## 张祖锦2023年数学专业真题分类70天之第17天
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369、 5、 (15 分) 求定积分 $\displaystyle \int\_0^\{n\pi\}\sqrt\{1+\sin 2x\}\mathrm\{ d\} x$, $\displaystyle n$ 是正整数. (云南大学2023年数学分析考研试题) [定积分 ]
[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\{原式\}=&\int\_0^\{n\pi\}\sqrt\{(\sin x+\cos x)^2\}\mathrm\{ d\} x =\int\_0^\{n\pi\} |\sin x+\cos x|\mathrm\{ d\} x\\\\ =&\sqrt\{2\}\int\_0^\{n\pi\}\left|\sin \left(x+\frac\{\pi\}\{4\}\right)\right|\mathrm\{ d\} x =\sqrt\{2\}\int\_\frac\{\pi\}\{4\}^\{n\pi+\frac\{\pi\}\{4\}\} |\sin t|\mathrm\{ d\} t\\\\ =&\sqrt\{2\}\left\[\int\_\frac\{\pi\}\{4\}^0+\int\_0^\{n\pi\}+\int\_\{n\pi\}^\{n\pi+\frac\{\pi\}\{4\}\} |\sin t|\mathrm\{ d\} t\right\]\\\\ =&\sqrt\{2\}\int\_0^\{n\pi\}|\sin t|\mathrm\{ d\} t =\sqrt\{2\}\cdot 2n\int\_0^\frac\{\pi\}\{2\}\sin t\mathrm\{ d\} t =2n\sqrt\{2\}. \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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370、 (2)、 若函数 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上连续, 则 $\displaystyle f(x)\equiv 0$ 是 $\displaystyle \int\_a^b |f(x)|\mathrm\{ d\} x=0$ 的 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$ 条件.
A. 充分必要
B. 充分且必要
C. 必要且充分
D. 既不充分也不必要 (长安大学2023年数学分析考研试题) [定积分 ]
[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) / $\displaystyle A$.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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371、 (3)、 设 $\displaystyle f(x)$ 为连续函数, 则下列函数必为偶函数的是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$.
A. $\displaystyle \int\_0^x f(t^2)\mathrm\{ d\} t$
B. $\displaystyle \int\_0^x f^2(t)\mathrm\{ d\} t$
C. $\displaystyle \int\_0^x t[f(t)-f(-t)]\mathrm\{ d\} t$
D. $\displaystyle \int\_0^x t[f(t)+f(-t)]\mathrm\{ d\} t$ (长安大学2023年数学分析考研试题) [定积分 ]
[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) / $\displaystyle D$. 设 $\displaystyle g(x)=\int\_0^x t[f(t)+f(-t)]\mathrm\{ d\} t$, 则
\begin\{aligned\} g(-x)=&\int\_0^\{-x\} t[f(t)+f(-t)]\mathrm\{ d\} t\\\\ \stackrel\{t=-s\}\{=\}&\int\_0^x (-s)[f(-s)+f(s)] (-\mathrm\{ d\} s)=g(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)/
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372、 (3)、 (10 分) 求曲线 $\displaystyle y=\sum\_\{n=1\}^\infty \frac\{(-1)^\{n+1\}x^\{2n\}\}\{(2n-1)!\}\ (0\leq x\leq \pi)$ 与 $\displaystyle x$ 轴所围成的图形绕 $\displaystyle x$ 轴旋转一周所得旋转体的体积. (长安大学2023年数学分析考研试题) [定积分 ]
[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\} y=x\sum\_\{n=1\}^\infty \frac\{(-1)^\{n-1\} x^\{2n-1\}\}\{(2n-1)!\}=x\sum\_\{k=0\}^\infty \frac\{(-1)^k x^\{2k+1\}\}\{(2k+1)!\}=x\sin x \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知体积
\begin\{aligned\} V=&\int\_0^\{2\pi\} \pi y^2\mathrm\{ d\} x =\pi \int\_0^\{2\pi\} x^2\sin^2x\mathrm\{ d\} x\\\\ =&\frac\{\pi\}\{2\}\int\_0^\{2\pi\} x^2(1-\cos 2x)\mathrm\{ d\} x=\frac\{\pi^2(8\pi^2-3)\}\{6\}. \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)/
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373、 11、 (10 分) 已知 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上可微, 当 $\displaystyle 0\leq x < 1$ 时, 有 $\displaystyle 0 < f(1) < f(x), f'(x)\neq f(x)$. 证明: 存在点 $\displaystyle \xi\in (0,1)$, 使得 $\displaystyle f(\xi)=\int\_0^\xi f(t)\mathrm\{ d\} t$. (中国矿业大学(北京)2023年数学分析考研试题) [定积分 ]
[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 $\displaystyle F(x)=\int\_0^x f(t)\mathrm\{ d\} t-f(x)$, 则
\begin\{aligned\} F(0)=-f(0) < 0, F(1)=\int\_0^1 f(t)\mathrm\{ d\} t-f(1)=\int\_0^1[f(x)-f(1)]\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\}
