## 张祖锦2023年数学专业真题分类70天之第18天
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392、 (4)、 求极限 $\displaystyle \lim\_\{n\to\infty\}\int\_0^1 x^n \sqrt\{4(3+x)\}\mathrm\{ d\} x$. (吉林大学2023年数学分析考研试题) [积分与极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \int\_0^1 x^n \sqrt\{4(3+x)\}\mathrm\{ d\} x =&\int\_0^\{1-\delta\} x^n \sqrt\{4(3+x)\}\mathrm\{ d\} x +\int\_\{1-\delta\}s^1 x^n \sqrt\{4(3+x)\}\mathrm\{ d\} x\\\\ \leq&(1-\delta)^n \sqrt\{4(3+1-\delta)\}\cdot (1-\delta) +\sqrt\{4(3+1)\}\cdot \delta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle n\to\infty$ 得
\begin\{aligned\} \varlimsup\_\{n\to\infty\}\int\_0^1 x^n \sqrt\{4(3+x)\}\mathrm\{ d\} x\leq 4\delta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再令 $\displaystyle \delta\to 0^+$ 得
\begin\{aligned\} &0\leq\varliminf\_\{n\to\infty\}\int\_0^1 x^n \sqrt\{4(3+x)\}\mathrm\{ d\} x\leq \varlimsup\_\{n\to\infty\}\int\_0^1 x^n \sqrt\{4(3+x)\}\mathrm\{ d\} x\leq 0\\\\ \Rightarrow& \lim\_\{n\to\infty\}\int\_0^1 x^n \sqrt\{4(3+x)\}\mathrm\{ d\} x=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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393、 4、 设 $\displaystyle f(x)$ 在任意有限区间可积, 且 $\displaystyle \lim\_\{x\to+\infty\}f(x)=l$, 证明:
\begin\{aligned\} \lim\_\{x\to+\infty\}\frac\{1\}\{x\}\int\_0^x f(t)\mathrm\{ d\} t=l. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(南京航空航天大学2023年数学分析考研试题) [积分与极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle \lim\_\{x\to+\infty\}f(x)=L$ 知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ X > 0,\mathrm\{ s.t.\} \forall\ x\geq X, |f(x)-l| < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle \lim\_\{x\to+\infty\}\frac\{1\}\{x\}\int\_0^X |f(t)-l|\mathrm\{ d\} t=0$ 知
\begin\{aligned\} \exists\ X' > X,\mathrm\{ s.t.\} \forall\ x\geq X', \frac\{1\}\{x\}\int\_0^X |f(t)-l|\mathrm\{ d\} t < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是当 $\displaystyle x\geq X'$ 时,
\begin\{aligned\} &\left|\frac\{1\}\{x\}\int\_0^x f(t)\mathrm\{ d\} t-l\right| \leq \frac\{1\}\{x\}\int\_0^x |f(t)-l|\mathrm\{ d\} t\\\\ \leq&\frac\{1\}\{x\}\int\_0^X |f(t)-l|\mathrm\{ d\} t +\frac\{1\}\{x\}\int\_X^x |f(t)-l|\mathrm\{ d\} t\\\\ < &\frac\{\varepsilon\}\{2\}+\frac\{1\}\{x\}\cdot \frac\{\varepsilon\}\{2\}(x-X) < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle \lim\_\{x\to+\infty\}\frac\{1\}\{x\}\int\_0^x f(t)\mathrm\{ d\} t=l$.跟锦数学微信公众号. [在线资料](
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394、 4、 设 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上连续可导, 且 $\displaystyle \int\_0^\infty f^2(x)\mathrm\{ d\} x$, $\displaystyle \int\_0^\infty |f'(x)|^2\mathrm\{ d\} x$ 都收敛. 证明: $\displaystyle \lim\_\{x\to+\infty\}f(x)=0$. (厦门大学2023年数学分析考研试题) [积分与极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle \int\_0^\infty f^2(x)\mathrm\{ d\} x$, $\displaystyle \int\_0^\infty |f'(x)|^2\mathrm\{ d\} x$ 都收敛及 Cauchy 收敛准则知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ X > 0,\mathrm\{ s.t.\} \forall\ y > x, \int\_x^y f^2(t)\mathrm\{ d\} t < \frac\{\varepsilon\}\{2\}, \int\_x^y |f'(t)|^2\mathrm\{ d\} t < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是当 $\displaystyle y > x\geq X$ 时,
\begin\{aligned\} &|f^2(x)-f^2(y)|=\left|\int\_x^y 2f(t)f'(t)\mathrm\{ d\} t\right|\\\\ \stackrel\{\tiny\mbox\{Schwarz\}\}\{\leq\}& 2\left\[\int\_x^y f^2(t)\mathrm\{ d\} t\right\]^\frac\{1\}\{2\} \left\[\int\_x^y |f'(t)|^2\mathrm\{ d\} t\right\]^\frac\{1\}\{2\} < 2\left(\frac\{\varepsilon\}\{2\}\right)^\frac\{1\}\{2\}\left(\frac\{\varepsilon\}\{2\}\right)^\frac\{1\}\{2\}=\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 Cauchy 收敛准则即知 $\displaystyle \lim\_\{x\to+\infty\}f^2(x)=\ell$ 存在. 进而
\begin\{aligned\} \ell=\lim\_\{x\to+\infty\}f^2(x)\xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\} \lim\_\{x\to+\infty\}f^2(\xi\_x) \xlongequal[\tiny\mbox\{分中值\}]\{\tiny\mbox\{第一积\}\} \lim\_\{x\to+\infty\}\int\_x^\{x+1\}f^2(t)\mathrm\{ d\} t=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就蕴含 $\displaystyle \lim\_\{x\to+\infty\}f^2(x)=0\Rightarrow \lim\_\{x\to+\infty\}f(x)=0$.跟锦数学微信公众号. [在线资料](
