## 张祖锦2023年数学专业真题分类70天之第19天
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415、 6、 设 $\displaystyle f(x)$ 在 $\displaystyle [0,1]$ 上有连续导数且 $\displaystyle f(1)=0$. 证明:
\begin\{aligned\} \int\_0^1 f^2(x)\mathrm\{ d\} x\leq 2\sqrt\{\int\_0^1 x^2f^2(x)\mathrm\{ d\} x\}\cdot \sqrt\{\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\}
(上海大学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\_0^1 f^2(x)\mathrm\{ d\} x\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} -\int\_0^1 2f(x)f'(x)\cdot x\mathrm\{ d\} x\\\\ =&2\int\_0^1 [-xf(x)]\cdot f'(x)\mathrm\{ d\} x \stackrel\{\tiny\mbox\{Schwarz\}\}\{\leq\} 2\sqrt\{\int\_0^1 x^2f^2(x)\mathrm\{ d\} x\}\cdot \sqrt\{\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\}
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)/
---
416、 8、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [0,+\infty)$ 上连续, 且存在正数 $\displaystyle \alpha$, 使得
\begin\{aligned\} f(x)\leq \alpha \int\_0^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\}
证明: 对任意的 $\displaystyle x\geq 0$, 有 $\displaystyle f(x)\leq 0$. (苏州大学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) / 设 $\displaystyle F(x)=\int\_0^x f(t)\mathrm\{ d\} t$, 则
\begin\{aligned\} &F'(x)=f(x)\leq \alpha F(x) \Rightarrow [\mathrm\{e\}^\{-\alpha x\}F(x)]'\leq 0\\\\ \Rightarrow&\mathrm\{e\}^\{-\alpha x\}F(x)\leq \mathrm\{e\}^\{-\alpha 0\}F(0)=0\Rightarrow f(x)\leq \alpha F(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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
417、 (5)、 设 $\displaystyle f$ 在 $\displaystyle [0,1]$ 上可积, 且
\begin\{aligned\} \mathscr\{A\}=\left\\{g\in R[0,1]; \int\_0^1 g^2(x)\mathrm\{ d\} x=1\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: $\displaystyle \left\[\int\_0^1 f^2(x)\mathrm\{ d\} x\right\]^\frac\{1\}\{2\}=\sup\_\{g(x)\in\mathscr\{A\}\}\int\_0^1 f(x)g(x)\mathrm\{ d\} x$. (西安交通大学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) / 对 $\displaystyle \forall\ g\in \mathscr\{A\}$,
\begin\{aligned\} \int\_0^1 f(x)g(x)\mathrm\{ d\} x\stackrel\{\tiny\mbox\{Schwarz\}\}\{\leq\} \sqrt\{\int\_0^1 f^2(x)\mathrm\{ d\} x\}\cdot \sqrt\{\int\_0^1 g^2(x)\mathrm\{ d\} x\}=\mbox\{左端\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \mbox\{右端\}\leq \mbox\{左端\}$. 再取 $\displaystyle g(x)=\frac\{f(x)\}\{\sqrt\{\int\_0^1 f^2(t)\mathrm\{ d\} t\}\}$, 则
\begin\{aligned\} g\in \mathscr\{A\}\Rightarrow \mbox\{左端\}=\frac\{\int\_0^1 f^2(x)\mathrm\{ d\} x\}\{\sqrt\{\int\_0^1 f^2(t)\mathrm\{ d\} t\}\} =\int\_0^1 f(x) g(x)\mathrm\{ d\} x\leq \mbox\{右端\}. \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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---
418、 3、 (15 分) 设 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 有连续的二阶导数, 且 $\displaystyle f(a)=f(b)=0$. 证明: (1)、 $\displaystyle \int\_a^b f(x)\mathrm\{ d\} x=\frac\{1\}\{2\}\int\_a^b (x-a)(x-b)f''(x)\mathrm\{ d\} x$; (2)、 $\displaystyle \left|\int\_a^b f(x)\mathrm\{ d\} x\right|\leq\frac\{1\}\{12\}(b-a)^3\max\_\{x\in [a,b]\}|f''(x)|$. (西北大学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) / 由
