本帖的题目均来自tian275461，版权由tian275461所有，未经许可，禁止转载。

中值定理：

设$f$是在$R$上有四阶连续可导的函数,$x\in [0,1]$,满足
\[\int_{0}^{1}f(x)dx+3f\left(\dfrac{1}{2}\right)=8\displaystyle\int_{\frac{1}{4}}^{\frac{3}{4}}f(x)dx\]
证明：存在$c\in(0,1)$,使得$f^{(4)}(c)=0$

证明：令
\[G(t)=\int_{-t}^{t}g(x)dx-8\int_{-\frac{t}{2}}^{\frac{t}{2}}g(x)dx\]
其中\[g(x)=f\left(x+\dfrac{1}{2}\right)-f\left(\dfrac{1}{2}\right)\]
易得
\[G(0)=0,G\left(\dfrac{1}{2}\right)=0\]
由\Rolle 定理有存在$t_{0}\in(0,1/2)$使得$G'(t_{0})=0$.由于
\[G'(t)=g(t)-4g(\frac{t}{2})-4g(-\frac{t}{2})+g(-t)\]
则$G'(0)=0,G'(t_{0})=0$,则由\Rolle 定理有：$G''(t_{1})=0$,又
\[G''(t)=g'(t)-2g'(\frac{t}{2})+2g'(-\frac{t}{2})-g'(-t)\]
显然$G''(0)=0$,故由中值定理有$G'''(t_{2})=0$\\
又
\[G'''(t)=(g''(t)-g''(\frac{t}{2}))-(g''(-\frac{t}{2})-g''(-t))\]
即
\[G'''(t_{2})=(g''(t_{2})-g''(\frac{t_{2}}{2}))-(g''(-\frac{t_{2}}{2})-g''(-t_{2}))\]
又由拉格朗日中值定理有存在$\theta_{+}\in(t_{2}/2,t_{2}),\theta_{-}\in(-t_{2},-\frac{t_{2}}{2})$,\\
\[(g''(t_{2})-g''(\frac{t_{2}}{2}))-(g''(-\frac{t_{2}}{2})-g''(-t_{2}))=g'''(\theta_{+})\dfrac{t_{2}}{2}-g'''(\theta_{-})\dfrac{t_{2}}{2}\]
注意$t_{2}\neq 0$即
\[g'''(\theta_{+})-g'''(\theta_{-})=0\]
再利用拉格朗日中值定理
\[g''''(\theta)=0 \]
即\[f^{(4)}\left(\theta+\dfrac{1}{2}\right)=0\]
将$\theta +\dfrac{1}{2}\longrightarrow \theta$,即有
\[f^{(4)}(\theta)=0\]

$\square$

级数求和:

\[\sum_{n=1}^{\infty}\dfrac{H_{n}}{n^3}\]

显然有
\[-n\displaystyle\int_{0}^{1}(1-x)^{n-1}\ln{x}dx=-\displaystyle\sum_{k=1}^{n}C_{n}^{k}\dfrac{(-1)^k}{k}=H_{n}\]
证明：考虑积分
\[\displaystyle\int_{0}^{1}\dfrac{1-(1-x)^n}{x}dx=\displaystyle\int_{0}^{1}\sum_{k=1}^{n}C_{n}^{k}(-1)^{k+1}x^{k-1}dx=\displaystyle\sum_{k=1}^{n}\dfrac{C_{n}^{k}(-1)^{k+1}}{k}\]
另外一方面
\[\displaystyle\int_{0}^{1}\dfrac{1-(1-x)^n}{x}dx=\displaystyle\int_{0}^{1}\dfrac{1-u^n}{1-u}du=H_{n}\]
所以
\[\displaystyle\sum_{n=1}^{\infty}\dfrac{H_{n}}{n^3}=-\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n^2}\displaystyle\int_{0}^{1}(1-x)^{n-1}\ln{x}dx=-\displaystyle\int_{0}^{1}\displaystyle\sum_{n=1}^{\infty}\dfrac{(1-x)^{n-1}}{n^2}\ln{x}dx\]
由于
\[\displaystyle\sum_{n=1}^{\infty}\dfrac{(1-x)^n}{n^2}=\dfrac{Li_{2}(1-x)}{1-x}\]
故
\[\displaystyle\sum_{n=1}^{\infty}\dfrac{H_{n}}{n^3}=-\displaystyle\int_{0}^{1}\dfrac{Li_{2}(1-x)\ln{x}}{1-x}dx=\dfrac{1}{2}(Li_{2}(1-x))^2|_{0}^{1}=\dfrac{1}{2}\left(\dfrac{\pi^2}{6}\right)^2\]
$\square$