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395、 5、 设 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上连续, 且 $\displaystyle f(x) > 0$. 证明:
\begin\{aligned\} \lim\_\{p\to 0^+\}\left\[\frac\{1\}\{b-a\}\int\_a^b f^p(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{p\} =\mathrm\{exp\}\left\\{\frac\{1\}\{b-a\}\int\_a^b \ln f(x)\mathrm\{ d\} x\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \mathrm\{exp\}(t)=\mathrm\{e\}^t$ 表示指数函数. (厦门大学2023年数学分析考研试题) [积分与极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由 Jensen 不等式
\begin\{aligned\} \phi'' > 0\Rightarrow \phi\left(\frac\{1\}\{b-a\}\int\_a^b g(t)\mathrm\{ d\} t\right)\leq \frac\{1\}\{b-a\}\int\_a^b \phi\circ g(t)\mathrm\{ d\} t \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 $\displaystyle (\ln x)''=-\frac\{1\}\{x^2\} < 0$ 知
\begin\{aligned\} &I\_p(f)\equiv\mathrm\{exp\}\left\\{\frac\{1\}\{p\}\ln \left\[\frac\{1\}\{b-a\}\int\_a^b f^p(x)\mathrm\{ d\} x\right\]\right\\}\\\\ \geq& \mathrm\{exp\}\left\[\frac\{1\}\{p\}\frac\{1\}\{b-a\}\int\_a^b \ln (f^p)\mathrm\{ d\} x\right\] =\mathrm\{exp\}\left\[\frac\{1\}\{b-a\}\int\_a^b \ln f\mathrm\{ d\} x\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 再者, 由对数不等式
\begin\{aligned\} \frac\{x\}\{1+x\} < \ln (1+x) < x, \forall\ x > -1, x\neq 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} I\_p(u)=&\mathrm\{exp\}\left\\{\frac\{1\}\{p\}\ln\left\[\frac\{1\}\{b-a\}\int\_a^b f^p\mathrm\{ d\} x\right\]\right\\} \geq \mathrm\{exp\}\left\[\frac\{1\}\{p\}\left(\frac\{1\}\{b-a\}\int\_a^b f^p\mathrm\{ d\} x-1\right)\right\]\\\\ =&\mathrm\{exp\}\left\[\frac\{1\}\{b-a\}\int\_a^b \frac\{f^p-1\}\{p\}\mathrm\{ d\} x\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 联合第 1,2 步及
\begin\{aligned\} \lim\_\{p\to 0^+\}\frac\{f^p-1\}\{p\}\xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{p\to 0^+\}\frac\{f^p\ln f\}\{1\}=\ln f \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即知
\begin\{aligned\} \lim\_\{p\to 0^+\}I\_p(u)=\mathrm\{exp\}\left\[\frac\{1\}\{b-a\}\int\_a^b \ln f\mathrm\{ d\} x\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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396、 4、 (15 分) $\displaystyle f(x)$ 在 $\displaystyle [a,+\infty)$ 上单调, $\displaystyle \int\_a^\infty f(x)\mathrm\{ d\} x$ 收敛. 证明: $\displaystyle \lim\_\{x\to+\infty\}xf(x)=0$. (山西大学2023年数学分析考研试题) [积分与极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 不妨设 $\displaystyle f$ 递减, 否则考虑 $\displaystyle -f$ 即可. (1)、 由 $\displaystyle f$ 递减知 $\displaystyle \lim\_\{x\to+\infty\}f(x)=\inf\_\{x > a\}f(x)\equiv l$ 存在. 事实上, 由下确界定义,
\begin\{aligned\} \{\color\{red\}\forall\ \varepsilon > 0, \exists\ X > a\},\mathrm\{ s.t.\} l\leq f(X) < l+\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \{\color\{red\}x > X\}&\Rightarrow l\leq f(x)\leq f(X) < l+\varepsilon\\\\ &\Rightarrow \{\color\{red\}|f(x)-l| < \varepsilon\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由 $\displaystyle \int\_a^\infty f(x)\mathrm\{ d\} x$ 收敛及 Cauchy 收敛准则知
\begin\{aligned\} &\forall\ \varepsilon > 0, \exists\ X > a,\mathrm\{ s.t.\} \varepsilon > \left|\int\_X^\{X+1\}f(x)\mathrm\{ d\} x\right|=|f(\xi\_X)|\left(\mbox\{积分中值定理\}\right)\\\\ \Rightarrow&\exists\ \xi\_n\mbox\{严\}\nearrow +\infty,\mathrm\{ s.t.\} |f(\xi\_n)| < \varepsilon\\\\ \Rightarrow&\lim\_\{n\to\infty\}\xi\_n=+\infty, \lim\_\{n\to\infty\}f(\xi\_n)=0\\\\ \Rightarrow&l=\lim\_\{x\to+\infty\}f(x)=\lim\_\{n\to\infty\}f(\xi\_n)=0\left(\mbox\{归结原则\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 由 $\displaystyle \int\_a^\infty f(x)\mathrm\{ d\} x$ 收敛及 Cauchy 收敛准则知
\begin\{aligned\} &\color\{red\}\forall\ \varepsilon > 0, \exists\ X > a,\mathrm\{ s.t.\} \forall\ x > X,\\\\ &\frac\{\varepsilon\}\{2\} > \int\_\frac\{x\}\{2\}^x f(t)\mathrm\{ d\} t \geq f(x)\left(x-\frac\{x\}\{2\}\right)=\frac\{xf(x)\}\{2\}\\\\ \Rightarrow&\varepsilon > xf(x)\geq 0\Rightarrow \{\color\{red\}|xf(x)| < \varepsilon\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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397、 (2)、 计算$\lim\_\{n\to\infty\}\int\_n^\{n+1\}\frac\{\sin x\}\{x\}\mathrm\{ d\} x$. (上海大学2023年数学分析考研试题) [积分与极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} \left|\int\_n^\{n+1\}\frac\{\sin x\}\{x\}\mathrm\{ d\} x\right| \leq \int\_n^\{n+1\}\frac\{\mathrm\{ d\} x\}\{x\}=\ln \left(1+\frac\{1\}\{n\}\right) \xrightarrow\{n\to\infty\}0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