由连续函数介值定理知 $\displaystyle \exists\ \xi\in (0,1),\mathrm\{ s.t.\} 0=F(\xi)\Rightarrow f(\xi)=\int\_0^\xi f(t)\mathrm\{ d\} t$.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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374、 3、 设 $\displaystyle f,g$ 均为定义在 $\displaystyle [a,b]$ 上的有界函数, 仅在 $\displaystyle [a,b]$ 上有限个点处 $\displaystyle f(x)\neq g(x)$. 证明: 若 $\displaystyle f$ 在 $\displaystyle [a,b]$ 上可积, 则 $\displaystyle g$ 在 $\displaystyle [a,b]$ 上也可积, 且 $\displaystyle \int\_a^b f(x)\mathrm\{ d\} x=\int\_a^b g(x)\mathrm\{ d\} x$. (中国矿业大学(徐州)2023年数学分析考研试题) [定积分 ]
[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 $\displaystyle \left\\{x\in [a,b]; f(x)\neq g(x)\right\\}=\left\\{c\_1,\cdots,c\_m\right\\}$. 由题设,
\begin\{aligned\} \exists\ M > 0,\mathrm\{ s.t.\} \forall\ x\in [a,b], |f(x)|\leq M, |g(x)|\leq M. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle f\in R[a,b]$ 知 $\displaystyle \forall\ \varepsilon > 0,\exists\ \delta\in \left(0,\frac\{\varepsilon\}\{8mM\}\right)$, 只要 $\displaystyle [a,b]$ 的分划 $\displaystyle T$ 满足 $\displaystyle \left\Vert T\right\Vert < \delta$, 就有 $\displaystyle \sum\_T \omega^f\_i\Delta x\_i < \frac\{\varepsilon\}\{2\}$, 其中 $\displaystyle \omega^f\_i$ 是 $\displaystyle f$ 在第 $\displaystyle i$ 个小区间上的振幅. 再设 $\displaystyle T'=T\cup \left\\{c\_1,\cdots,c\_m\right\\}$, 则将 $\displaystyle T'$ 分成两部分 $\displaystyle T''\cup T'''$, 其中 $\displaystyle T''$ 中的小区间都是 $\displaystyle T$ 中的小区间, $\displaystyle T'''=T'\backslash T''$ 至多有 $\displaystyle 2m$ 个分点. 于是
\begin\{aligned\} &\sum\_\{T'\}\omega^g\_\{T'\}\Delta x\_i' =\sum\_\{T''\}\omega^g\_\{i''\}\Delta x\_\{i''\} +\sum\_\{T'''\}\omega^g\_\{i'''\}\Delta x\_\{i'''\}\\\\ =&\sum\_\{T''\}\omega^f\_\{i''\}\Delta x\_\{i''\} +\sum\_\{T'''\}\omega^g\_\{i'''\}\Delta x\_\{i'''\} \leq \sum\_\{T\}\omega^f\_\{i\}\Delta x\_\{i\} +\sum\_\{T'''\}\omega^g\_\{i'''\}\Delta x\_\{i'''\}\\\\ < &\frac\{\varepsilon\}\{2\}+2M\cdot \delta \cdot 2m < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故而 $\displaystyle g\in R[a,b]$, 且令 $\displaystyle c\_0=a, c\_\{m+1\}=b$ 后,
\begin\{aligned\} \int\_a^b f(x)\mathrm\{ d\} x=\sum\_\{i=0\}^m \int\_\{c\_i\}^\{c\_\{i+1\}\}f(x)\mathrm\{ d\} x =\sum\_\{i=0\}^m \int\_\{c\_i\}^\{c\_\{i+1\}\}g(x)\mathrm\{ d\} x=\int\_a^b 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\}
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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375、 (2)、 求将曲线
\begin\{aligned\} (x-b)^2+y^2=a^2\left(0 < a\leq b\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle y$ 轴旋转一周所得旋转曲面的面积. (中南大学2023年数学分析考研试题) [定积分 ]
[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) / 旋转体表面积为
\begin\{aligned\} S=&2\int\_0^a 2\pi \left(b+\sqrt\{a^2-y^2\}\right) \sqrt\{1+\left\[ \frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} y\} \left(b+\sqrt\{a^2-y^2\}\right) \right\]^2\}\mathrm\{ d\} y\\\\ &+2\int\_0^a 2\pi \left(b-\sqrt\{a^2-y^2\}\right) \sqrt\{1+\left\[ \frac\{\mathrm\{ d\} \}\{\mathrm\{ d\} y\} \left(b-\sqrt\{a^2-y^2\}\right) \right\]^2\}\mathrm\{ d\} y\\\\ =&2\pi a(2a+b\pi)+2\pi a(b\pi-2a)=4\pi^2ab. \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)/
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376、 (2)、 (8 分) $\displaystyle \int\_0^\frac\{\pi\}\{2\}\frac\{\cos x\}\{\sin x+\cos x\}\mathrm\{ d\} x$. (中山大学2023年数学分析考研试题) [定积分 ]
[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\{原式\}&\stackrel\{\frac\{\pi\}\{2\}-x=t\}\{=\}\int\_0^\frac\{\pi\}\{2\} \frac\{\sin t\}\{\cos t+\sin 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\}
将上式取算术平均, 我们发现
\begin\{aligned\} \mbox\{原式\}&=\frac\{1\}\{2\}\left\[\int\_0^\frac\{\pi\}\{2\}\frac\{\cos x\}\{\sin x+\cos x\}\mathrm\{ d\} x +\int\_0^\frac\{\pi\}\{2\} \frac\{\sin t\}\{\cos t+\sin t\}\mathrm\{ d\} t \right\]\\\\ &=\frac\{1\}\{2\}\int\_0^\frac\{\pi\}\{2\}1\mathrm\{ d\} x=\frac\{\pi\}\{4\}. \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)/
