{k=0}^{n}f\left(\frac{k}{n}\right)C_{n}^{k}x^{k}(1-x)^{n-k}\qquad (n=1,2,\cdots)} \]
我们将证明：若$n>\max\bigg\{\frac{1}{\delta^{4}},\left(\frac{4M}{\varepsilon}\right)^{2}\bigg\}$,则所有$x\in[0,1]$都满足$|f(x)-B_{n}(x)|<\varepsilon$.证明如下
\[ f(x)-B_{n}(x)=\sum_{k=0}^{n}\left(f(x)-f\left(\frac{k}{n}\right)\right)C_{n}^{k}x^{k}(1-x)^{n-k}=\sum_{(a)}+\sum_{(b)} \]
其中$\sum_{(a)}$表示对满足$\left|\frac{k}{n}-x\right|<\frac{1}{n^{\frac{1}{4}}}$求和，而$\sum_{(b)}$表示对满足$\left|\frac{k}{n}-x\right|\geq \frac{1}{n^{\frac{1}{4}}}$求和。
若$\left|\frac{k}{n}-x\right|\geq \frac{1}{n^{\frac{1}{4}}}$,则$|k-nx|\geq n^{\frac{3}{4}}$,故$(k-nx)^{2}\geq n^{\frac{3}{2}}$.由此
\begin{align*}
\left|\sum_{(b)}\right|&\leq \sum_{(b)}\left[|f(x)|+\left|f\left(\frac{k}{n}\right)\right|\right]C_{n}^{k}x^{k}(1-x)^{n-k}\\
&\leq 2M\sum_{(b)}C_{n}^{k}x^{k}(1-x)^{n-k}\\
&\leq \frac{2M}{n^{\frac{3}{2}}}\sum_{(b)}C_{n}^{k}(k-nx)^2x^{k}(1-x)^{n-k}\\
&\leq \frac{2M}{n^{\frac{3}{2}}}nx(1-x)\\
&\leq \frac{2M}{n^{\frac{1}{2}}}<\frac{\varepsilon}{2}
 \end{align*}
 若$\left|\frac{k}{n}-x\right|<\frac{1}{n^{\frac{1}{4}}}$,则$\left|\frac{k}{n}-x\right|<\delta$,故$\left|f\left(\frac{k}{n}\right)-f(x)\right|<\frac{\varepsilon}{2}$,由此
 \[ \left|\sum_{a}\right|\leq\sum_{a}\left|f(x)-f\left(\frac{k}{n}\right)\right|C_{n}^{k}x^{k}(1-x)^{n-k}<\frac{\varepsilon}{2} \]
 因此，马上有
 \[ |f(x)-B_{n}(x)|<\varepsilon \]
 $B_{i}^{n}(t)$称为Bernstein基底
 \[ B_{i}^{n}=C_{n}^{i}t^{i}(1-t)^{n-i} \]