即知结论成立.跟锦数学微信公众号. [在线资料](
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398、 11、 设 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上连续, 且 $\displaystyle \int\_0^\infty f(x)\mathrm\{ d\} x$ 收敛. (1)、 给出例子说明 $\displaystyle \lim\_\{x\to+\infty\}f(x)$ 不存在; (2)、 若 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上一致连续, 证明: $\displaystyle \lim\_\{x\to+\infty\}f(x)=0$. (太原理工大学2023年数学分析考研试题) [积分与极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、
\begin\{aligned\} f(x)=\left\\{\begin\{array\}\{llllllllllll\}1-n^2|x-n|,&n-\frac\{1\}\{n^2\}\leq x\leq n+\frac\{1\}\{n^2\},\\\\ 0,&\mbox\{其它 $\displaystyle x$\}\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
满足
\begin\{aligned\} \int\_1^\infty f(x)\mathrm\{ d\} x=\frac\{1\}\{2\}+\sum\_\{n=2\}^\infty \frac\{1\}\{n^2\} < \infty, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但由 $\displaystyle \lim\_\{n\to\infty\}f(n)=1, \lim\_\{n\to\infty\}f\left(n-\frac\{1\}\{n^2\}\right)=0$ 知 $\displaystyle \lim\_\{x\to+\infty\}f(x)$ 不存在. (2)、 由 $\displaystyle f$ 在 $\displaystyle [0,+\infty)$ 上一致连续知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ \delta > 0,\mathrm\{ s.t.\} \forall\ 0 < x < x'\leq x+\delta, |f(x')-f(x)| < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle \int\_0^\infty f(x)\mathrm\{ d\} x$ 收敛及 Cauchy 收敛准则知 $\displaystyle \exists\ X > 0,\mathrm\{ s.t.\} \forall\ x\geq X$,
\begin\{aligned\} \frac\{\varepsilon\delta\}\{2\} > \left|\int\_x^\{x+\delta\}f(t)\mathrm\{ d\} t\right|\xlongequal[\tiny\mbox\{分中值\}]\{\tiny\mbox\{第一积\}\} |f(\xi\_x)\delta| \Rightarrow |f(\xi\_x)| < \frac\{\varepsilon\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是当 $\displaystyle x\geq X$ 时,
\begin\{aligned\} |f(x)|\leq |f(x)-f(\xi\_x)|+|f(\xi\_x)| < \frac\{\varepsilon\}\{2\}+\frac\{\varepsilon\}\{2\}=\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这就证明了 $\displaystyle \lim\_\{x\to+\infty\}f(x)=0$.跟锦数学微信公众号. [在线资料](
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---
399、 4、 (10 分) 证明: $\displaystyle \lim\_\{n\to\infty\}\int\_0^\frac\{\pi\}\{4\}\sin^\{n+\frac\{1\}\{3\}\}x\mathrm\{ d\} x=0$. (西南交通大学2023年数学分析考研试题) [积分与极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle 0\leq \int\_0^\frac\{\pi\}\{4\}\sin^\{n+\frac\{1\}\{3\}\}x\mathrm\{ d\} x\leq \left(\frac\{1\}\{\sqrt\{2\}\}\right)^\{n+\frac\{1\}\{3\}\} \cdot \frac\{\pi\}\{4\}\xrightarrow\{n\to\infty\}0$ 及迫敛性知结论成立.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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400、 (1-2)、 使用积分第二中值定理证明: $\displaystyle \lim\_\{x\to+\infty\}\frac\{1\}\{x\}\int\_0^x \sqrt\{t\}\sin t\mathrm\{ d\} t=0$. (长安大学2023年数学分析考研试题) [积分与极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{左端\}\xlongequal[\tiny\mbox\{分中值\}]\{\tiny\mbox\{第二积\}\} \lim\_\{x\to+\infty\}\frac\{1\}\{x\}\cdot \sqrt\{x\}\int\_\{\xi\_x\}^x \sin t\mathrm\{ d\} t\xlongequal[\tiny\mbox\{无穷小\}]\{\tiny\mbox\{有界 $\displaystyle \cdot$\}\} 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
401、 7、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle \left\[0,\frac\{\pi\}\{2\}\right\]$ 上黎曼可积, 求极限 $\displaystyle \lim\_\{n\to\infty\}\int\_0^\frac\{\pi\}\{2\}\frac\{f(x)\}\{1+x\}\sin^nx\mathrm\{ d\} x$. (郑州大学2023年数学分析考研试题) [积分与极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle I\_n=\int\_0^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x$, 则
\begin\{aligned\} |I\_n|&=\left|\int\_0^\{\frac\{\pi\}\{2\}-\delta\}\sin^nx\mathrm\{ d\} x+\int\_\{\frac\{\pi\}\{2\}-\delta\}^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\right|\\\\ &\leq \int\_0^\{\frac\{\pi\}\{2\}-\delta\}x^n\mathrm\{ d\} x+\int\_\{\frac\{\pi\}\{2\}-\delta\}^\frac\{\pi\}\{2\}1\mathrm\{ d\} x\\\\ &\leq \frac\{1\}\{n+1\}\left(\frac\{\pi\}\{2\}-\delta\right)^\{n+1\}+\delta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle n\to\infty$ 得
\begin\{aligned\} \varlimsup\_\{n\to\infty\}|I\_n|\leq \delta. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再令 $\displaystyle \delta\to 0^+$ 得