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377、 (3)、 曲线 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}x=a(t-\sin t),\\\\ y=a(1-\cos t)\end\{array\}\right.\ (0\leq t\leq 2\pi)$ 的弧长为 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (重庆师范大学2023年数学分析考研试题) [定积分 ]
[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) / 弧长
\begin\{aligned\} l=\int\_0^\{2\pi\}\sqrt\{x'^2(t)+y'^2(t)\}\mathrm\{ d\} t =\int\_0^\{2\pi\} \sqrt\{2\}a\sqrt\{1+\cos t\}\mathrm\{ d\} t =8a. \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)/
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378、 (4)、 已知 $\displaystyle f$ 在 $\displaystyle [0,\pi]$ 上二阶连续可导, 且
\begin\{aligned\} f(\pi)=2, \int\_0^\pi \left\[f(x)+f''(x)\right\]\sin x\mathrm\{ d\} x=5. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle f(0)$. (重庆师范大学2023年数学分析考研试题) [定积分 ]
[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\} 5=&\int\_0^\pi f(x)\sin x\mathrm\{ d\} x+\int\_0^\pi \sin x\mathrm\{ d\} f'(x)\\\\ =&\int\_0^\pi f(x)\sin x\mathrm\{ d\} x-\int\_0^\pi \cos xf'(x)\mathrm\{ d\} x\\\\ =&\int\_0^\pi f(x)\sin x\mathrm\{ d\} x-\int\_0^\pi \cos x\mathrm\{ d\} f(x)\\\\ =&-\int\_0^\pi f(x)\mathrm\{ d\} \cos x-\int\_0^\pi \cos x\mathrm\{ d\} f(x)\\\\ =&-\int\_0^\pi \mathrm\{ d\} (f(x)\cos x) =-[-f(\pi)-f(0)]=f(0)+2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle f(0)=3$.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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379、 10、 函数 $\displaystyle f(x)$ 在区间 $\displaystyle [a,b]$ 上可积, 证明: $\displaystyle \lim\_\{p\to\infty\}\int\_a^b f(x)\cos(px)\mathrm\{ d\} x=0$. (北京工业大学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 $\displaystyle I(p)\equiv \int\_a^b f(x)\cos(px)\mathrm\{ d\} x$ 关于 $\displaystyle p$ 是偶函数知只要证明 $\displaystyle \lim\_\{p\to+\infty\}I(p)=0$. 由 $\displaystyle f\in R[a,b]$ 知 $\displaystyle f$ 有界:
\begin\{aligned\} \exists\ M > 0,\mathrm\{ s.t.\} \forall\ a\leq x\leq b, |f(x)|\leq M. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
记 $\displaystyle n=[\sqrt\{p\}]$, 则 $\displaystyle \lim\_\{p\to+\infty\}n=+\infty$. 将 $\displaystyle [a,b]$ $\displaystyle n$ 等分, 并记分点
\begin\{aligned\} x\_i=a+\frac\{i\}\{n\}(b-a), i=0,1,\cdots,n, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
记 $\displaystyle \omega\_i$ 为 $\displaystyle f$ 在 $\displaystyle [x\_\{i-1\},x\_i]$ 上的振幅. 于是
\begin\{aligned\} f\in R[a,b]\Rightarrow \lim\_\{n\to\infty\}\sum\_\{i=1\}^n \omega\_i\Delta x\_i=0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \Delta x\_i=x\_i-x\_\{i-1\}$. 注意到
\begin\{aligned\} \left|\int\_\{x\_\{i-1\}\}^\{x\_i\} \cos p x\mathrm\{ d\} x\right|=\frac\{1\}\{p\}|\sin (p x\_i)-\sin (p x\_\{i-1\})|\leq \frac\{2\}\{p\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
我们有
\begin\{aligned\} &\left|\int\_a^b f(x)\sin p x\mathrm\{ d\} x\right| =\left|\sum\_\{i=1\}^n\int\_\{x\_\{i-1\}\}^\{x\_i\} f(x)\cos p x\mathrm\{ d\} x\right|\\\\ =&\left|\sum\_\{i=1\}^n \int\_\{x\_\{i-1\}\}^\{x\_i\} [f(x)-f(x\_\{i-1\})]\cos p x\mathrm\{ d\} x +\int\_\{x\_\{i-1\}\}^\{x\_i\} f(x\_\{i-1\})\cos p x\mathrm\{ d\} x\right|\\\\ \leq&\sum\_\{i=1\}^n \omega\_i\Delta x\_i +\frac\{2n\}\{p\}M =\sum\_\{i=1\}^n \omega\_i\Delta x\_i+\frac\{2[\sqrt\{p\}]\}\{p\}M\\\\ \leq& \sum\_\{i=1\}^n \omega\_i\Delta x\_i+\frac\{2\}\{\sqrt\{p\}\}M\xrightarrow\{p\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\}
这就证明了 $\displaystyle \lim\_\{p\to+\infty\}I(p)=0$. 结论得证.跟锦数学微信公众号. [在线资料](
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---