\begin\{aligned\} &\int\_a^b (x-a)(x-b)f''(x)\mathrm\{ d\} x=\int\_a^b (x-a)(x-b)\mathrm\{ d\} f'(x)\\\\ &\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} -\int\_a^b (2x-a-b)f'(x)\mathrm\{ d\} x =-\int\_a^b (2x-a-b)\mathrm\{ d\} f(x) =\int\_a^b 2f(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\} \int\_a^b (x-a)(x-b)f''(x)\mathrm\{ d\} x&\xlongequal[\tiny\mbox\{分中值\}]\{\tiny\mbox\{第一积\}\} f''(\xi)\int\_a^b (x-a)(x-b)\mathrm\{ d\} x =\frac\{1\}\{6\}(a-b)^3f''(\xi) \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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
419、 9、 (20 分) 设 $\displaystyle D$ 是由简单光滑闭曲线 $\displaystyle L$ 围成的区域, $\displaystyle f(x,y)$ 在 $\displaystyle \overline\{D\}$ 上具有连续的偏导数, 记 $\displaystyle d=\max\_\{(x,y)\in D\}\sqrt\{x^2+y^2\}$. (1)、 证明:
\begin\{aligned\} \iint\_D f(x,y)\mathrm\{ d\} x\mathrm\{ d\} y=\int\_L xf(x,y)\mathrm\{ d\} y-\iint\_D x\frac\{\partial f\}\{\partial x\}\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\}
(2)、 若对任意的 $\displaystyle (x,y)\in L$, 有 $\displaystyle f(x,y)=0$, 证明:
\begin\{aligned\} \iint\_D f^2(x,y)\mathrm\{ d\} x\mathrm\{ d\} y\leq d^2\iint\_D \left\[\left(\frac\{\partial f\}\{\partial x\}\right)^2+\left(\frac\{\partial f\}\{\partial y\}\right)^2\right\]\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\}
(西北大学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) / (1)、
\begin\{aligned\} \int\_L xf\mathrm\{ d\} y\xlongequal\{\tiny\mbox\{Green\}\} \iint\_D (xf)\_x\mathrm\{ d\} x\mathrm\{ d\} y =\iint\_D [f+xf\_x]\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\}
(2)、 若 $\displaystyle \iint\_D f^2=0$, 则结论自明. 若 $\displaystyle \iint\_D f^2 > 0$, 则由题设知 $\displaystyle f|\_L=0$. 在第 1 步中 $\displaystyle f\to f^2$ 得
\begin\{aligned\} \iint\_D f^2=&-\iint\_D x(f^2)\_x =-2\iint\_D xf\cdot f\_x\stackrel\{\tiny\mbox\{Schwarz\}\}\{\leq\} 2\sqrt\{\iint\_D x^2f^2\cdot \iint\_D f\_x^2\}. \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\} \left\[\iint\_D f^2\right\]^2\leq 2\iint\_D x^2f^2\cdot \iint\_D |\nabla f|^2, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle |\nabla f|^2=\left(\frac\{\partial f\}\{\partial x\}\right)^2+\left(\frac\{\partial f\}\{\partial y\}\right)^2$. 同理,
\begin\{aligned\} \left\[\iint\_D f^2\right\]^2\leq 2\iint\_D y^2f^2\cdot \iint\_D |\nabla f|^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
上述两式相加即得
\begin\{aligned\} 2\left\[\iint\_D f^2\right\]^2\leq 2\iint\_D |\nabla f|^2 \iint\_D (x^2+y^2)f^2 \leq 2d^2 \iint\_D |\nabla f|^2 \cdot \iint\_D f^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
化简即得结论.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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
420、 8、 解答如下问题: (1)、 假设 $\displaystyle f$ 与 $\displaystyle g$ 都是区间 $\displaystyle [a,b]$ 上的连续函数, 证明:
\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\}