\begin\{aligned\} \varlimsup\_\{n\to\infty\}|I\_n|\leq 0\Rightarrow \lim\_\{n\to\infty\}I\_n=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
最后由
\begin\{aligned\} 0\leq \left|\int\_0^\frac\{\pi\}\{2\}\frac\{f(x)\}\{1+x\}\sin^nx\mathrm\{ d\} x\right| \leq \max\_\{x\in [a,b]\}\left|\frac\{f(x)\}\{1+x\}\right|\cdot I\_n\xrightarrow\{n\to\infty\}0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原式 $\displaystyle =0$.跟锦数学微信公众号. [在线资料](
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---
402、 8、 (15 分) 设 $\displaystyle f$ 是 $\displaystyle [a,b]$ 上定义的连续函数. (1)、 计算 $\displaystyle \int\_0^\{2\pi\}|\sin x|\mathrm\{ d\} x$. (2)、 计算
\begin\{aligned\} \lim\_\{n\to\infty\}\int\_a^b |\sin nx|\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 确定常数 $\displaystyle c$ 使得以下极限成立, 并证明之:
\begin\{aligned\} \lim\_\{n\to\infty\}\int\_a^b f(x)|\sin nx|\mathrm\{ d\} x=c\int\_a^b f(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中国科学技术大学2023年数学分析考研试题) [积分与极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \int\_0^\{2\pi\}|\sin x|\mathrm\{ d\} x=4\int\_0^\frac\{\pi\}\{2\}\sin x\mathrm\{ d\} x=4$. (1)、 我们将给出一般的结果. 设 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上黎曼可积; $\displaystyle g$ 是 $\displaystyle T$-周期函数, 在 $\displaystyle [0,T]$ 上可积, 则
\begin\{aligned\} \lim\_\{n\to\infty\}\int\_a^b f(x)g(nx)\mathrm\{ d\} x =\frac\{1\}\{T\} \int\_a^b f(x)\mathrm\{ d\} x\cdot \int\_0^T g(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
通过引进 $\displaystyle g=g^+-g^-, g^+=\max\left\\{g,0\right\\}, g^-=-\min\left\\{g,0\right\\}$, 而不妨设 $\displaystyle g\geq 0$ [为后文利用积分中指定理提供方便]. 再者,
\begin\{aligned\} b > a\Rightarrow& \exists\ m\in\mathbb\{Z\}\_+,\mathrm\{ s.t.\} a+(m-1)T\leq b < a+mT\\\\ \Rightarrow&\mbox\{将 $\displaystyle f$ 在 $\displaystyle [b,a+mT]$ 上零延拓\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而仅需证明
\begin\{aligned\} &\lim\_\{n\to\infty\}\int\_a^\{a+mT\} f(x)g(nx)\mathrm\{ d\} x =\frac\{1\}\{T\} \int\_a^\{a+mT\}f(x)\mathrm\{ d\} x \cdot \int\_0^T g(x)\mathrm\{ d\} x,\ \forall\ m\in\mathbb\{Z\}\_+\\\\ \Leftrightarrow& \lim\_\{n\to\infty\}\int\_c^\{c+T\} f(x)g(nx)\mathrm\{ d\} x =\frac\{1\}\{T\} \int\_c^\{c+T\}f(x)\mathrm\{ d\} x \cdot \int\_0^T g(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
事实上, 设 $\displaystyle x^n\_k=c+\frac\{k\}\{n\}T, M^n\_k=\sup\_\{[x^n\_\{k-1\},x^n\_k]\}f, m^n\_k=\inf\_\{[x^n\_\{k-1\},x^n\_k]\}f$, 则
\begin\{aligned\} &\quad \int\_c^\{c+T\}f(x)g(nx)\mathrm\{ d\} x =\sum\_\{k=1\}^n \int\_\{x^n\_\{k-1\}\}^\{x^n\_k\} f(x) g(nx)\mathrm\{ d\} x\\\\ &=\sum\_\{k=1\}^n f(x^n\_k) \int\_\{x^n\_\{k-1\}\}^\{x^n\_k\}g(nx)\mathrm\{ d\} x +\sum\_\{k=1\}^n \int\_\{x^n\_\{k-1\}\}^\{x^n\_k\} \left\[f(x)-f(x^n\_k)\right\] g(nx)\mathrm\{ d\} x\\\\ &=\sum\_\{k=1\}^n f(x^n\_k) \int\_\{nc+(k-1)T\}^\{nc+kT\} g(t)\frac\{\mathrm\{ d\} t\}\{n\} +\sum\_\{k=1\}^n \int\_\{x^n\_\{k-1\}\}^\{x^n\_k\} \left\[f(x)-f(x^n\_k)\right\] g(nx)\mathrm\{ d\} x\\\\ &=\frac\{1\}\{T\}\int\_0^T g(t)\mathrm\{ d\} t\cdot \sum\_\{k=1\}^n f(x^n\_k) \frac\{T\}\{n\} +\sum\_\{k=1\}^n \int\_\{x^n\_\{k-1\}\}^\{x^n\_k\} \left\[f(x)-f(x^n\_k)\right\] g(nx)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle n\to\infty$, 并注意到
\begin\{aligned\} \lim\_\{n\to\infty\}\sum\_\{k=1\}^n f(x^n\_k) \frac\{T\}\{n\}=\int\_c^\{c+T\}f(x)\mathrm\{ d\} x, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} \left|\sum\_\{k=1\}^n \int\_\{x^n\_\{k-1\}\}^\{x^n\_k\} \left\[f(x)-f(x^n\_k)\right\] g(nx)\mathrm\{ d\} x\right| &\leq \sum\_\{k=1\}^n (M^n\_k-m^n\_k)\int\_\{x^n\_\{k-1\}\}^\{x^n\_k\} |g(nx)|\mathrm\{ d\} x\\\\ &= \sum\_\{k=1\}^n (M^n\_k-m^n\_k) \int\_\{nc+(k-1)T\}^\{nc+kT\} |g(t)|\frac\{\mathrm\{ d\} t\}\{n\}\\\\ &=\frac\{1\}\{T\} \int\_0^T |g(t)|\mathrm\{ d\} t\cdot \sum\_\{k=1\}^n(M^n\_k-m^n\_k) \frac\{T\}\{n\}\\\\ &\to 0\left(n\to\infty, \mbox\{因为 $\displaystyle f$ 可积\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
我们得到
\begin\{aligned\} \lim\_\{n\to\infty\}\int\_c^\{c+T\} f(x)g(nx)\mathrm\{ d\} x =\frac\{1\}\{T\} \int\_c^\{c+mT\}f(x)\mathrm\{ d\} x \cdot \int\_0^T g(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 在第 1 步中取 $\displaystyle g(x)=|\sin x|, T=\pi$, 则