积分求值:
\[\displaystyle\int_{0}^{\infty}\dfrac{1}{(x^4+(1+2\sqrt{2})x^2+1)(x^{100}-x^{98}+\cdots+1)}dx\]
解 设
\[I=\displaystyle\int_{0}^{\infty}\dfrac{1}{(x^4+(1+2\sqrt{2})x^2+1)(x^{100}-x^{98}+\cdots+1)}dx\]
令$x=\dfrac{1}{y}$
则
\[I=\displaystyle\int_{0}^{\infty}\dfrac{x^{102}}{(x^4+(1+2\sqrt{2})x^2+1)(x^{100}-x^{98}+\cdots+1)}dx\]
注意
\[x^{100}-x^{98}+\cdots+1=\dfrac{1+x^{102}}{1+x^2}\]
所以
\[2I=\displaystyle\int_{0}^{\infty}\dfrac{1+x^2}{x^4+(1+2\sqrt{2})x^2+1}dx\]
即
\[I=\frac{1}{2}\displaystyle\int_{0}^{\infty}\dfrac{1+\dfrac{1}{x^2}}{x^2+\dfrac{1}{x^2}+1+2\sqrt{2}}dx=\dfrac{\pi}{2(1+\sqrt{2})}\]

$\square$

积分不等式:

1.求所有的连续可导函数$f:[0,1]\longrightarrow (0,\infty)$,满足$f(1)=ef(0)$,且
\[\displaystyle\int_{0}^{1}\dfrac{dx}{(f(x))^2}+\displaystyle\int_{0}^{1}(f'(x))^2dx\le 2\]
解：注意
\begin{align*}
&0\le\displaystyle\int_{0}^{1}\left(f'(x)-\dfrac{1}{f(x)}\right)^2dx=\displaystyle\int_{0}^{1}(f'(x))^2dx-2\displaystyle\int_{0}^{1}\dfrac{f'(x)}{f(x)}dx+\displaystyle\int_{0}^{1}\dfrac{dx}{(f(x))^2}\\
&=\left(\displaystyle\int_{0}^{1}(f'(x))^2dx+\displaystyle\int_{0}^{1}\dfrac{dx}{(f(x))^2}dx\right)-2\displaystyle\int_{0}^{1}(\ln{f(x)})'dx\\
&=\left(\displaystyle\int_{0}^{1}(f'(x))^2dx+\displaystyle\int_{0}^{1}\dfrac{dx}{(f(x))^2}dx\right)-2\ln{\dfrac{f(1)}{f(0)}}\\
&\le 0
\end{align*}
所以
\[f(x)f'(x)=1\]
\[\Longrightarrow f(x)=\sqrt{2x+C},C>0\]
由于\[\dfrac{f(1)}{f(0)}=e\Longrightarrow C=\dfrac{2}{e^2-1}\]
故
\[f(x)=\sqrt{2x+\dfrac{2}{e^2-1}}\]
$\square$

