陈洪葛的博客

\[\displaystyle\lim\limits_{n\to\infty}\frac{1}{(2n-1)^{2011}}\sum_{k=0}^{n-1}\int_{2k\pi}^{(2k+1)\pi}x^{2010}\sin^3 x\cos^2 x \mathrm{d}x\]

(tian_275461)
解:
根据推广的积分第一中值定理，对每个正整数~$n~\exists\qquad \theta_n\in(0,1)$~使得
\begin{equation*}
    \int_{2n\pi}^{(2n+1)\pi}x^{2010}\sin^3 x\cos^2 x \mathrm{d}x=
    ((2n+\theta_n)\pi)^{2010}\int_{2n\pi}^{(2n+1)\pi}\sin^3 x\cos^2 x \mathrm{d}x
\end{equation*}
由此得
\begin{align*}
&\quad\int_{2n\pi}^{(2n+1)\pi}x^{2010}\sin^3 x\cos^2 x \mathrm{d}x\\
&=\left((2n\pi)^{2010}+o(n^{2010})\right)\int_{2n\pi}^{2n\pi+\pi}\sin^3 x\cos^2 x\mathrm{d}x\\
&=\left((2n\pi)^{2010}+o(n^{2010})\right)\left(\frac{\cos 5x}{80}-\frac{\cos 3x}{48}-\frac{\cos x}{8}\right)\Bigg|_{2n\pi}^{(2n+1)\pi}\\
&=\frac{4}{15}((2n\pi)^{2010}+o(n^{2010}))\qquad (n\to\infty)
\end{align*}
另外
\begin{equation*}
    (2n+1)^{2011}-(2n-1)^{2011}=4022(2n)^{2010}+o(n^{2010})\qquad (n\to\infty)
\end{equation*}
根据~Stolz~定理
\begin{align*}
&\quad\lim\limits_{n\to\infty}\frac{1}{(2n-1)^{2011}}\sum_{k=0}^{n-1}\int_{2k\pi}^{(2k+1)\pi}x^{2010}\sin^3 x\cos^2 x \mathrm{d}x\\
&=\lim_{n\to\infty}\frac{\displaystyle\int_{2n\pi}^{(2n+1)\pi}x^{2010}\sin^3 x\cos^2 x \mathrm{d}x}{ (2n+1)^{2011}-(2n-1)^{2011}}\\
&=\frac{2}{30165}\lim_{n\to \infty}\frac{(2n\pi)^{2010}+o(n^{2010})}{(2n)^{2010}+o(n^{2010})}\\
&=\frac{2\pi^{2010}}{30165}
\end{align*}

此题的更一般结果为
\begin{equation*}
    \lim_{n\to\infty}\frac{1}{(2n-1)^{p+1}}\sum_{k=0}^{n-1}\int_{2k\pi}^{(2k+1)\pi}x^p\sin^3 x\cos^2 x \mathrm{d}x
=\frac{2\pi^p}{15(p+1)}(p>0)
\end{equation*}

求极限
\[ \displaystyle\lim_{n\to\infty}\frac{1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}}{e^n}\]