\begin\{aligned\} \lim\_\{n\to\infty\}\int\_a^b f(x)|\sin nx|\mathrm\{ d\} x =&\frac\{1\}\{\pi\}\int\_a^b f(x)\mathrm\{ d\} x\cdot \int\_0^\pi |\sin x|\mathrm\{ d\} x\\\\ =&\frac\{2\}\{\pi\}\int\_a^b f(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
特别地,
\begin\{aligned\} \lim\_\{n\to\infty\}\int\_a^b |\sin nx|\mathrm\{ d\} x=\frac\{2(b-a)\}\{\pi\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
403、 5、 (15 分) 设 $\displaystyle f(x), g(x)$ 为定义在 $\displaystyle [a,b]$ 上的连续函数, $\displaystyle f(x)\geq 0, g(x) > 0$. 记 $\displaystyle M=\max\_\{x\in [a,b]\}f(x)$. 证明:
\begin\{aligned\} \lim\_\{n\to\infty\}\left\[\int\_a^b f^n(x)g(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{n\}=M. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中国人民大学2023年数学分析考研试题) [积分与极限 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} \left\[\int\_a^b f^n(x)g(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{n\}\leq \max\_\{a\leq x\leq b\}f(x)\cdot \left\[\int\_a^b g(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{n\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 [$\{\lim\_\{n\to\infty\}a^\frac\{1\}\{n\}=1,\ a > 0\}$]
\begin\{aligned\} \varlimsup\_\{n\to\infty\}\left\[\int\_a^b f^n(x)g(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{n\} \leq \max\_\{a\leq x\leq b\}f(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle f$ 在 $\displaystyle \xi$ 处取得最大值, 则由保号性知对 $\displaystyle \forall\ \varepsilon > 0$,$\exists\ \delta > 0,\mathrm\{ s.t.\}$
\begin\{aligned\} x\in [c,d]=[\xi,\xi+\delta]\mbox\{ 或 \}[\xi-\delta,\xi]\Rightarrow f(x)\geq \max\_\{a\leq x\leq b\}f(x)-\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而
\begin\{aligned\} \left\[\int\_a^b f^n(x)g(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{n\} &\geq \left\[\int\_c^d f^n(x)g(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{n\}\\\\ &\geq \left\[\max\_\{a\leq x\leq b\}f(x)-\varepsilon\right\]\cdot \left\[\int\_a^b g(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{n\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle n\to\infty$, 有
\begin\{aligned\} \varliminf\_\{n\to\infty\}\left\[\int\_a^b f^n(x)g(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{n\} \geq \max\_\{a\leq x\leq b\}f(x)-\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再令 $\displaystyle \varepsilon\to 0+$, 有
\begin\{aligned\} \varliminf\_\{n\to\infty\}\left\[\int\_a^b f^n(x)g(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{n\}\geq \max\_\{a\leq x\leq b\}f(x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
综上即知 $\displaystyle \lim\_\{n\to\infty\} \left\[\int\_a^b f^n(x)g(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{n\}=\max\_\{a\leq x\leq b\}f(x)$.跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
404、 (4)、 设 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上连续, $\displaystyle \int\_0^1 f(x)\mathrm\{ d\} x=0$. 求证:
\begin\{aligned\} \left\[\int\_0^1 xf(x)\mathrm\{ d\} x\right\]^2\leq\frac\{1\}\{12\}\int\_0^1 f^2(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(大连理工大学2023年数学分析考研试题) [积分法与不等式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 先证明一个结论. 设 $\displaystyle f$ 是 $\displaystyle [0,1]$ 上的连续可微函数, 满足 $\displaystyle f(0)=f(1)=0$. 则
\begin\{aligned\} \left\[\int\_0^1 f(x)\mathrm\{ d\} x\right\]^2\leq \frac\{1\}\{12\}\int\_0^1 |f'(x)|^2\mathrm\{ d\} x, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且等号成立当且仅当 $\displaystyle f(x)=Ax(1-x)$, 其中 $\displaystyle A$ 是常数. 事实上, 设
\begin\{aligned\} F(x)=\int\_0^x f'(t)\mathrm\{ d\} t,\quad G(x)=\int\_1^x f'(t)\mathrm\{ d\} t, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} F(x)=f(x)=G(x),\quad F(0)=0=G(1). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} \int\_0^1 f(x)\mathrm\{ d\} x&=\int\_0^1 F(x)\mathrm\{ d\} x =xF(x)|\_0^1-\int\_0^1 xF'(x)\mathrm\{ d\} x\\\\ &=F(1)-\int\_0^1 xf'(x)\mathrm\{ d\} x =\int\_0^1 (1-x)f'(x)\mathrm\{ d\} x,\\\\ \int\_0^1 f(x)\mathrm\{ d\} x&=\int\_0^1 G(x)\mathrm\{ d\} x =xG(x)|\_0^1-\int\_0^1 xG'(x)\mathrm\{ d\} x\\\\ &=-\int\_0^1 xf'(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
两式相加得