380、 6、 (10 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上有定义, 对任意 $\displaystyle b > 0$, $\displaystyle f(x)$ 在 $\displaystyle [0,b]$ 上可积且 $\displaystyle \lim\_\{x\to+\infty\}f(x)=a$. 若 $\displaystyle \varphi(t)$ 在 $\displaystyle [0,+\infty)$ 上非负连续且 $\displaystyle \int\_0^\infty \varphi(t)\mathrm\{ d\} t=1$. 证明:
\begin\{aligned\} \lim\_\{t\to 0^+\}t\int\_0^\infty \varphi(tx)f(x)\mathrm\{ d\} x=a. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(北京科技大学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) / 即要证
\begin\{aligned\} &\lim\_\{t\to 0^+\}t\int\_0^\infty \varphi(tx)f(x)\mathrm\{ d\} x=a=a\int\_0^\infty \varphi(s)\mathrm\{ d\} s \stackrel\{s=tx\}\{=\}at\int\_0^\infty \varphi(tx)\mathrm\{ d\} x\\\\ \Leftrightarrow&\lim\_\{t\to 0^+\}t\int\_0^\infty\varphi(tx)[f(x)-a]\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\}
写出
\begin\{aligned\} &\left|t\int\_0^\infty \varphi(tx)[f(x)-a]\mathrm\{ d\} x\right|\\\\ \leq& t\int\_0^X \varphi(tx)|f(x)-a|\mathrm\{ d\} x +t\int\_X^\infty \varphi(tx)|f(x)-a|\mathrm\{ d\} x\\\\ \leq&\sup\_\{x\in [0,X]\}|f(x)-a|\cdot t\int\_0^X \varphi(tx)\mathrm\{ d\} x +\sup\_\{x\in [X,+\infty)\}|f(x)-a|\cdot t\int\_X^\infty \varphi(tx)\mathrm\{ d\} x\\\\ \stackrel\{tx=s\}\{\leq\}&\sup\_\{x\in [0,X]\}|f(x)-a|\cdot \int\_0^\{tX\}\varphi(s)\mathrm\{ d\} s +\sup\_\{x\in [X,+\infty)\}|f(x)-a|\cdot t\int\_0^\infty \varphi(tx)\mathrm\{ d\} x\\\\ =&\sup\_\{x\in [0,X]\}|f(x)-a|\cdot \int\_0^\{tX\}\varphi(s)\mathrm\{ d\} s +\sup\_\{x\in [X,+\infty)\}|f(x)-a|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle t\to 0^+$ 得
\begin\{aligned\} \varlimsup\_\{t\to 0^+\}\left|t\int\_0^\infty \varphi(tx)[f(x)-a]\mathrm\{ d\} x\right|\leq \sup\_\{x\in [X,+\infty)\}|f(x)-a|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再令 $\displaystyle X\to+\infty$ 得
\begin\{aligned\} &0\leq \varliminf\_\{t\to 0^+\}\left|t\int\_0^\infty \varphi(tx)[f(x)-a]\mathrm\{ d\} x\right| \leq \varlimsup\_\{t\to 0^+\}\left|t\int\_0^\infty \varphi(tx)[f(x)-a]\mathrm\{ d\} x\right|\leq 0\\\\ \Rightarrow&\lim\_\{t\to 0^+\}\left|t\int\_0^\infty \varphi(tx)[f(x)-a]\mathrm\{ d\} x\right|=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故有结论.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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---
381、 4、 (15 分) 已知 $\displaystyle a\_n=\int\_0^1 \frac\{x^n\}\{1+x\}\mathrm\{ d\} x$. 证明:
\begin\{aligned\} a\_n=\frac\{1\}\{2n\}+\frac\{1\}\{4n^2\}+o\left(\frac\{1\}\{n^2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(北京师范大学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\} &\lim\_\{n\to\infty\}\left(n^2a\_n-\frac\{n\}\{2\}\right) =\lim\_\{n\to\infty\}\left\[n^2\int\_0^1 \frac\{x^n\}\{1+x\}\mathrm\{ d\} x-\frac\{n\}\{2\}\right\]\\\\ =&\lim\_\{n\to\infty\} n\left\[n\int\_0^1 x^\{n-1\}\frac\{x\}\{1+x\}\mathrm\{ d\} x-\frac\{1\}\{2\}\right\] =\lim\_\{n\to\infty\}n \int\_0^1 nx^\{n-1\}\left(\frac\{x\}\{1+x\}-\frac\{1\}\{2\}\right)\mathrm\{ d\} x\\\\ =&\lim\_\{n\to\infty\}n \int\_0^1 \left(\frac\{x\}\{1+x\}-\frac\{1\}\{2\}\right) \mathrm\{ d\} x^n \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \lim\_\{n\to\infty\} n\int\_0^1 \frac\{x^n\}\{(1+x)^2\}\mathrm\{ d\} x\\\\ =&\lim\_\{n\to\infty\}(n+1)\int\_0^1 \frac\{x^n\}\{(1+x)^2\}\mathrm\{ d\} x =\lim\_\{n\to\infty\}\frac\{1\}\{(1+x)^2\}\mathrm\{ d\} x^\{n+1\}\\\\ \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\}&\frac\{1\}\{4\}+2\int\_0^1 \frac\{x^\{n+1\}\}\{(1+x)^3\}\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\}
由