（高等数学中的若干问题解析）
解：\begin{equation*}
    e^n=1+n+\frac{n^2}{2!}+\cdots +\frac{n^n}{n!}+\frac{1}{n!}\int_0^n e^x(n-x)^n\mathrm{d}x
\end{equation*}
原命题等价于
\begin{equation*}
  \lim_{n\to\infty}\frac{e^{-n}}{n!}\int_0^n e^x(n-x)^n\mathrm{d}x=\frac{1}{2}\quad\text{而}
     n!=\sqrt{2n\pi}(\frac{n}{e})^ne^{\frac{\theta}{12n}},\theta\in(0,1)
\end{equation*}
\begin{equation*}
   \Leftrightarrow \lim_{n\to\infty}\sqrt{n}\int_0^1[e^x(1-x)]^n\mathrm{d}x=\sqrt{\frac{\pi}{2}}
\end{equation*}
注意到~$e^{-\frac{x^2}{2}}\geq(1-x)e^x (x\geq 0)$
\begin{equation*}
   \therefore\qquad \overline{\lim_{n\to\infty}}\sqrt{n}\int_0^1[e^x(1-x)]^n\mathrm{d}x\leq
    \overline{\lim_{n\to\infty}}\int_0^1\sqrt{n}e^{-\frac{nx^2}{2}}\mathrm{d}x=\sqrt{\frac{\pi}{2}}
\end{equation*}
考虑
\begin{equation*}
  f(x)=(1-x)e^x-e^{-\frac{ax^2}{2}}(x\geq 0,a\geq 1),f'(x)=xe^x(ae^{-\frac{ax^2}{2}-x}-1)
\end{equation*}
$\displaystyle\because\quad \lim_{x\to 0^+}(ae^{-\frac{ax^2}{2}-x}-1)=a-1>0$，故存在$x_a\in(0,1)$，使得
\[ae^{-\frac{ax^2}{2}-x}-1>0\]
\begin{align*}
   (1-x)e^x\geq e^{-\frac{ax^2}{2}}(x\in[0,x_a])
    &\Rightarrow\mathop{\underline{\lim}}_{n\to\infty}\sqrt{n}\int_0^1[e^x(1-x)]^n\mathrm{d}x \\
    &\geq\mathop{\underline{\lim}}_{n\to\infty}\int_0^{x_a}\sqrt{n}e^{-\frac{nax^2}{2}}\mathrm{d}x\\
    &=\sqrt{\frac{\pi}{2a}}
\end{align*}
因为$a$是任意的，所以
\begin{equation*}
 \mathop{\underline{\lim}}_{n\to\infty}\sqrt{n}\int_0^1[e^x(1-x)]^n\mathrm{d}x\geq\sqrt{\frac{\pi}{2}}
\end{equation*}
综上得
\begin{equation*}
   \lim_{n\to\infty}\sqrt{n}\int_0^1[e^x(1-x)]^n\mathrm{d}x=\sqrt{\frac{\pi}{2}}
\end{equation*}

已知$f(x)$在$[0,1]$上二阶可导，求证：$\displaystyle\int_0^1\left |f' (x)\right |\mathrm{d}x\le 9\int_0^1\left |f(x)\right |\mathrm{d}x +\int_0^1\left |f''(x)\right |\mathrm{d}x$
证明:
对任意$0<\xi<\dfrac{1}{3}$和$\dfrac{2}{3}<\eta<1$,则存在$\lambda\in(\xi,\eta)$,使得
\begin{equation*}
 |f'(\lambda)|=\left|\frac{f(\eta)-f(\xi)}{\eta-\xi}\right|\leq 3|f(\xi)|+3|f(\eta)|
\end{equation*}
因此对任意的$x\in(0,1)$成立
\begin{equation*}
    |f'(x)|=\left|f'(\lambda)+\int_\lambda^x f''(t)dt\right|\leq 3|f(\xi)|+3|f(\eta)|+\int_0^1|f''(t)|dt
\end{equation*}
分别对$\xi$在$(0,\dfrac{1}{3})$上和对$\eta$在$(\dfrac{2}{3},1)$上积分以上不等式，得
\begin{align*}
  \frac{1}{9}|f'(x)|&\leq \int_0^{\frac{1}{3}}|f(\xi)|d\xi+\int_{\frac{2}{3}}^1 |f(\eta)|d\eta+
  \frac{1}{9}\int_0^1|f''(t)|dt\\
  &\leq \int_0^1|f(t)|dt+\frac{1}{9}\int_0^1|f''(t)|dt
\end{align*}
于是
\begin{equation*}
    |f'(x)|\leq 9\int_0^1|f(t)|dt+\int_0^1|f''(t)|dt,\hspace{10mm}x\in[0,1]
\end{equation*}
对上式两边在$[0,1]$积分,得到
\begin{equation*}
    \int_0^1\left |f' (x)\right |\mathrm{d}x\le 9\int_0^1\left |f(x)\right |\mathrm{d}x +\int_0^1\left |f''(x)\right |\mathrm{d}x
\end{equation*}
$\square$
问题似乎可以加强到
\[\displaystyle\int_0^1\left |f' (x)\right |\mathrm{d}x\le 4\int_0^1\left |f(x)\right |\mathrm{d}x +\int_0^1\left |f''(x)\right |\mathrm{d}x\]