\begin\{aligned\} 2\int\_0^1 f(x)\mathrm\{ d\} x =\int\_0^1 (1-2x)f'(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
利用 Cauchy-Schwarz 不等式知
\begin\{aligned\} &4\left\[\int\_0^1 f(x)\mathrm\{ d\} x\right\]^2\leq \left\[\int\_0^1 (1-2x)f'(x)\mathrm\{ d\} x\right\]^2\\\\ \leq& \int\_0^1 (1-2x)^2\mathrm\{ d\} x\cdot \int\_0^1|f''(x)|^2\mathrm\{ d\} x =\frac\{1\}\{3\}\int\_0^1 |f''(x)|^2\mathrm\{ d\} x, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
且等号成立当且仅当
\begin\{aligned\} f'(x)=A(1-2x)\Leftrightarrow f(x)=Ax(1-x). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 回到题目. 设 $\displaystyle F(x)=\int\_0^x f(t)\mathrm\{ d\} t$, 则 $\displaystyle F(0)=F(1)=0$,
\begin\{aligned\} \mbox\{左端\}=&\left\[\int\_0^1 x\mathrm\{ d\} F(x)\right\]^2 \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\}\left\[\int\_0^1 F(x)\mathrm\{ d\} x\right\]^2\\\\ \stackrel\{\mbox\{第 1 步\}\}\{\leq\}&\frac\{1\}\{12\}\int\_0^1 [F'(x)]^2\mathrm\{ d\} x =\frac\{1\}\{12\}\int\_0^1 f^2(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
405、 10、 已知 $\displaystyle f(x)$ 在 $\displaystyle (0,1)$ 内有连续导函数, 且 $\displaystyle f(0)=f(1)=0$. 求证:
\begin\{aligned\} \left|\int\_0^1 f(x)\mathrm\{ d\} x\right|\leq\frac\{1\}\{4\}\max\_\{0\leq x\leq 1\}|f'(x)|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(东北大学2023年数学分析考研试题) [积分法与不等式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle M=\max\_\{0\leq x\leq 1\}|f'(x)|$, 则
\begin\{aligned\} \left|\int\_0^1 f(x)\mathrm\{ d\} x\right| \leq& \left|\int\_0^\frac\{1\}\{2\}f(x)\mathrm\{ d\} x\right| +\left|\int\_\frac\{1\}\{2\}^1 f(x)\mathrm\{ d\} x\right|\\\\ =&\left|\int\_0^\frac\{1\}\{2\}\int\_0^x f'(t)\mathrm\{ d\} t\mathrm\{ d\} x\right| +\left|\int\_\frac\{1\}\{2\}^1 \int\_x^1 f'(t)\mathrm\{ d\} t\mathrm\{ d\} x\right|\\\\ =&\left|\int\_0^\frac\{1\}\{2\} f'(t)\left(\frac\{1\}\{2\}-t\right)\mathrm\{ d\} t\right| +\left|\int\_\frac\{1\}\{2\}^1 f'(t)\left(t-\frac\{1\}\{2\}\right)\mathrm\{ d\} t\right|\\\\ \leq& M\left\[\int\_0^\frac\{1\}\{2\} \left(\frac\{1\}\{2\}-t\right)\mathrm\{ d\} t +\int\_\frac\{1\}\{2\}^t \left(t-\frac\{1\}\{2\}\right)\mathrm\{ d\} t\right\] =\frac\{1\}\{4\}M. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
406、 3、 (20 分) 设函数 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上可微, 当 $\displaystyle x\in (0,1)$ 时, $\displaystyle 0 < f'(x) < 1$, $\displaystyle f(0)=0$. 证明:
\begin\{aligned\} \left\[\int\_0^1 f(x)\mathrm\{ d\} x\right\]^2 > \int\_0^1 f^3(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(福州大学2023年数学分析考研试题) [积分法与不等式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} F(t)&\equiv \int\_0^t f^3(x)\mathrm\{ d\} x-\left\[\int\_0^t f(x)\mathrm\{ d\} x\right\]^2,\\\\ F'(t)=&f^3(t)-2\int\_0^t f(x)\mathrm\{ d\} x\cdot f(t)=f(t)\left\[f^2(t)-2\int\_0^t f(x)\mathrm\{ d\} x\right\],\\\\ G(t)&\equiv f^2(t)-2\int\_0^t f(x)\mathrm\{ d\} x,\\\\ G'(t)=&2f(t)f'(t)-2f(t) =2f(t)[f'(t)-1]\leq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
逐步往上推, 对 $\displaystyle 0\leq t\leq 1$ 有
\begin\{aligned\} G(t)\leq 0, F'(t)\leq 0, F(t)\leq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
407、 5、 (0-5)、 证明施瓦茨 (Schwarz) 不等式:
\begin\{aligned\} \left\[\int\_a^b f(x)g(x)\mathrm\{ d\} x\right\]^2\leq \int\_a^b f^2(x)\mathrm\{ d\} x \cdot \int\_a^b g^2(x)\mathrm\{ d\} x, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle f,g$ 在 $\displaystyle [a,b]$ 上可积. (湖南大学2023年数学分析考研试题) [积分法与不等式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 若 $\displaystyle \int\_a^b g^2(x)\mathrm\{ d\} x=0$, 则结论自明. 若 $\displaystyle \int\_a^b g^2(x)\mathrm\{ d\} x > 0$, 则由关于 $\displaystyle \lambda\in\mathbb\{R\}$ 的二次函数
\begin\{aligned\} 0\leq&\int\_a^b \left\[f(x)+\lambda g(x)\right\]^2\mathrm\{ d\} x\\\\ =&\int\_a^b f^2(x)\mathrm\{ d\} x+2\lambda \int\_a^b f(x)g(x)\mathrm\{ d\} x+\lambda^2\int\_a^b g^2(x)\mathrm\{ d\} x \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知判别式 $\displaystyle \Delta\leq 0$. 整理即得结论.跟锦数学微信公众号. [在线资料](
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---
408、 (0-6)、 利用施瓦茨不等式证明:
\begin\{aligned\} \left\[\int\_a^b f(x)\sin kx\mathrm\{ d\} x\right\]^2 +\left\[\int\_a^b f(x)\cos kx\mathrm\{ d\} x\right\]^2\leq 1, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle f$ 为 $\displaystyle [a,b]$ 上的非负连续函数, 且 $\displaystyle \int\_a^b f(x)\mathrm\{ d\} x=1$, $\displaystyle k$ 是实数. (湖南大学2023年数学分析考研试题) [积分法与不等式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} &\left\[\int\_a^b f(x)\cos kx\mathrm\{ d\} x\right\]^2+\left\[\int\_a^b f(x)\sin kx\mathrm\{ d\} x\right\]^2\\\\ =&\left\[\int\_a^b \sqrt\{f(x)\}\cdot\sqrt\{f(x)\} \cos kx\mathrm\{ d\} x\right\]^2+\left\[\int\_a^b \sqrt\{f(x)\}\cdot\sqrt\{f(x)\}\sin kx\mathrm\{ d\} x\right\]^2\\\\ \stackrel\{\tiny\mbox\{Schwarz\}\}\{\leq\}& \int\_a^bf(x)\mathrm\{ d\} x\cdot \int\_a^bf(x)\cos^2kx\mathrm\{ d\} x +\int\_a^bf(x)\mathrm\{ d\} x\cdot \int\_a^bf(x)\sin^2kx\mathrm\{ d\} x\\\\ =&\left\[\int\_0^1 f(x)\mathrm\{ d\} x\right\]^2=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
409、 4、 (15 分) 设函数 $\displaystyle f$ 在有界区间 $\displaystyle [a,b]$ 上连续, 且对任意的 $\displaystyle x\in [a,b]$ 有 $\displaystyle f(x) > 0$. 证明:
\begin\{aligned\} \ln \left\[\frac\{1\}\{b-a\}\int\_a^b f(x)\mathrm\{ d\} x\right\]\geq \frac\{1\}\{b-a\}\int\_a^b \ln f(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(华中师范大学2023年数学分析考研试题) [积分法与不等式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle I=\frac\{1\}\{b-a\}\int\_a^b f(x)\mathrm\{ d\} x$, 则由 $\displaystyle g(t)=\ln t\Rightarrow g''(t)=-\frac\{1\}\{t^2\} < 0$ 知 $\displaystyle g$ 是凹函数,
\begin\{aligned\} g(t)\geq g(t\_0)+g'(t\_0)(t-t\_0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle t=f(x), t\_0=I$, 则
\begin\{aligned\} \ln f(x)\geq \ln I+\frac\{1\}\{I\}\left\[f(x)-A\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
关于 $\displaystyle x$ 在 $\displaystyle [a,b]$ 上积分得
\begin\{aligned\} \int\_a^b \ln f(x)\mathrm\{ d\} x\geq (b-a)\ln I. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
410、 (4)、 设 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上二阶可导, 且 $\displaystyle f''(x)\geq 0$. 证明:
\begin\{aligned\} f\left(\frac\{a+b\}\{2\}\right)\leq \frac\{1\}\{b-a\}\int\_a^b f(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(暨南大学2023年数学分析考研试题) [积分法与不等式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle f$ 凸知
\begin\{aligned\} a\leq x\leq b&\Rightarrow f(x)\geq f\left(\frac\{a+b\}\{2\}\right)+f'\left(\frac\{a+b\}\{2\}\right)\left(x-\frac\{a+b\}\{2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
关于 $\displaystyle x$ 在 $\displaystyle [a,b]$ 上积分即得第一个不等式. 再者,
\begin\{aligned\} a\leq x\leq b&\Rightarrow x=\frac\{b-x\}\{b-a\}a+\frac\{x-a\}\{b-a\}b\\\\ &\Rightarrow f(x)\leq \frac\{b-x\}\{b-a\}f(a)+\frac\{x-a\}\{b-a\}f(b). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
关于 $\displaystyle x$ 在 $\displaystyle [a,b]$ 上积分即得第二个不等式.跟锦数学微信公众号. [在线资料](
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---
411、 5、 证明: $\displaystyle \int\_0^\frac\{\pi\}\{2\}\frac\{\sin x\}\{1+x^2\}\mathrm\{ d\} x < \int\_0^\frac\{\pi\}\{2\}\frac\{\cos x\}\{1+x^2\}\mathrm\{ d\} x$. (南京航空航天大学2023年数学分析考研试题) [积分法与不等式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \int\_0^\frac\{\pi\}\{2\}\frac\{\sin x-\cos x\}\{1+x^2\}\mathrm\{ d\} x =&\sqrt\{2\}\int\_0^\frac\{\pi\}\{2\}\frac\{\sin \left(x-\frac\{\pi\}\{4\}\right)\}\{1+x^2\}\mathrm\{ d\} x =\sqrt\{2\}\int\_\{-\frac\{\pi\}\{4\}\}^\frac\{\pi\}\{4\} \frac\{\sin t\}\{1+\left(\frac\{\pi\}\{4\}+t\right)^2\}\mathrm\{ d\} t\\\\ =&\sqrt\{2\}\int\_0^\frac\{\pi\}\{4\} \sin t\left\[ \frac\{1\}\{1+\left(\frac\{\pi\}\{4\}+t\right)^2\} - \frac\{1\}\{1+\left(\frac\{\pi\}\{4\}-t\right)^2\}\right\]\mathrm\{ d\} t < 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
412、 8、 (10 分) 令 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上连续可微且单调递减, 广义积分 $\displaystyle \int\_0^\infty x^3f(x)\mathrm\{ d\} x$ 收敛. 证明:
\begin\{aligned\} \left\[\int\_0^\infty x^2f(x)\mathrm\{ d\} x\right\]^2\leq \frac\{8\}\{9\}\int\_0^\infty xf(x)\mathrm\{ d\} x\cdot \int\_0^\infty x^3f(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(南开大学2023年数学分析考研试题) [积分法与不等式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由 $\displaystyle f$ 递减知 $\displaystyle \lim\_\{x\to+\infty\}f(x)=\inf\_\{x > a\}f(x)\equiv l$ 存在. 事实上, 由下确界定义,