(2)、 对任意的 $\displaystyle \varepsilon > 0$, 存在常数 $\displaystyle C\_\varepsilon > 0$, 使得对任意的 $\displaystyle f(x)\in C^1[a,b]$, 有
\begin\{aligned\} \sup\_\{x\in [a,b]\}|f(x)|^2\leq C\_\varepsilon\int\_a^b f^2(x)\mathrm\{ d\} x+\varepsilon \int\_a^b |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\}
(湘潭大学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)、 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. (2)、
\begin\{aligned\} \int\_a^b f(x)\mathrm\{ d\} x\xlongequal[\tiny\mbox\{分中值\}]\{\tiny\mbox\{第一积\}\} f(\xi)(b-a). \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\} |f(\xi)|^2=\frac\{1\}\{(b-a)^2\}\left\[\int\_a^b f(x)\mathrm\{ d\} x\right\]\stackrel\{\tiny\mbox\{Schwarz\}\}\{\leq\} \int\_a^b 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\}
对 $\displaystyle \forall\ x\in [a,b]$,
\begin\{aligned\} &f^2(x)\leq f^2(\xi)+|f^2(x)-f^2(\xi)| =f^2(\xi)+\left|\int\_\xi^x [f^2(t)]'\mathrm\{ d\} t\right|\\\\ \leq&f^2(\xi)+\int\_\{\min\left\\{\xi,x\right\\}\}^\{\max\left\\{\xi,x\right\\}\} |2f(t)f'(t)|\mathrm\{ d\} t\\\\ \leq& f^2(\xi)+\int\_\{\min\left\\{\xi,x\right\\}\}^\{\max\left\\{\xi,x\right\\}\} \left|\frac\{\sqrt\{2\}f(t)\}\{\sqrt\{\varepsilon\}\}\right| \cdot|\sqrt\{2\varepsilon\}f'(t)|\mathrm\{ d\} t\\\\ \stackrel\{\tiny\mbox\{Schwarz\}\}\{\leq\}&f^2(\xi)+\int\_\{\min\left\\{\xi,x\right\\}\}^\{\max\left\\{\xi,x\right\\}\} \frac\{1\}\{2\}\left\[\frac\{2f^2(t)\}\{\varepsilon\}+2\varepsilon|f'(t)|^2\right\]\mathrm\{ d\} t\\\\ \leq&\int\_a^b f^2+\frac\{1\}\{\varepsilon^2\}\int\_a^b f^2+\varepsilon \int\_a^b |f'|^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)/
---
421、 6、 (15 分) 求证:
\begin\{aligned\} 3\sqrt\{\mathrm\{e\}\}\leq \int\_\{\mathrm\{e\}\}^\{4\mathrm\{e\}\}\frac\{\ln x\}\{\sqrt\{x\}\}\mathrm\{ d\} x\leq 6. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中国海洋大学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\} I\equiv&\int\_\mathrm\{e\}^\{4\mathrm\{e\}\}\frac\{\ln x\}\{\sqrt\{x\}\}\mathrm\{ d\} x =2\int\_\mathrm\{e\}^\{4\mathrm\{e\}\} \frac\{\ln \sqrt\{x\}\}\{\sqrt\{x\}\}\mathrm\{ d\} x \stackrel\{\ln \sqrt\{x\}=t\}\{=\}2\int\_\frac\{1\}\{2\}^\{\frac\{1\}\{2\}+\ln 2\}\frac\{t\}\{\mathrm\{e\}^t\}\cdot 2\mathrm\{e\}^\{2t\}\mathrm\{ d\} t\\\\ =&4\int\_\frac\{1\}\{2\}^\{\frac\{1\}\{2\}+\ln 2\}t\mathrm\{e\}^t\mathrm\{ d\} t \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \cdots =2\sqrt\{\mathrm\{e\}\}(4\ln 2-1). \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\} &2\sqrt\{\mathrm\{e\}\}(4\ln 2-1)\geq 3\sqrt\{\mathrm\{e\}\} \Leftrightarrow 4\ln 2-1\geq \frac\{3\}\{2\}\\\\ \Leftrightarrow& 4\ln 2\geq \frac\{5\}\{2\} \Leftrightarrow 8\ln 2\geq 5 \Leftrightarrow 256\geq \mathrm\{e\}^5 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
成立知 $\displaystyle I\geq 3\sqrt\{\mathrm\{e\}\}$. 再者,
