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此题由tian275461提供。题目非常困难，仅供观赏(水平一般的同学请勿模仿 )。

求积分
\[ I=\int_{0}^{1}{\frac{\ln{x}}{x^2-x-1}dx} \]
解（tian275461）

由方程$x^2-x-1=0$ 的两个根，为了简单起见，我们记$ r_{1}=\varphi=\dfrac{1+\sqrt{5}}{2},r_{2}=\dfrac{1-\sqrt{5}}{2}=1-\varphi$,
且$r_{1}-r_{2}=\sqrt{5},\varphi^{2}=\varphi+1 , \dfrac{\varphi-1}{\varphi}=\dfrac{1}{\varphi^{2}} $
则有

\[ \begin{align*}
I&=\int_{0}^{1}{\frac{\ln{x}}{x^2-x-1}dx}\\
&=\frac{1}{r_{1}-r_{2}}\int_{0}^{1}{\ln{x}\left(\frac{1}{x-\varphi}-\frac{1}{x-(1-\varphi)} \right)dx}\\
&=\dfrac{1}{\sqrt{5}}\int_{0}^{1}{\ln{x}\left(\frac{1}{x-\varphi}-\frac{\varphi}{\varphi x+1}\right)dx}\\
&=\dfrac{1}{\sqrt{5}}\int_{0}^{1}{\frac{\ln{x}}{x-\varphi}dx}-\frac{\varphi}{\sqrt{5}}\int_{0}^{1}{\frac{\ln{y}}{\varphi y+1}dy}\\
&=\frac{1}{\sqrt{5}}\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{\varphi u}}{u-1}du}-\frac{1}{\sqrt{5}}\int_{0}^{\varphi}{\frac{\ln{\frac{u}{\varphi}}}{u+1}du}\\
&=\frac{\ln{\varphi}}{\sqrt{5}}\left(\int_{0}^{\frac{1}{\varphi}}{\frac{1}{u-1}du}+\int_{0}^{\varphi}{\frac{1}{u+1}du} \right)-\frac{1}{\sqrt{5}}\left(\int_{0}^{\varphi}{\frac{\ln{u}}{1+u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du} \right)\\
&=\frac{\ln{\varphi}}{\sqrt{5}}\ln{\frac{\varphi^{2}-1}{\varphi}}-\frac{1}{\sqrt{5}}\left(\int_{0}^{\varphi}{\frac{\ln{u}}{1+u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du} \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{0}^{\varphi}{\frac{\ln{u}}{1+u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du} \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{1}^{1+\varphi}{\frac{\ln{(u-1)}}{u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}  \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{1}^{\varphi^{2}}{\frac{\ln{(u-1)}}{u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}  \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{\frac{1}{\varphi^{2}}}^{1}{\frac{\ln{(1-u)}-\ln{u}}{u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}  \right)\\
&=-\frac{1}{\sqrt{5}}\left(\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}-\ln{(1-u)}}{u}du}+\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}  \right)\\
&=-\frac{2}{\sqrt{5}}\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}+\frac{1}{\sqrt{5}}\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{(1-u)}}{1-u}du}\\
&=-\frac{2}{\sqrt{5}}\int_{0}^{\frac{1}{\varphi}}{\frac{\ln{u}}{1-u}du}-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=-\frac{2}{\sqrt{5}}\int_{\frac{1}{\varphi^{2}}}^{1}{\frac{\ln{(1-u)}}{u}du}-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=-\frac{2}{\sqrt{5}}\left(\int_{0}^{1}{\frac{\ln{(1-u)}}{u}du}-\int_{0}^{\frac{1}{\varphi^{2}}}{\frac{\ln{(1-u)}}{u}du} \right)-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=\frac{\pi^{2}}{3\sqrt{5}}-\frac{2}{\sqrt{5}}\Li_{2}{\left(\frac{1}{\varphi^{2}}\right)}-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=\frac{\pi^{2}}{3\sqrt{5}}-\frac{2}{\sqrt{5}}\left(\frac{\pi^{2}}{15}-\ln^{2}{\varphi}
\right)-\frac{2}{\sqrt{5}}\ln^{2}{\varphi}\\
&=\frac{\pi^{2}}{5\sqrt{5}}
\end{align*} \]