求极限:

\[ \lim_{y\rightarrow+\infty}{\left(\ln^{2}{y}-2\int_{0}^{y}{\frac{\ln{y}}{\sqrt{x^2+1}}dx}\right)} \]

(Proposed by tian275461, Solution by sos440)

\[ \begin{align*} I&=\ln^{2}y-2\int_{0}^{y}{\frac{\ln{x}}{x}\cdot\frac{x}{\sqrt{x^2+1}}dx}\\
&=\ln^{2}{y}-\frac{y}{\sqrt{y^2+1}}\cdot\ln^{2}{y}+\int_{0}^{y}{\frac{\ln^{2}{x}}{(x^2+1)^{\frac{3}{2}}}dx}                                          \end{align*}\]
notice that
\[ \lim_{y\rightarrow+\infty}{\left(\ln^{2}{y}-\frac{y}{\sqrt{y^2+1}}\cdot\ln^{2}{y}\right)}=0 \]
So,just find the value of the following integral
\[ \int_{0}^{\infty}{\frac{\ln^{2}{x}}{(x^2+1)^{\frac{3}{2}}}dx} \]
First note that
\[\begin{align*}
I(p, q) := \int_{0}^{\infty} \frac{x^{p}}{(1+x^2)^{q}} \, dx
&= \int_{0}^{\frac{\pi}{2}} \frac{\tan^{p} \theta}{\sec^{2q} \theta} \, \sec^2 \theta \, d\theta \qquad (x = \tan\theta)\\
&= \int_{0}^{\frac{\pi}{2}} \sin^{p} \theta \cos^{2q-p-2} \theta \, d\theta \\
&= \frac{1}{2}\beta\left( \frac{p+1}{2}, \frac{2q-p-1}{2} \right) \\
&= \frac{1}{2}\frac{\Gamma \left( \frac{p+1}{2} \right) \Gamma \left( \frac{2q-p-1}{2} \right)}{\Gamma(q)}.
\end{align*}\]
Thus
\[ \int_{0}^{\infty} \frac{\log^2 x}{(1+x^2)^{3/2}} \, dx = \frac{\partial^2 I}{\partial p^2}\left( 0, \tfrac{3}{2} \right) \]
and the rest is a mere calculation. Note first that
\[ I\left(p, \tfrac{3}{2} \right) = \frac{\Gamma \left(\frac{1+p}{2}\right) \Gamma \left(1 - \frac{p}{2}\right) }{\Gamma \left(\frac{1}{2}\right) }. \]
Differentiating, we have
\[ \frac{\partial I}{\partial p}\left(p, \tfrac{3}{2} \right) = \frac{\Gamma \left(\frac{1+p}{2}\right) \Gamma \left(1 - \frac{p}{2}\right) }{2\Gamma \left(\frac{1}{2}\right) } \left[ \psi_{0}\left( \frac{1+p}{2} \right) - \psi_{0}\left( 1 - \frac{p}{2} \right) \right]. \]
Differentiating once again, we have
\[ \frac{\partial^2 I}{\partial p^2} \left(p, \tfrac{3}{2} \right) = \frac{\Gamma \left(\frac{1+p}{2}\right) \Gamma \left(1 - \frac{p}{2}\right) }{4\Gamma \left(\frac{1}{2}\right) } \left[ \left\{\psi_{0}\left( \frac{1+p}{2} \right) - \psi_{0}\left( 1 - \frac{p}{2} \right) \right\}^2 + \left\{\psi_{1}\left( \frac{1+p}{2} \right) + \psi_{1}\left( 1 - \frac{p}{2} \right) \right\} \right] . \]
Plugging $p = 0$, we have
\[ \frac{\partial^2 I}{\partial p^2} \left(0, \tfrac{3}{2} \right) = \frac{1}{4} \left[ \left\{\psi_{0}\left( \tfrac{1}{2} \right) - \psi_{0}(1) \right\}^2 + \left\{\psi_{1}\left( \tfrac{1}{2} \right) + \psi_{1}(1) \right\} \right] . \]
In view of the expansion formula
\[ \psi_{0}(s) = -\gamma + \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+s} \right), \]
we obtain
\[\begin{align*}
\psi_{0}\left( \tfrac{1}{2} \right) - \psi_{0}(1)
&= \sum_{n=0}^{\infty} \left( \frac{1}{n+1} - \frac{1}{n+\frac{1}{2}} \right) \\
&= -2 \sum_{n=0}^{\infty} \left( \frac{1}{2n+1} - \frac{1}{2n+2} \right) \\
&= -2 \ln{2}
\end{align*}\]
and likewise
\[\begin{align*}
\psi_{1}\left( \tfrac{1}{2} \right) + \psi_{1}(1)
&= \sum_{n=0}^{\infty} \left( \frac{1}{\left( n+\frac{1}{2} \right)^2} + \frac{1}{(n+1)^2} \right) \\
&= 4 \sum_{n=0}^{\infty} \left( \frac{1}{(2n+1)^2} + \frac{1}{(2n+2)^2} \right) \\
&= 4 \zeta(2) .
\end{align*}\]
Therefore we have
\[ \frac{\partial^2 I}{\partial p^2} \left(0, \tfrac{3}{2} \right) = \zeta(2) + \ln^{2}{2}.\]