\begin\{aligned\} &2\sqrt\{\mathrm\{e\}\}(4\ln 2-1)\leq 6 \Leftrightarrow 4\ln 2-1\leq\frac\{3\}\{\sqrt\{\mathrm\{e\}\}\}\\\\ \Leftrightarrow& \ln 16\leq \frac\{3\}\{\sqrt\{\mathrm\{e\}\}\}+1 \Leftrightarrow 16\leq \mathrm\{e\}^\{\frac\{3\}\{\sqrt\{\mathrm\{e\}\}\}+1\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
成立. 事实上,
\begin\{aligned\} \sqrt\{\mathrm\{e\}\}&=\sum\_\{k=0\}^4 \frac\{1\}\{k!\}\left(\frac\{1\}\{2\}\right)^k+\sum\_\{k=5\}^\infty \frac\{1\}\{k!\}\left(\frac\{1\}\{2\}\right)^k\\\\ &\leq \sum\_\{k=0\}^4 \frac\{1\}\{k!\}\left(\frac\{1\}\{2\}\right)^k+\frac\{1\}\{5!\}\sum\_\{k=5\}^\infty \left(\frac\{1\}\{2\}\right)^k =\frac\{1583\}\{960\} < \frac\{33\}\{20\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
蕴含
\begin\{aligned\} \mathrm\{e\}^\{\frac\{3\}\{\sqrt\{\mathrm\{e\}\}\}+1\} > \mathrm\{e\}^\{\frac\{20\}\{11\}+1\}=\mathrm\{e\}^\frac\{31\}\{11\} > \sum\_\{k=0\}^6 \frac\{1\}\{k!\}\left(\frac\{31\}\{11\}\right)^k =\frac\{20823009997\}\{1275523920\} > 16. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle I\leq 6$.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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---
422、 7、 (15 分) 设 $\displaystyle p(x)$ 是定义在 $\displaystyle [a,b]$ 上的非负连续函数, $\displaystyle f(x),g(x)$ 是定义在 $\displaystyle [a,b]$ 上的连续单调递增函数. 证明:
\begin\{aligned\} \int\_a^b p(x)f(x)\mathrm\{ d\} x\cdot \int\_a^b p(x)g(x)\mathrm\{ d\} x \leq \int\_a^b p(x)\mathrm\{ d\} x\cdot \int\_a^b p(x)f(x)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\}
(中国人民大学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) / 由题设,
\begin\{aligned\} p(x)p(y)[f(x)-f(y)][g(x)-g(y)]\geq 0, \forall\ x,y\in [a,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 $\displaystyle [a,b]$ 上积分得
\begin\{aligned\} &p(y)\int\_a^b p(x)f(x)g(x)\mathrm\{ d\} x -p(y)g(y)\int\_a^b p(x)f(x)\mathrm\{ d\} x\\\\ &-p(y)f(y)\int\_a^b p(x)g(x)\mathrm\{ d\} x +p(y)f(y)g(y)\int\_a^b p(x)\mathrm\{ d\} x\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
关于 $\displaystyle y$ 在 $\displaystyle [a,b]$ 上积分得
\begin\{aligned\} &\int\_a^b p(y)\mathrm\{ d\} y\cdot\int\_a^b p(x)f(x)g(x)\mathrm\{ d\} x -\int\_a^b p(y)g(y)\mathrm\{ d\} y\cdot\int\_a^b p(x)f(x)\mathrm\{ d\} x\\\\ &-\int\_a^b p(y)f(y)\mathrm\{ d\} y\cdot\int\_a^b p(x)g(x)\mathrm\{ d\} x +\int\_a^b p(y)f(y)g(y)\mathrm\{ d\} y\cdot\int\_a^b p(x)\mathrm\{ d\} x\geq 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
化简即得结论.跟锦数学微信公众号. [在线资料](
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---
423、 2、 (20 分) 解答如下问题. (1)、 证明柯西不等式, 即若 $\displaystyle f(x),g(x)$ 在 $\displaystyle [a,b]$ 上可积, 则有
\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\}
(2)、 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上有连续导数, 且 $\displaystyle f(a)=0$. 证明: (2-1)、
\begin\{aligned\} \max\_\{x\in [a,b]\}|f(x)|\leq \sqrt\{(b-a)\int\_a^b |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\}
(2-2)、 $\displaystyle f(x)$ 在 $\displaystyle [a,b]$ 上有连续导数, 且 $\displaystyle f(a)=0$. 证明:
\begin\{aligned\} \int\_a^b f^2(x)\mathrm\{ d\} x\leq \frac\{(b-a)^2\}\{2\}\int\_a^b |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\}
(中南大学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) / (1)、 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. (2)、 (2-1)、
\begin\{aligned\} |f(x)|=&|f(x)-f(a)|=\left|\int\_a^x f'(t)\mathrm\{ d\} t\right| \leq \int\_a^b |f'(t)|\mathrm\{ d\} t\\\\ \stackrel\{\tiny\mbox\{Schwarz\}\}\{\leq\}&\sqrt\{\int\_a^b 1^2\mathrm\{ d\} t\cdot \int\_a^b |f'(t)|^2\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\}
(2-2)、
\begin\{aligned\} &\int\_a^b f^2(x)\mathrm\{ d\} x =\int\_a^b \left\[\int\_a^x f'(t)\mathrm\{ d\} t\right\]^2\mathrm\{ d\} x\\\\ \stackrel\{\tiny\mbox\{Schwarz\}\}\{\leq\}& \int\_a^b \left\[\int\_a^x 1^2\mathrm\{ d\} t\cdot \int\_a^x f'^2(t)\mathrm\{ d\} t\right\]\mathrm\{ d\} x\\\\ \leq&\int\_a^b f'^2(t)\mathrm\{ d\} t\cdot \int\_a^b (x-a)\mathrm\{ d\} x =\frac\{(b-a)^2\}\{2\} \int\_a^b [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\}
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---
424、 5、 (20 分) 设无穷积分 $\displaystyle \int\_a^\infty f(x)\mathrm\{ d\} x$ 收敛. (1)、 证明: 若 $\displaystyle f$ 在 $\displaystyle [a,+\infty)$ 上一致连续, 则 $\displaystyle \lim\_\{x\to+\infty\}f(x)=0$; (2)、 若去掉’一致连续‘能否推出’$\lim\_\{x\to+\infty\}f(x)=0$‘? 若可以, 请证明, 否则举出反例. (安徽大学2023年高等代数考研试题) [广义积分 ]
[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) / (1)、 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\}
这就证明了 $\displaystyle \lim\_\{x\to+\infty\}f(x)=0$. (2)、 若去掉’一致连续‘不能推出’$\lim\_\{x\to+\infty\}f(x)=0$‘. 比如
\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 $\displaystyle \lim\_\{n\to\infty\}f(n)=1$ 知 $\displaystyle \lim\_\{x\to+\infty\}f(x)=0$ 不成立.跟锦数学微信公众号. [在线资料](
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---
425、 8、 证明无穷积分 $\displaystyle \int\_\{-\infty\}^\{+\infty\}\mathrm\{e\}^\{-ax^2\}\mathrm\{ d\} x\ (a > 0)$ 收敛, 并求其值.(北京工业大学2023年数学分析考研试题) [广义积分 ]
[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 $\displaystyle \lim\_\{x\to+\infty\}\frac\{\mathrm\{e\}^\{-ax^2\}\}\{\frac\{1\}\{x^2\}\}=\lim\_\{t\to+\infty\}\frac\{t\}\{\mathrm\{e\}^\{at\}\}\xlongequal\{\tiny\mbox\{L'Hospital\}\} \cdots=0$ 及比较判别法知原广义积分收敛. 进一步,
\begin\{aligned\} \mbox\{原式\}^2=&\iint\_\{\mathbb\{R\}^2\} \mathrm\{e\}^\{-a(x^2+y^2)\}\mathrm\{ d\} x\mathrm\{ d\} y =\int\_0^\infty \mathrm\{e\}^\{-ar^2\}\cdot 2\pi r\mathrm\{ d\} r \stackrel\{ar^2=s\}\{=\}\frac\{\pi\}\{a\}\int\_0^\infty \mathrm\{e\}^\{-s\}\mathrm\{ d\} s=\frac\{\pi\}\{a\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
蕴含原式 $\displaystyle =\sqrt\{\frac\{\pi\}\{a\}\}$.跟锦数学微信公众号. [在线资料](
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---
426、 8、 计算积分 $\displaystyle \int\_0^\infty \frac\{\cos x-\cos 2x\}\{x\}\mathrm\{e\}^\{-x\}\mathrm\{ d\} x$. (北京邮电大学2023年数学分析考研试题) [广义积分 ]