$\square$

最后一步用了dilogarithm函数的性质，详细可以参见http://mathworld.wolfram.com/Dilogarithm.html

______________________________________________________________________________

简单一点的做法

我们有
\[ \Li_{2}(x)=-\int_{0}^{x}\frac{\ln(1-t)}{t}dt \]
则
\begin{align*}
\int_{0}^{1}\frac{\ln{x}}{x-a}&=-\frac{1}{a}\cdot\int_{0}^{1}\frac{\ln{x}}{1-\frac{x}{a}}dx\\
&= -\frac{1}{a}\cdot\left(\int_{0}^{1}\frac{\ln\left(\frac{x}{a}\right)}{1-\frac{x}{a}}dx+\ln{a}\int_{0}^{1}\frac{1}{1-\frac{x}{a}}dx\right)\\
&=-\int_{0}^{\frac{1}{a}}\frac{\ln{t}}{1-t}dt-\frac{\ln{a}}{a}\int_{0}^{1}\frac{1}{1-\frac{x}{a}}dx\\
&=-\int_{0}^{\frac{1}{a}}\frac{\ln(1-t)}{t}dt\\
&=\Li_{2}\left(\frac{1}{a}\right)
\end{align*}
注意到
\[ x^2-x-1=\left(x-\frac{1+\sqrt{5}}{2}\right)\left(x-\frac{1-\sqrt{5}}{2}\right) \]
\[ I=\int_{0}^{1}\frac{\ln{x}}{x^2-x-1}dx=\frac{1}{\sqrt{5}}\left(\int_{0}^{1}\frac{\ln{x}}{x-\frac{\sqrt{5}+1}{2}}dx-\int_{0}^{1}\frac{\ln{x}}{x-\frac{1-\sqrt{5}}{2}}dx \right)=\frac{1}{\sqrt{5}}\left(\Li_{2}\left(\frac{\sqrt{5}-1}{2}\right)-\Li_{2}\left(-\frac{\sqrt{5}+1}{2}\right)\right)\]
记$\phi=\frac{\sqrt{5}+1}{2}$,则
\[ I=\frac{1}{\sqrt{5}}\left(\Li_{2}\left(\frac{1}{\phi}\right)-\Li_{2}(-\phi)\right) \]
我们还知道公式
\[ \Li_{2}\left(\frac{1}{\phi}\right)=\frac{1}{10}\pi^2-\ln^{2}\phi \]
\[ \Li_{2}(-\phi)=-\frac{1}{10}\pi^2-\ln^2\phi \]
则
\[ I=\frac{1}{5\sqrt{5}}\pi^2 \]

设$f(x)\in C^{(2)}(0,1) $,且$\displaystyle \lim_{x \rightarrow 1^{-}}{f(x)}=0$.若存在$M>0$，使得$(1-x)^{2}|f''(x)|\leq M (0<x<1)$则
\[ \lim_{x\rightarrow 1^{-}}{(1-x)f'(x)}=0 \]
证明：
对 $t,x\in(0,1): t>x$,用Taylor公式
\[ f(t)=f(x)+f'(x)(t-x)+f''(\xi)\frac{(t-x)^{2}}{2}, x<\xi<t \]
并取$t=x+(1-x)\delta, (0<\delta<\frac{1}{2}) $,我们有
\[ f(t)-f(x)=\delta (1-x)f'(x)+\frac{\delta^{2}}{2}f''(\xi)(1-x)^{2} \]
\[ \Leftrightarrow (1-x)f'(x)=\frac{f(t)-f(x)}{\delta}-\frac{\delta}{2}f''(\xi)(1-x)^{2} \]
\[ |f'(x)(1-x)|\leq \frac{|f(t)-f(x)|}{\delta}+\frac{\delta}{2}|f''(\xi)|(1-x)^{2} \]
注意到\[ \xi=x+(t-x)\theta   , 0<\theta<1 \]
\[\Rightarrow (1-\xi)^{2}=(1-x)^{2}(1-\delta\theta)^{2}>\frac{1}{4}(1-x)^{2} \]
（这里是由于$ 0<\delta\theta<\frac{1}{2}$ )
及条件 $(1-x)^{2}|f''(x)|\leq M (0<x<1)$
\[ \frac{\delta}{2}|f''(\xi)|(1-x)^{2}=|f''(\xi)|(1-\xi)^{2}\cdot\frac{(1-x)^{2}}{(1-\xi)^{2}}\cdot\frac{\delta}{2}<2M\delta   \]
\[ \Rightarrow |f'(x)(1-x)|\leq \frac{|f(t)-f(x)|}{\delta}+2M\delta \]
现在，对$\forall \varepsilon $,取 $ \delta=\frac{\varepsilon}{4M} $
对上述$ \delta\varepsilon $存在$ \eta>0 $
对$ \forall 0<1-x<\eta $
有$ |f(t)-f(x)|<\frac{\delta\varepsilon}{2}$
这样，对$ \forall 0<1-x<\eta $，就有
\[ \Rightarrow |f'(x)(1-x)|<\varepsilon \]
故得 \[ \lim_{x\rightarrow 1^{-}}{(1-x)f'(x)}=0 \]

