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
)。