[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^\infty \mathrm\{e\}^\{-x\}\mathrm\{ d\} x\int\_1^2 \sin xt\mathrm\{ d\} t =\int\_1^2 \mathrm\{ d\} t\int\_0^\infty \mathrm\{e\}^\{-x\}\sin xt\mathrm\{ d\} x\\\\ &\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \int\_1^2\frac\{t\mathrm\{ d\} t\}\{1+t^2\}=\frac\{1\}\{2\}\ln\frac\{5\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
第 2 步可以交换积分次序是因为 $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-x\}\sin xt\mathrm\{ d\} x$ 关于 $\displaystyle t\in [1,2]$ 是一致收敛的. 这可由 Dirichlet 判别法得到.跟锦数学微信公众号. [在线资料](
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---
427、 (2)、 证明无穷积分 $\displaystyle \int\_0^\infty \frac\{\sin x\}\{x\}\mathrm\{ d\} x$ 条件收敛. (大连理工大学2023年数学分析考研试题) [广义积分 ]
[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 $\displaystyle \lim\_\{x\to 0\}\frac\{\sin x\}\{x\}=1$ 知 $\displaystyle x=0$ 不是瑕点. 由 Dirichlet 判别法知原广义积分收敛. 又由
\begin\{aligned\} \int\_1^\infty \left|\frac\{\sin x\}\{x\}\right|\mathrm\{ d\} x \geq\int\_1^\infty \frac\{\sin^2x\}\{x\}\mathrm\{ d\} x =\frac\{1\}\{2\}\int\_1^\infty\frac\{1-\cos 2x\}\{x\}\mathrm\{ d\} x=+\infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原广义积分条件收敛.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)/
---
428、 (5)、 求定积分 $\displaystyle \int\_0^1 \frac\{x^2-x\}\{\ln x\}\mathrm\{ d\} x$. (电子科技大学2023年数学分析考研试题) [广义积分 ]
[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\_0^1 \int\_1^2 x^t\mathrm\{ d\} t\mathrm\{ d\} x =\int\_1^2 \int\_0^1 x^t\mathrm\{ d\} x\mathrm\{ d\} t\\\\ =&\int\_1^2 \frac\{1\}\{t+1\}\mathrm\{ d\} t=\ln \frac\{3\}\{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](
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---
429、 (2)、 设函数 $\displaystyle f\in C^1[0,+\infty)$, $\displaystyle f(0) > 0$ 且 $\displaystyle f'(x)\geq 0, x\in [0,+\infty)$. 证明: 若反常积分 $\displaystyle \int\_0^\infty \frac\{1\}\{f(x)+f'(x)\}\mathrm\{ d\} x$ 收敛, 则反常积分 $\displaystyle \int\_0^\infty \frac\{1\}\{f(x)\}\mathrm\{ d\} x$ 也收敛. (电子科技大学2023年数学分析考研试题) [广义积分 ]
[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) / (2-1)、 由 $\displaystyle f(0) > 0, f'(x)\geq 0$ 知
\begin\{aligned\} \forall\ x\geq 0, f(x) > 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2-2)、 对 $\displaystyle \forall\ X > 0$,
\begin\{aligned\} 0&\leq \int\_0^X \frac\{1\}\{f(x)\}\mathrm\{ d\} x-\int\_0^X \frac\{1\}\{f(x)+f'(x)\}\mathrm\{ d\} x\\\\ &=\int\_0^X \frac\{f'(x)\}\{f(x)\left\[f(x)+f'(x)\right\]\}\mathrm\{ d\} x \leq \int\_0^X \frac\{f'(x)\}\{f^2(x)\}\mathrm\{ d\} x\\\\ &=\left.-\frac\{1\}\{f(x)\}\right|\_0^X < \frac\{1\}\{f(0)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle X\to+\infty$ 得
\begin\{aligned\} \int\_0^\{+\infty\}\frac\{\mathrm\{ d\} x\}\{f(x)+f'(x)\} \leq \int\_0^\{+\infty\}\frac\{\mathrm\{ d\} x\}\{f(x)\} \leq \int\_0^\{+\infty\}\frac\{\mathrm\{ d\} x\}\{f(x)+f'(x)\}+\frac\{1\}\{f(0)\} < +\infty. \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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---
430、 8、 证明积分 $\displaystyle \int\_1^\infty \frac\{\sin x\}\{x\}\mathrm\{ d\} x$ 条件收敛. (东北大学2023年数学分析考研试题) [广义积分 ]