\begin\{aligned\} &\int\_0^1 \frac\{x^\{n+1\}\}\{(1+x)^3\}\mathrm\{ d\} x=\int\_0^\{1-\delta\}+\int\_\{1-\delta\}^1\cdots \leq (1-\delta)^\{n+1\}+\delta\\\\ \stackrel\{n\to\infty\}\{\Rightarrow\}&\varlimsup\_\{n\to\infty\}\int\_0^1 \frac\{x^\{n+1\}\}\{(1+x)^3\}\mathrm\{ d\} x\leq \delta\\\\ \stackrel\{\delta\to 0^+\}\{\Rightarrow\}&0\leq \varliminf\_\{n\to\infty\}\int\_0^1 \frac\{x^\{n+1\}\}\{(1+x)^3\}\mathrm\{ d\} x \leq\varlimsup\_\{n\to\infty\}\int\_0^1 \frac\{x^\{n+1\}\}\{(1+x)^3\}\mathrm\{ d\} x\leq 0\\\\ \Rightarrow&\lim\_\{n\to\infty\}\int\_0^1 \frac\{x^\{n+1\}\}\{(1+x)^3\}\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\}
即知
\begin\{aligned\} &\lim\_\{n\to\infty\}\left(n^2a\_n-\frac\{n\}\{2\}\right)=\frac\{1\}\{4\} \Rightarrow na\_n-\frac\{n\}\{2\}=\frac\{1\}\{4\}+o(1)\\\\ \Rightarrow& a\_n=\frac\{1\}\{2n\}+\frac\{1\}\{4n^2\}+o\left(\frac\{1\}\{n^2\}\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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---
382、 3、 证明题. (1)、 设 $\displaystyle f: \mathbb\{R\}^n\to\mathbb\{R\}$ 连续, $\displaystyle \lim\_\{|x|\to+\infty\}f(x)=0$, 其中
\begin\{aligned\} x=(x\_1,\cdots,x\_n), |x|=\sqrt\{x\_1^2+\cdots+x\_n^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求证:
\begin\{aligned\} \lim\_\{n\to\infty\}\int\_\{[0,1]^n\} f(kx)\mathrm\{ d\} x\_1\cdots \mathrm\{ d\} x\_n=0. \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 $\displaystyle \lim\_\{|x|\to+\infty\}f(x)=0$ 知
\begin\{aligned\} \exists\ X > 0,\mathrm\{ s.t.\} \forall\ x: |x|\geq X, |f(x)|\leq 1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又由 $\displaystyle f$ 在 $\displaystyle |x|\leq X$ 上连续知
\begin\{aligned\} \exists\ M > 0,\mathrm\{ s.t.\} \forall\ x: |x|\leq X, |f(x)|\leq M. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是 $\displaystyle f$ 在 $\displaystyle \mathbb\{R\}^n$ 上有界, 且 $\displaystyle A=M+1$ 是一个界. 写出
\begin\{aligned\} \left|\int\_\{[0,1]^n\}f(kx)\mathrm\{ d\} x\right| \leq& \int\_\{[0,\delta]^n\}|f(kx)|\mathrm\{ d\} x +\int\_\{[0,1]^n: \exists\ i,\mathrm\{ s.t.\} x\_i > \delta\} |f(kx)|\mathrm\{ d\} x\\\\ \leq&A\delta^n+ +\int\_\{[0,1]^n: \exists\ i,\mathrm\{ s.t.\} x\_i > \delta\} |f(kx)|\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 $\displaystyle k\to\infty$, 并注意到 $\displaystyle kx\_i\geq k\delta\to+\infty$, 得
\begin\{aligned\} \varlimsup\_\{k\to\infty\}\left|\int\_\{[0,1]^n\}f(kx)\mathrm\{ d\} x\right|\leq A\delta^n. \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\_\{k\to\infty\}\left|\int\_\{[0,1]^n\}f(kx)\mathrm\{ d\} x\right|\leq \varlimsup\_\{k\to\infty\}\left|\int\_\{[0,1]^n\}f(kx)\mathrm\{ d\} x\right| \leq 0\\\\ \Rightarrow& \lim\_\{k\to\infty\}\int\_\{[0,1]^n\}f(kx)\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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
383、 6、 (15 分) 证明: $\displaystyle \lim\_\{n\to\infty\}\int\_0^\frac\{\pi\}\{2\}\sin^\frac\{1\}\{n\}x\mathrm\{ d\} x=\frac\{\pi\}\{2\}$. (合肥工业大学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) / 对 $\displaystyle \forall\ 0 < \delta < 1$,
\begin\{aligned\} &\frac\{\pi\}\{2\}\geq \int\_0^\frac\{\pi\}\{2\}\sin^\frac\{1\}\{n\}x\mathrm\{ d\} x \geq \int\_\delta^\frac\{\pi\}\{2\}\sin^\frac\{1\}\{n\}x\mathrm\{ d\} x \geq \left(\frac\{\pi\}\{2\}-\delta\right)\sin^\frac\{1\}\{n\}\delta\\\\ \stackrel\{n\to\infty\}\{\Rightarrow\}&\frac\{\pi\}\{2\}\geq \varlimsup\_\{n\to\infty\}\int\_0^\frac\{\pi\}\{2\}\sin^\frac\{1\}\{n\}x\mathrm\{ d\} x \geq \varliminf\_\{n\to\infty\}\int\_0^\frac\{\pi\}\{2\}\sin^\frac\{1\}\{n\}x\mathrm\{ d\} x\geq \frac\{\pi\}\{2\}-\delta\\\\ \stackrel\{\delta\to 0^+\}\{\Rightarrow\}&\frac\{\pi\}\{2\}\geq \varlimsup\_\{n\to\infty\}\int\_0^\frac\{\pi\}\{2\}\sin^\frac\{1\}\{n\}x\mathrm\{ d\} x \geq \varliminf\_\{n\to\infty\}\int\_0^\frac\{\pi\}\{2\}\sin^\frac\{1\}\{n\}x\mathrm\{ d\} x\geq \frac\{\pi\}\{2\}\\\\ \Rightarrow&\lim\_\{n\to\infty\}\int\_0^\frac\{\pi\}\{2\}\sin^\frac\{1\}\{n\}x\mathrm\{ d\} x=\frac\{\pi\}\{2\}. \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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---