问题1
设函数$f$在$[0,+\infty)$上一致连续，且$\forall x\in[0,1]$有
\[ \lim_{n\to+\infty}f(x+n)=0 \qquad (n\in \mathbf{N}) \]
证明:
\[ \lim_{x\to+\infty}f(x)=0 \]
证明
由于$f$在$[0,+\infty)$上一致连续，故对$\forall \varepsilon>0$,$\exists \delta>0$,当$x,y$ 满足$ |x-y|<\delta$时，有
\[ |f(x)-f(y)|<\varepsilon \]
这样，我们可以把区间$[0,1]$分成$m$等份，记座$\Delta_{1},\Delta_{2},\cdots,\Delta_{n} $分点为$x_{0}=0<x_{1}=\frac{1}{m}<x_{2}=\frac{2}{m}<\cdots<1=\frac{m}{m}=x_{m} $,其中$m$满足$\frac{1}{m}<\delta$.
那么，对于这$m+1$个分点$x_{i},(i=0,1,\cdots,m)$,由于
\[ \lim_{n\to+\infty}f(x+n)=0\]
则有
\[  \lim_{n\to+\infty}f(x_{i}+n)=0 \qquad (i=0,1,\cdots,m)\]
换一种说法就是对$\forall \varepsilon>0$,及每一个$x_{i} (i=0,1,\cdots,m)$,存在$ N_{i}\in \mathbf{N},(i=0,1,\cdots,m)$, 当$n> N_{i}$时有
\[ |f(x_{i}+n)|<\varepsilon \]
这样，我们可以取$N=\max\{N_{0},N_{1},\cdots N_{m}\}$,于是当$n>N$时，有
\[ |f(x_{i}+n)|<\varepsilon\qquad (i=0,1,\cdots,m) \]
这样，对$ \forall x>N+1$,$x=[x]+t_{0}$,其中$t_{0}\in [0,1) $,这样既有$[x]>N$,又有$t_{0}$落入某个$\Delta_{i}$，不妨设$t_{0}$落入第$k$个区间,显然有$|t_{0}-x_{k}|<\frac{1}{m}<\delta$,这样根据一致连续性有
\[ |f(x_{k}+[x])-f(x)|=|f(x_{k}+[x])-f([x]+t_{0})|<\varepsilon \]
而由于$[x]>N$,则又有
\[ |f(x_{k}+[x])|<\varepsilon\]
马上看到
\[ |f(x)|\leq |f(x)-f(x_{k}+[x])|+|f(x_{k}+[x])|<2\varepsilon\]
由此知
\[ \lim_{x\to+\infty}f(x)=0 \]

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问题2.
设函数$f$在$[0,+\infty)$上连续，$\forall \alpha>0$,有$ \displaystyle \lim_{n\to+\infty}f(n\alpha)=0$. 证明：
\[ \lim_{x\to+\infty}f(x)=0 \]
这个问题显然比上面的问题要复杂不少，去掉一致连续性条件弱了好多。
证明
$\forall\varepsilon>0$,显然$ A_{n}=\{\alpha||f(n\alpha)|\leq \varepsilon\}$是个闭集，（由$f$的连续性不难验证，$A_{n}^{c}$是开集），$\displaystyle B_{k}=\bigcap_{n>k}{A_{n}}$也为闭集。
$\forall x\in(0,+\infty)$,由于$\displaystyle \lim_{n\to +\infty}f(nx)=0$,知,$\exists k\in \mathbf{N}$, 当$n>k$时$|f(nx)|<\varepsilon$,于是$x\in B_{k}$,因此
\[ (0,+\infty)=\bigcup_{k=1}^{\infty} B_{k} \]
由著名的Baire定理知，$ \exists [a,b]\subset B_{k_{0}}$.取$l\geq \max\bigg\{k_{0}+1,\frac{a}{b-a}\bigg\}$,当$n\geq l$时，有$n(b-a)\geq a$从而$nb\geq (n+1)a$.于是，对$\forall x>la,$即 $\displaystyle x\in[la,+\infty)\subset \bigcup_{n>k_{0}}[na,nb]$,存在某个$n>k_{0}$ 使得$x\in[na,nb]$,从而存在$\displaystyle \alpha\in[a,b]\subset B_{k_{0}}=\bigcap_{n>k_{0}}A_{n}$,$\alpha\in A_{n},n>k_{0}$,使得$x=n\alpha$,所以
\[ |f(x)|=|f(n\alpha)|\leq \varepsilon \]
即
\[ \lim_{x\to+\infty}f(x)= 0 \]

著名的Baire定理是这么说的：设$E,E_{m}\in \mathbf{R^{n}}$,$\displaystyle E=\bigcup_{m\in\Gamma}E_{m}$, 其中$\Gamma$为至多可数集。$E_{m}$的内点集$ \text{int}E_{m}=\varnothing$,$E_{m}$为闭集，$m\in\Gamma$,则$\text{int}E=\varnothing$.对刚刚的问题，我们知道若每一个$B_{k}$都没有内点，那么会得到$(0,+\infty)$没内点，显然是个矛盾，故存在某个$B_{k_{0}}$有内点，显然就存在一个$[a,b]\subset B_{k_{0}}$.