[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 Dirichlet 判别法知原广义积分收敛. 又由
\begin\{aligned\} \int\_1^\infty \left|\frac\{\sin x\}\{x\}\right|\mathrm\{ d\} x \geq\int\_1^\infty \frac\{\sin^2x\}\{x\}\mathrm\{ d\} x =\frac\{1\}\{2\}\int\_1^\infty\frac\{1-\cos 2x\}\{x\}\mathrm\{ d\} x=+\infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原广义积分条件收敛.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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---
431、 (5)、 计算 $\displaystyle \int\_0^1 \left(\frac\{1\}\{t\}-\left\[\frac\{1\}\{t\}\right\]\right)\mathrm\{ d\} t$. (东北师范大学2023年数学分析考研试题) [广义积分 ]
[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\}&=\sum\_\{k=1\}^\infty \int\_\frac\{1\}\{k+1\}^\frac\{1\}\{k\}\left(\frac\{1\}\{t\}-k\right)\mathrm\{ d\} t =\sum\_\{k=1\}^\infty \left\[\ln (k+1)-\ln k-k\left(\frac\{1\}\{k\}-\frac\{1\}\{k+1\}\right)\right\]\\\\ &=\sum\_\{k=1\}^\infty \left\[\ln(k+1)-\ln k-\frac\{1\}\{k+1\}\right\]\\\\ &=\lim\_\{n\to\infty\}\sum\_\{k=1\}^n \left\[\ln (k+1)-\ln k-\frac\{1\}\{k+1\}\right\]\\\\ &=\lim\_\{n\to\infty\}\left\[\ln(n+1)-\sum\_\{k=2\}^n \frac\{1\}\{k\}\right\]\\\\ &=-\lim\_\{n\to\infty\}\left(\sum\_\{k=1\}^n \frac\{1\}\{k\}-\ln n\right) +1+\lim\_\{n\to\infty\}\ln \left(1+\frac\{1\}\{n\}\right) =1-\gamma, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \gamma\approx 0.577215665$ 是 Euler 常数.跟锦数学微信公众号. [在线资料](
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---
432、 4、 设 $\displaystyle [a,+\infty)$ 上非负连续函数 $\displaystyle f$ 可导, 且具有连续导函数, 若存在 $\displaystyle r > 1$, 使
\begin\{aligned\} \lim\_\{x\to+\infty\}\frac\{xf'(x)\}\{f(x)\}\leq -r. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
证明: 反常积分 $\displaystyle \int\_a^\infty f(x)\mathrm\{ d\} x$ 收敛. (复旦大学2023年分析(第6,7,8,9,10题没做)考研试题) [广义积分 ]
[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) / 由题设, $\displaystyle \exists\ X > a,\mathrm\{ s.t.\} \forall\ x\geq X$,
\begin\{aligned\} &\frac\{xf'(x)\}\{f(x)\} < \frac\{-1-r\}\{2\} \Rightarrow \left\[\ln f(x)\right\]'\leq \frac\{2\}\{1-r\}[x^\frac\{1-r\}\{2\}]'\\\\ \Rightarrow&\ln f(x)-\ln f(X)\leq \frac\{2\}\{1-r\}(x^\frac\{1-r\}\{2\}-X^\frac\{1-r\}\{2\})\\\\ \Rightarrow&f(x)\leq f(X) \mathrm\{e\}^\{\frac\{2\}\{1-r\}X^\frac\{1-r\}\{2\}\} \mathrm\{e\}^\{-\frac\{2\}\{r-1\}x^\frac\{1-r\}\{2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由比较判别法即知 $\displaystyle \int\_a^\infty f(x)\mathrm\{ d\} x$ 收敛.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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433、 (4)、 证明 $\displaystyle \int\_\{-\infty\}^\{+\infty\}\frac\{\sin x\}\{x\}\mathrm\{ d\} x$ 条件收敛. (广西大学2023年数学分析考研试题) [广义积分 ]
[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 $\displaystyle \lim\_\{x\to 0\}\frac\{\sin x\}\{x\}=1$ 知 $\displaystyle 0$ 不是瑕点. 又由 $\displaystyle \int\_0^A \frac\{\sin x\}\{x\}\mathrm\{ d\} x=\int\_\{-A\}^0\frac\{\sin x\}\{x\}\mathrm\{ d\} x$ 知我们仅需证明 $\displaystyle \int\_0^\{+\infty\}\frac\{\sin x\}\{x\}\mathrm\{ d\} x$ 条件收敛. 由 Dirichlet 判别法知 $\displaystyle \int\_0^\{+\infty\}\frac\{\sin x\}\{x\}\mathrm\{ d\} x$ 收敛. 再者,