384、 (3)、 非负函数 $\displaystyle f(x)$ 满足 $\displaystyle \int\_a^\infty f(x)\mathrm\{ d\} x$ 收敛, 则 $\displaystyle \lim\_\{x\to+\infty\}f(x)=0$. (河海大学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) / $\displaystyle \times$. 比如对函数
\begin\{aligned\} f(x)=\left\\{\begin\{array\}\{llllllllllll\}1-2^n|x-n|,&|x-n|\leq\frac\{1\}\{2^n\},\\\\ 0,&\mbox\{其他\} \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\}
有
\begin\{aligned\} \int\_0^\{+\infty\}f(x)\mathrm\{ d\} x=\sum\_\{n=1\}^\infty\frac\{1\}\{2^n\}=1,\quad \lim\_\{n\to\infty\}f(n)=1\neq 0=\lim\_\{n\to\infty\}f\left(n\pm\frac\{1\}\{2\}\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](
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---
385、 (4)、 解答如下问题: (4-1)、 证明: $\displaystyle \forall\ \delta\in \left(0,\frac\{\pi\}\{2\}\right), \lim\_\{n\to\infty\}\frac\{\displaystyle\int\_0^\{\frac\{\pi\}\{2\}-\delta\}\sin^nx\mathrm\{ d\} x\}\{\displaystyle\int\_0^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\}=0$; (4-2)、 设 $\displaystyle f(x)\in C\left\[0,\frac\{\pi\}\{2\}\right\]$, 证明: $\displaystyle \lim\_\{n\to\infty\}\frac\{\displaystyle\int\_0^\frac\{\pi\}\{2\}f(x)\sin^nx\mathrm\{ d\} x\}\{\displaystyle\int\_0^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\}=f\left(\frac\{\pi\}\{2\}\right)$. (河海大学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) / (4-1)、 由
\begin\{aligned\} 0& < \frac\{\displaystyle\int\_0^\{\frac\{\pi\}\{2\}-\delta\}\sin^nx\mathrm\{ d\} x\}\{\displaystyle\int\_0^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\} < \frac\{\displaystyle\int\_0^\{\frac\{\pi\}\{2\}-\delta\}\sin^nx\mathrm\{ d\} x\}\{\displaystyle\int\_\{\frac\{\pi\}\{2\}-\delta\}^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\}\\\\ &\leq \frac\{\displaystyle\sin^n\left(\frac\{\pi\}\{2\}-\delta\right)\cdot \left(\frac\{\pi\}\{2\}-\delta\right)\}\{\displaystyle\sin^n\left(\frac\{\pi\}\{2\}-\frac\{\delta\}\{2\}\right)\cdot \frac\{\delta\}\{2\}\} =\frac\{\pi-2\delta\}\{\delta\}\left\[\frac\{\displaystyle\sin\left(\frac\{\pi\}\{2\}-\delta\right)\}\{\displaystyle\sin\left(\frac\{\pi\}\{2\}-\frac\{\delta\}\{2\}\right)\}\right\]^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\}
及迫敛性知结论成立. (4-2)、 由
\begin\{aligned\} &\left|\frac\{\displaystyle\int\_0^\frac\{\pi\}\{2\}f(x)\sin^nx\mathrm\{ d\} x\}\{\displaystyle\int\_0^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\}-f\left(\frac\{\pi\}\{2\}\right)\right| =\left|\frac\{\displaystyle\int\_0^\frac\{\pi\}\{2\}\left\[f(x)-f\left(\frac\{\pi\}\{2\}\right)\right\]\sin^nx\mathrm\{ d\} x\}\{\displaystyle\int\_0^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\}\right|\\\\ =&\left|\frac\{\displaystyle\int\_0^\{\frac\{\pi\}\{2\}-\delta\}+\int\_\{\frac\{\pi\}\{2\}-\delta\}^\frac\{\pi\}\{2\}\left\[f(x)-f\left(\frac\{\pi\}\{2\}\right)\right\]\sin^nx\mathrm\{ d\} x\}\{\displaystyle\int\_0^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\}\right|\\\\ \leq& 2\max\_\{\left\[0,\frac\{\pi\}\{2\}\right\]\}|f|\cdot \frac\{\displaystyle\int\_0^\{\frac\{\pi\}\{2\}-\delta\}\sin^nx\mathrm\{ d\} x\}\{\displaystyle\int\_0^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\} +\max\_\{\left\[\frac\{\pi\}\{2\}-\delta,\frac\{\pi\}\{2\}\right\]\}\left|f(x)-f\left(\frac\{\pi\}\{2\}\right)\right| \frac\{\displaystyle\int\_\{\frac\{\pi\}\{2\}-\delta\}^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\}\{\displaystyle\int\_0^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\}\\\\ \leq& 2\max\_\{\left\[0,\frac\{\pi\}\{2\}\right\]\}|f|\cdot \frac\{\displaystyle\int\_0^\{\frac\{\pi\}\{2\}-\delta\}\sin^nx\mathrm\{ d\} x\}\{\displaystyle\int\_0^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\} +\max\_\{\left\[\frac\{\pi\}\{2\}-\delta,\frac\{\pi\}\{2\}\right\]\}\left|f(x)-f\left(\frac\{\pi\}\{2\}\right)\right|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle n\to\infty$, 并利用第 1 步知