好久没更新博客了，这段时间比较忙。以后争取有空多贴点好的问题，多谢大家光临我的博客。

Problem

If $a,b,c$ are positive real numbers such that $a+b+c=3$,then
\[ \sqrt{a^3+3b}+\sqrt{b^3+3c}+\sqrt{c^3+3a}\geq 6 \]

Proof. By the Cauchy-Schwarz inequality,we have
\[ (a^3+3b)(a+3b)\geq (a^2+3b)^2 \]
Thus,it suffices to show that
\[ \sum\frac{a^2+3b}{\sqrt{a+3b}}\geq 6 \]
By Holder inequality,we have
\[\left(\sum\frac{a^2+3b}{\sqrt{a+3b}}\right)^{2}\left[\sum(a^2+3b)(a+3b)\right]\geq \left[\sum(a^2+3b)\right]^3=\left(\sum{a^2}+9\right)^3 \]
Therefore,it is enough to show that
\[ \left(\sum a^2+9\right)^3\geq 36\sum(a^2+3b)(a+3b) \]
Let $p=a+b+c=3$ and $q=ab+bc+ca, q\leq 3$.We have
\[ \sum a^2+9=p^2-2q+9=2(9-q) \]
\begin{align*}
\sum(a^2+3b)(a+3b)&=\sum a^3+3\sum a^2b+9\sum a^2+3\sum ab\\
&=(p^3-3pq+3abc)+3\sum a^2b+9(p^2-2q)+3q \\
&=108-24q+3\left(abc+\sum a^2b \right) 
\end{align*}
Since
\[ a^2b+b^2c+c^2a+abc\leq \frac{4}{27}(a+b+c)^3 \]
We get
\[ \sum{(a^2+3b)(a+3b)}\leq 24(5-q) \]
Thus.it suffices to show that
\[ (9-a)^3\geq 108(5-a) \]
Which is equivalent to the obvious inequality
\[ (3-q)^2(21-q)\geq 0 \]
The equality holds for $a=b=c=1 $.     

 Problem

设$a,b,c \in \mathbf{R}$.且$ab+bc+ca=11$.求
\[ \sqrt{a^2+21}+\sqrt{2b^2+14}+\sqrt{c^2+91} \]
的最小值。

Solution

由Cauchy-Schwarz

\[ [(a^2+1)+20](5+20)\geq \left[\sqrt{5(a^2+1)}+20\right]^{2}   \]
 \[      [ 2(b^2+1)+12](4+12)\geq \left[\sqrt{8(b^2+1)}+12\right]^2   \]
 \[      [(c^2+1)+90](10+90)\geq \left[\sqrt{10(c^2+1)}+90\right]^2 \]
得到
\[ \sqrt{a^2+21}\geq \sqrt{\frac{1}{5}\left(a^2+1\right)}+4 \]
\[ \sqrt{2b^2+14}\geq \sqrt{\frac{1}{2}\left(b^2+1\right)}+3 \]
\[ \sqrt{c^2+91} \geq \sqrt{\frac{1}{10}\left(c^2+1\right)}+9 \]
\begin{align*}
\sqrt{a^2+21}+\sqrt{2b^2+14}+\sqrt{c^2+91}&\geq 16 +\sqrt{\frac{1}{5}\left(a^2+1\right)}+\sqrt{\frac{1}{2}\left(b^2+1\right)}+\sqrt{\frac{1}{10}\left(c^2+1\right)}\\
&\geq 16+3\sqrt[6]{\frac{1}{100}(a^2+1)(b^2+1)(c^2+1)}
\end{align*}
注意到恒等式
\[ (a^2+1)(b^2+1)(c^2+1)=(ab+bc+ca-1)^2+(a+b+c-abc)^2\geq (ab+bc+ca-1)^2\ge 100 \]
故
\[ 16+3\sqrt[6]{\frac{1}{100}(a^2+1)(b^2+1)(c^2+1)}\geq 19 \]
\[ \sqrt{a^2+21}+\sqrt{2b^2+14}+\sqrt{c^2+91}\geq 19 \]
等号成立当$a=2,b=1,c=3 $                                                        $\square$