$\square$

求证：
\[ \int_{0}^{\frac{\pi}{2}}{x\ln{\sin{x}}\ln{\cos{x}}dx}=\frac{(\pi\ln{2})^{2}}{8}-\frac{\pi^{4}}{192} \]
证明：
首先，我们设
\[ A=\int_{0}^{\frac{\pi}{2}}{x\ln{\sin{x}}\ln{\cos{x}}dx} \]
很显然，
\[ A=\frac{\pi}{4}\int_{0}^{\frac{\pi}{2}}{\ln{\sin{x}}\ln{\cos{x}}dx}\]
所以，只需要求
\[ B=\int_{0}^{\frac{\pi}{2}}{\ln{\sin{x}}\ln{\cos{x}}dx}\]
由傅里叶级数不难得到
\[ \ln{(2\cos{\frac{x}{2}})}=\sum_{n=1}^{\infty}{(-1)^{n-1}\frac{\cos{nx}}{n}}, \qquad -\pi<x<\pi \]
由$2x$替换$x$,得到
\[ \ln{(2\cos{x})}=\sum_{n=1}^{\infty}{(-1)^{n-1}\frac{\cos{2nx}}{n}}, \qquad -\frac{\pi}{2}<x<\frac{\pi}{2} \]
而另一方面
\begin{align*}
\int_{0}^{\frac{\pi}{2}}{\cos{2nx}\ln{\sin{x}}dx}&=\int_{0}^{\frac{\pi}{2}}{\ln{\sin{x}}d\frac{\sin{2nx}}{2n}}\\
&=\frac{1}{2n}\sin{2nx}\cdot\ln{\sin{x}}|_{0}^{\frac{\pi}{2}}-\frac{1}{2n}\int_{0}^{\frac{\pi}{2}}{\frac{\cos{x}\cdot\sin{2nx}}{\sin{x}}dx}\\
&=-\frac{1}{4n}\int_{0}^{\frac{\pi}{2}}{\frac{\sin{(2n+1)x}+\sin{(2n-1)x}}{\sin{x}}dx}\\
&=-\frac{\pi}{4n}
\end{align*}
所以
\begin{align*}
\int_{0}^{\frac{\pi}{2}}{\ln{2\cos{x}}\cdot\ln{\sin{x}}dx}&=\sum_{n=1}^{\infty}{(-1)^{n}\frac{\pi}{4n^2}}\\
&=B+\ln{2}\cdot\int_{0}^{\frac{\pi}{2}}{\ln{\sin{x}}dx}
\end{align*}
马上看到
\[ B=\frac{\pi}{2}\ln^{2}{2}-\frac{1}{48}{\pi^{3}} \]
\[ A= \frac{(\pi\ln{2})^{2}}{8}-\frac{\pi^{4}}{192} \]
Done!

$ \square$

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

说实话，那个

\[ \ln{(2\cos{\frac{x}{2}})}=\sum_{n=1}^{\infty}{(-1)^{n-1}\frac{\cos{nx}}{n}}, \qquad -\pi<x<\pi \]

真心很有用，用这个可以算出下面结果

\[ \int_{0}^{\frac{\pi}{2}}{\ln^{2}{\sin{x}}dx}=\int_{0}^{\frac{\pi}{2}}{\ln^{2}{\cos{x}}dx}=\frac{\pi^3}{24}+\frac{\pi}{2}\ln^{2}{2} \]

只要考虑傅里叶级数正交就好。

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
\[\displaystyle\lim_{\varepsilon\to 0}\dfrac{2\ln{(1+2\varepsilon)}-2\ln{2}-2\ln{\varepsilon}}{\ln{(1+\varepsilon)}-\ln{\varepsilon}}=2\displaystyle\lim_{\varepsilon\to 0}\dfrac{-\dfrac{\ln{(1+2\varepsilon)}}{\ln{\varepsilon}}+\dfrac{\ln{2}}{\ln{\varepsilon}}+1}{-\dfrac{\ln{(1+\varepsilon)}}{\ln{\varepsilon}}+1}=2\]

$\square$

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