\begin\{aligned\} 0\leq \varlimsup\_\{n\to\infty\}\left|\frac\{\displaystyle\int\_0^\frac\{\pi\}\{2\}f(x)\sin^nx\mathrm\{ d\} x\}\{\displaystyle\int\_0^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\}-f\left(\frac\{\pi\}\{2\}\right)\right| \leq \max\_\{\left\[\frac\{\pi\}\{2\}-\delta,\frac\{\pi\}\{2\}\right\]\}\left|f(x)-f\left(\frac\{\pi\}\{2\}\right)\right|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再令 $\displaystyle \delta\to 0^+$, 并利用 $\displaystyle f$ 的连续性知
\begin\{aligned\} 0\leq \varlimsup\_\{n\to\infty\}\left|\frac\{\displaystyle\int\_0^\frac\{\pi\}\{2\}f(x)\sin^nx\mathrm\{ d\} x\}\{\displaystyle\int\_0^\frac\{\pi\}\{2\}\sin^nx\mathrm\{ d\} x\}-f\left(\frac\{\pi\}\{2\}\right)\right| \leq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故有结论.跟锦数学微信公众号. [在线资料](
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---
386、 11、 设 $\displaystyle \alpha > 0$. 试证:
\begin\{aligned\} \lim\_\{n\to\infty\}n \left\[\frac\{1\}\{n\}\sum\_\{k=1\}^n \left(\frac\{k\}\{n\}\right)^\alpha-\frac\{1\}\{1+\alpha\}\right\]=\frac\{1\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(黑龙江大学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 $\displaystyle f(x)=x^\alpha$, 则
\begin\{aligned\} \mbox\{左端\}=&\lim\_\{n\to\infty\} n\left\[\frac\{1\}\{n\} \sum\_\{k=1\}^n f\left(\frac\{k\}\{n\}\right)-\int\_0^1 f(x)\mathrm\{ d\} x\right\]\\\\ =&\lim\_\{n\to\infty\} n \sum\_\{k=1\}^n \int\_\frac\{k-1\}\{n\}^\frac\{k\}\{n\} \left\[f\left(\frac\{k\}\{n\}\right)-f(x)\right\]\mathrm\{ d\} x =\lim\_\{n\to\infty\} n\sum\_\{k=1\}^n \int\_\frac\{k-1\}\{n\}^\frac\{k\}\{n\} \int\_x^\frac\{k\}\{n\}f'(t)\mathrm\{ d\} t\mathrm\{ d\} x\\\\ =&\lim\_\{n\to\infty\} n\sum\_\{k=1\}^n \int\_\frac\{k-1\}\{n\}^\frac\{k\}\{n\} \int\_\frac\{k-1\}\{n\}^t f'(t)\mathrm\{ d\} x\mathrm\{ d\} t =\lim\_\{n\to\infty\} n\sum\_\{k=1\}^n \int\_\frac\{k-1\}\{n\}^\frac\{k\}\{n\} f'(t)\left(t-\frac\{k-1\}\{n\}\right)\mathrm\{ d\} t\\\\ \xlongequal[\tiny\mbox\{分中值\}]\{\tiny\mbox\{第一积\}\}&\lim\_\{n\to\infty\} n\sum\_\{k=1\}^n f'(\xi\_k) \int\_\frac\{k-1\}\{n\}^\frac\{k\}\{n\} \left(t-\frac\{k-1\}\{n\}\right)\mathrm\{ d\} t\\\\ =&\lim\_\{n\to\infty\}\frac\{1\}\{2n\}\sum\_\{k=1\}^n f'(\xi\_k) =\frac\{1\}\{2\}\int\_0^1 f'(t)\mathrm\{ d\} t=\frac\{f(1)-f(0)\}\{2\}=\frac\{1\}\{2\}. \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)/
---
387、 1、 设 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上一致连续, 且 $\displaystyle \int\_0^\infty f(x)\mathrm\{ d\} x$ 收敛. 求证: $\displaystyle \lim\_\{x\to+\infty\}f(x)=0$. (湖南大学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 $\displaystyle f(x)$ 在 $\displaystyle [a,+\infty)$ 上一致连续知
\begin\{aligned\} \{\color\{red\}\forall\ \varepsilon > 0\}, \exists\ \delta > 0,\mathrm\{ s.t.\} \forall\ x',x''\geq a, |x'-x''|\leq \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 $\displaystyle \int\_a^\{+\infty\}f(x)\mathrm\{ d\} x$ 收敛及 Cauchy 收敛准则知
\begin\{aligned\} \{\color\{red\}\exists\ X > a\},\mathrm\{ s.t.\} \forall\ x > X, \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. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故当 $\displaystyle \{\color\{red\}x > X\}$ 时,
\begin\{aligned\} \{\color\{red\}|f(x)|\}&\leq |f(x)-f(\xi\_x)| +|f(\xi\_x)| \{\color\{red\} < \}\frac\{\varepsilon\}\{2\}+\frac\{\varepsilon\}\{2\}=\{\color\{red\}\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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