2.设$f,g:[0,1]\longrightarrow (0,+\infty)$是连续的,且$f,\dfrac{g}{f}$递增的,求证：
\[\displaystyle\int_{0}^{1}\left(\dfrac{\displaystyle\int_{0}^{x}f(t)dt}{\displaystyle\int_{0}^{x}g(t)dt}\right)dx\le 2\displaystyle\int_{0}^{1}\dfrac{f(x)}{g(x)}dx\]
并说明右边系数$2$是最佳的.
证明：由切比雪夫不等式有
\[\left(\dfrac{1}{x}\displaystyle\int_{0}^{x}f(t)dt\right)\left(\dfrac{1}{x}\displaystyle\int_{0}^{x}\dfrac{g(t)}{f(t)}dt\right)\le\dfrac{1}{x}\displaystyle\int_{0}^{x}g(t)dt\]
即
\[\dfrac{\displaystyle\int_{0}^{x}f(t)dt}{\displaystyle\int_{0}^{x}g(t)dt}\le\dfrac{x}{\displaystyle\int_{0}^{x}\dfrac{g(t)}{f(t)}dt}\]
另外由Cauchy-Schwarz有
\[\left(\displaystyle\int_{0}^{x}\dfrac{g(t)}{f(t)}dt\right)\left(\displaystyle\int_{0}^{x}\dfrac{t^2f(t)}{g(t)}dt\right)\ge\left(\displaystyle\int_{0}^{x}tdt\right)^2=\dfrac{x^4}{4}\]
即
\[\dfrac{1}{\displaystyle\int_{0}^{x}\dfrac{g(t)}{f(t)}dt}\le\dfrac{4}{x^4}\displaystyle\int_{0}^{x}\dfrac{t^2f(t)}{g(t)}dt\]
所以有
\[\dfrac{\displaystyle\int_{0}^{x}f(t)dt}{\displaystyle\int_{0}^{x}g(t)dt}\le\dfrac{4}{x^3}\displaystyle\int_{0}^{x}\dfrac{t^2f(t)}{g(t)}dt\]
故有
\begin{align*}
\displaystyle\int_{0}^{1}\dfrac{\displaystyle\int_{0}^{x}f(t)dt}{\displaystyle\int_{0}^{x}g(t)dt}dx&\le\displaystyle\int_{0}^{1}\left(\displaystyle\int_{0}^{x}\dfrac{4t^2f(t)}{x^3g(t)}dt\right)dx
=\displaystyle\int_{0}^{1}\left(\displaystyle\int_{t}^{1}\dfrac{4t^2f(t)}{x^3g(t)}dx\right)dt\\
&=\displaystyle\int_{0}^{1}\dfrac{4t^2f(t)}{g(t)}\left(\displaystyle\int_{t}^{1}\dfrac{dx}{x^3}\right)dt\\
&=2\displaystyle\int_{0}^{1}\dfrac{f(t)}{g(t)}(1-t^2)dt\\
&\le 2\displaystyle\int_{0}^{1}\dfrac{f(t)}{g(t)}dt
\end{align*}
另外一方面:我们令\[f(t)=1,g(t)=t+\varepsilon,\varepsilon>0\]
则\[\displaystyle\int_{0}^{1}\dfrac{\displaystyle\int_{0}^{x}f(t)dt}{\displaystyle\int_{0}^{x}g(t)dt}dx
=\displaystyle\int_{0}^{1}\dfrac{x}{\dfrac{1}{2}x^2+\varepsilon x}dx=2\ln{(1+2\varepsilon)}-2\ln{2}-2\ln{\varepsilon}\]
\[\displaystyle\int_{0}^{1}\dfrac{f(t)}{g(t)}dt=\displaystyle\int_{0}^{1}\dfrac{dt}{t+\varepsilon}=\ln{(1+\varepsilon)}-\ln{\varepsilon}\]
所以
\[\displaystyle\lim_{\varepsilon\to 0}\dfrac{2\ln{(1+2\varepsilon)}-2\ln{2}-2\ln{\varepsilon}}{\ln{(1+\varepsilon)}-\ln{\varepsilon}}=2\displaystyle\lim_{\varepsilon\to 0}\dfrac{-\dfrac{\ln{(1+2\varepsilon)}}{\ln{\varepsilon}}+\dfrac{\ln{2}}{\ln{\varepsilon}}+1}{-\dfrac{\ln{(1+\varepsilon)}}{\ln{\varepsilon}}+1}=2\]

$\square$