$\square$

设$f:[0,1]\rightarrow R$是连续函数，且$\displaystyle \int_{0}^{1}{f^{3}(x)dx}=0$,求证:
\[ \int_{0}^{1}{f^{4}(x)dx}\geq \frac{27}{4}\left(\int_{0}^{1}{f(x)dx} \right)^{4} \]

(tian276461)
证明:令
 \[ I_{n}=\int_{0}^{1}{f^{n}(x)dx} \]
由Cauchy-Schwarz得
\[ I_{2}\geq I_{1}^{2} \]
又由Cauchy-Schwarz得
\[ \left(\int_{0}^{1}{(r+f^{2}(x))\cdot f(x)dx} \right)^{2}\leq \int_{0}^{1}{(r+f^{2}(x))^{2}dx}\cdot\int_{0}^{1}{f^{2}(x)dx} \]
展开得
\[ (I_{2}-I^{2}_{1})r^2+2I_{2}^{2}r+I_{2}I_{4}\geq 0 \]
上式恒成立，所以 $ \Delta\leq 0 $
\[ \Rightarrow I_{4}\geq \frac{I^{3}_{2}}{I_{2}-I_{1}^{2}} \]
所以只要证明
\[ \frac{I^{3}_{2}}{I_{2}-I^{2}_{1}}\geq \frac{27}{4}I^{4}_{1} \]
由AM-GM
\[ (I_{2}-I_{1}^{2})I^{4}_{1}=\frac{1}{2}(2I_{2}-2I_{1}^{2})\cdot I^{2}_{1}\cdot I^{2}_{1}\leq \frac{4}{27}I_{2}^{3}\]
故有
\[ \int_{0}^{1}{f^{4}(x)dx}\geq \frac{27}{4}\left(\int_{0}^{1}{f(x)dx} \right)^{4} \]
$\square$