388、 2、 设函数 $\displaystyle f(x)$ 在区间 $\displaystyle [a,b]$ 上连续, 有积分中值公式有
\begin\{aligned\} \int\_a^x f(t)\mathrm\{ d\} t=f(\xi)(x-a)\left(a\leq \xi\leq x\leq b\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若导数 $\displaystyle f'\_+(a)$ 存在且非零, 求 $\displaystyle \lim\_\{x\to a^+\}\frac\{\xi-a\}\{x-a\}$. (湖南大学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\{原式\}=&\lim\_\{x\to a^+\}\frac\{f(\xi)-f(a)\}\{x-a\}\cdot \frac\{1\}\{\frac\{f(\xi)-f(a)\}\{\xi-a\}\}\\\\ =&\lim\_\{x\to a^+\}\frac\{\frac\{1\}\{x-a\}\int\_a^x f(t)\mathrm\{ d\} t-f(a)\}\{x-a\}\cdot \frac\{1\}\{f'\_+(a)\}\\\\ =&\lim\_\{x\to a^+\}\frac\{\int\_a^x f(t)\mathrm\{ d\} t-f(a)(x-a)\}\{(x-a)^2\}\cdot \frac\{1\}\{f'\_+(a)\}\\\\ \xlongequal\{\tiny\mbox\{L'Hospital\}\}& \lim\_\{x\to a^+\}\frac\{f(x)-f(a)\}\{2(x-a)\}\cdot \frac\{1\}\{f'\_+(a)\}\\\\ =&\frac\{f'\_+(a)\}\{2\}\cdot\frac\{1\}\{f'\_+(a)\}=\frac\{1\}\{2\}. \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)/
---
389、 10、 设 $\displaystyle f(x,y)$ 在 $\displaystyle [0,\pi]\times [0,\pi]$ 上连续, 且恒取正值, 试求
\begin\{aligned\} \lim\_\{n\to\infty\}\iint\_\{0\leq x\leq \pi\atop 0\leq y\leq \pi\} (\sin x)[f(x,y)]^\frac\{1\}\{n\}\mathrm\{ d\} x\mathrm\{ d\} y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(湖南大学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 $\displaystyle 0 < m=\min\_\{[0,\pi]^2\}f, M=\max\_\{[0,\pi]^2\}f$, 则由
\begin\{aligned\} \lim\_\{n\to\infty\}m^\frac\{1\}\{n\}=1=\lim\_\{n\to\infty\}M^\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
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ N,\mathrm\{ s.t.\} \forall\ n\geq N, |m^\frac\{1\}\{n\}-1| < \varepsilon, |M^\frac\{1\}\{n\}-1| < \varepsilon. \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 $\displaystyle n\geq N$ 时, $\displaystyle \forall\ 0\leq x,y\leq \pi$,
\begin\{aligned\} &-\varepsilon < m^\frac\{1\}\{n\}-1\leq [f(x,y)]^\frac\{1\}\{n\}-1\leq M^\frac\{1\}\{n\}-1 < \varepsilon\\\\ \Rightarrow&\left|\sin x[f(x,y)]^\frac\{1\}\{n\}-\sin x\right| \leq |\sin x|\cdot \varepsilon\leq \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 $\displaystyle \sin x [f(x,y)]^\frac\{1\}\{n\}\rightrightarrows \sin x$, 于 $\displaystyle [0,\pi]^2$. 于是
\begin\{aligned\} \mbox\{原式\}=\iint\_\{[0,\pi]^2\} \sin x\mathrm\{ d\} x\mathrm\{ d\} y =\pi \int\_0^\pi \sin x\mathrm\{ d\} x=2\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)/
---
390、 6、 求极限 $\displaystyle \lim\_\{n\to\infty\}\int\_0^1 \frac\{x^n\}\{1+\sqrt\{x\}\}\mathrm\{ d\} x$. (华南理工大学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) / 由 $\displaystyle 0\leq \left|\int\_0^1 \frac\{x^n\}\{1+\sqrt\{x\}\}\mathrm\{ d\} x\right|\leq \int\_0^1 x^n\mathrm\{ d\} x=\frac\{1\}\{n+1\}$ 及迫敛性知原式 $\displaystyle =0$.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)/
---
391、 7、 设 $\displaystyle f$ 是 $\displaystyle [0,1]$ 上的正值连续函数, 求证:
\begin\{aligned\} \lim\_\{n\to\infty\}\left\[\int\_0^1 (a+x)^nf(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{n\}=1+a. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(华中科技大学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)、 先证一个结论. 设 $\displaystyle f(x)\geq 0, g(x) > 0$, 二函数在 $\displaystyle [a,b]$ 上连续. 则
\begin\{aligned\} \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). \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, 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)$. (2)、 由第 1 步即知
\begin\{aligned\} \lim\_\{n\to\infty\}\left\[\int\_0^1 (a+x)^nf(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{n\}=\max\_\{0\leq x\leq 1\}(a+x)=1+a. \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 08:49:59