\begin\{aligned\} \{\color\{red\}\forall\ \varepsilon > 0, \exists\ X > a\},\mathrm\{ s.t.\} l\leq f(X) < l+\varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \{\color\{red\}x > X\}&\Rightarrow l\leq f(x)\leq f(X) < l+\varepsilon\\\\ &\Rightarrow \{\color\{red\}|f(x)-l| < \varepsilon\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由 $\displaystyle \int\_a^\infty f(x)\mathrm\{ d\} x$ 收敛及 Cauchy 收敛准则知
\begin\{aligned\} &\forall\ \varepsilon > 0, \exists\ X > a,\mathrm\{ s.t.\} \varepsilon > \left|\int\_X^\{X+1\}f(x)\mathrm\{ d\} x\right|=|f(\xi\_X)|\left(\mbox\{积分中值定理\}\right)\\\\ \Rightarrow&\exists\ \xi\_n\mbox\{严\}\nearrow +\infty,\mathrm\{ s.t.\} |f(\xi\_n)| < \varepsilon\\\\ \Rightarrow&\lim\_\{n\to\infty\}\xi\_n=+\infty, \lim\_\{n\to\infty\}f(\xi\_n)=0\\\\ \Rightarrow&l=\lim\_\{x\to+\infty\}f(x)=\lim\_\{n\to\infty\}f(\xi\_n)=0\left(\mbox\{归结原则\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(3)、 对 $\displaystyle p=1,2,3$,
\begin\{aligned\} \int\_0^\infty x^p f(x)\mathrm\{ d\} x =\frac\{1\}\{p+1\}\int\_0^\infty f(x)\mathrm\{ d\} x^\{p+1\} \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} -\frac\{1\}\{p+1\}\int\_0^\infty f'(x)x^\{p+1\}\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故仅需证明
\begin\{aligned\} &\left\[\frac\{1\}\{3\}\int\_0^\infty x^3f'(x)\mathrm\{ d\} x\right\]^2 \leq \frac\{8\}\{9\}\cdot \frac\{1\}\{2\} \int\_0^\infty x^2f'(x)\mathrm\{ d\} x \cdot\frac\{1\}\{4\}\int\_0^\infty x^4f'(x)\mathrm\{ d\} x\\\\ \Leftrightarrow&\left\[\int\_0^\infty x^3f'(x)\right\]^2 \leq \int\_0^\infty x^2f'(x)\mathrm\{ d\} x\cdot \int\_0^\infty x^4f'(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
事实上,
\begin\{aligned\} &\left\[\int\_0^\infty x^3f'(x)\right\]^2 =\left\[\int\_0^\infty x\sqrt\{-f'(x)\}\cdot x^2 \sqrt\{-f'(x)\}\mathrm\{ d\} x\right\]^2\\\\ \stackrel\{\tiny\mbox\{Schwarz\}\}\{\leq\}&\int\_0^\infty x^2[-f'(x)]\mathrm\{ d\} x\cdot \int\_0^\infty x^4[-f'(x)]\mathrm\{ d\} x\\\\ =&\int\_0^\infty x^2f'(x)\mathrm\{ d\} x\cdot \int\_0^\infty x^4f'(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
413、 3、 (30 分) 证明与分析题. (1)、 (15 分) 设 $\displaystyle f(x,y)$ 在 $\displaystyle D=\left\\{(x,y); x^2+y^2\leq 1\right\\}$ 上非负连续, 且 $\displaystyle |f(x,y)|\leq 1$. 证明:
\begin\{aligned\} \left\[\iint\_D f(x,y)\cos(x+y)\mathrm\{ d\} x\mathrm\{ d\} y\right\]^2 +\left\[\iint\_D f(x,y)\sin(x+y)\mathrm\{ d\} x\mathrm\{ d\} y\right\]^2\leq \pi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(山东大学2023年数学分析考研试题) [积分法与不等式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{左端\}\stackrel\{\tiny\mbox\{Schwarz\}\}\{\leq\}& \iint\_D f^2\mathrm\{ d\} x\mathrm\{ d\} y\cdot \iint\_D \cos^2(x+y)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &+\iint\_D f^2\mathrm\{ d\} x\mathrm\{ d\} y\cdot \iint\_D \sin^2(x+y)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&\iint\_D f^2\mathrm\{ d\} x\mathrm\{ d\} y\leq \iint\_D 1\mathrm\{ d\} x\mathrm\{ d\} y=\pi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
414、 6、 设 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上可微, $\displaystyle 0 < f'(x) < 1$, 且 $\displaystyle f(0)=0$. 证明: 对任意的 $\displaystyle a\in (0,1)$, 有
\begin\{aligned\} \left\[\int\_0^a f(x)\mathrm\{ d\} x\right\]^2 > \int\_0^a f^3(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(上海财经大学2023年数学分析考研试题) [积分法与不等式 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle f' > 0$ 知 $\displaystyle f\mbox\{严\}\nearrow $, 而 $\displaystyle \forall\ 0 < x < 1, f(x) > 0$. 又由 $\displaystyle f'(x) < 1$ 知 $\displaystyle f(x)=f(x)-f(0)\xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Lagrange\}\} f'(\xi)x < x < 1$. 以下所涉及式子都大于 $\displaystyle 0$, 而有
\begin\{aligned\} &\frac\{\displaystyle \left\[\int\_0^a f(x)\mathrm\{ d\} x\right\]^2\}\{\displaystyle \int\_0^a f^3(x)\mathrm\{ d\} x\} \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Cauchy\}\} \frac\{\displaystyle 2\int\_0^\xi f(x)\mathrm\{ d\} x\cdot f(\xi)\}\{f^3(\xi)\}\\\\ =&\frac\{\displaystyle 2\int\_0^\xi f(x)\mathrm\{ d\} x\}\{f^2(\xi)\} \xlongequal[\tiny\mbox\{中值\}]\{\tiny\mbox\{Cauchy\}\} \frac\{2f'(\eta)\}\{2f(\eta)f'(\eta)\} =\frac\{1\}\{f(\eta)\} > 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
发表于 2023-3-5 08:50:34