\begin\{aligned\} \int\_0^\infty \left|\frac\{\sin x\}\{x\}\right|\mathrm\{ d\} x \geq \int\_0^\infty \frac\{\sin^2x\}\{x\}\mathrm\{ d\} x =\int\_0^\infty \frac\{1-\cos 2x\}\{2x\}\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\}
作为一个发散广义积分与收敛广义积分的差, 是发散的. 故 $\displaystyle \int\_0^\{+\infty\}\frac\{\sin x\}\{x\}\mathrm\{ d\} x$ 条件收敛, 结论得证.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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434、 11、 求 $\displaystyle \int\_0^\infty \frac\{\arctan(ax)-\arctan(bx)\}\{x\}\mathrm\{ d\} x\ (a > b > 0)$. (哈尔滨工程大学2023年数学分析考研试题) [广义积分 ]
[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) / (1)、 我们有如下一般的 Froullani 积分. 设 $\displaystyle f(x)$ 在 $\displaystyle (0,+\infty)$ 上连续, 且极限 $\displaystyle f(0+)=\lim\_\{x\to 0^+\}f(x)$ 与 $\displaystyle f(+\infty)=\lim\_\{x\to+\infty\}f(x)$ 存在, 则
\begin\{aligned\} \int\_0^\infty \frac\{f(ax)-f(bx)\}\{x\}\mathrm\{ d\} x=[f(0+)-f(+\infty)]\ln \frac\{b\}\{a\},\left(b > 0 > 0\right). \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, 对 $\displaystyle \forall\ 0 < \varepsilon < A < +\infty$,
\begin\{aligned\} &\int\_\varepsilon^A \frac\{f(ax)-f(bx)\}\{x\}\mathrm\{ d\} x =\int\_\varepsilon^A \frac\{f(ax)\}\{ax\}\mathrm\{ d\} (ax) -\int\_\varepsilon^A \frac\{f(bx)\}\{bx\}\mathrm\{ d\} (bx)\\\\ =&\int\_\{a\varepsilon\}^\{aA\}\frac\{f(t)\}\{t\}\mathrm\{ d\} t -\int\_\{b\varepsilon\}^\{bA\} \frac\{f(t)\}\{t\}\mathrm\{ d\} t =\int\_\{a\varepsilon\}^\{b\varepsilon\}\frac\{f(t)\}\{t\}\mathrm\{ d\} t -\int\_\{aA\}^\{bA\} \frac\{f(t)\}\{t\}\mathrm\{ d\} t\\\\ \xlongequal[\tiny\mbox\{分中值\}]\{\tiny\mbox\{第一积\}\}&f(\xi)\int\_\{a\varepsilon\}^\{b\varepsilon\}\frac\{1\}\{t\}\mathrm\{ d\} t -f(\eta)\int\_\{aA\}^\{bA\}\frac\{1\}\{t\}\mathrm\{ d\} t =[f(\xi)-f(\eta)] \ln \frac\{b\}\{a\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle \varepsilon\to 0^+, A\to+\infty$, 并注意到 $\displaystyle \lim\_\{x\to 0^+\}f(x)=f(0+), \lim\_\{x\to+\infty\}f(x)=f(+\infty)$, 我们有
\begin\{aligned\} \int\_0^\{+\infty\} \frac\{f(ax)-f(bx)\}\{x\}\mathrm\{ d\} x =[f(0+)-f(+\infty)]\ln \frac\{b\}\{a\}. \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 $\displaystyle f(x)=\arctan x$, 则由第 1 步知
\begin\{aligned\} I=\left\[0-\frac\{\pi\}\{2\}\right\] \ln \frac\{b\}\{a\}=-\frac\{\pi\}\{2\}\ln \frac\{b\}\{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)/
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435、 9、 (15 分) 试证
\begin\{aligned\} \int\_0^\infty \frac\{\mathrm\{ d\} x\}\{1+x^4\}=\int\_0^\infty \frac\{x^2\}\{1+x^4\}\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\}
并求积分值. (合肥工业大学2023年数学分析考研试题) [广义积分 ]
[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\} I&\equiv \int\_0^\infty \frac\{\mathrm\{ d\} x\}\{1+x^4\} \stackrel\{t=\frac\{1\}\{x\}\}\{=\}\int\_0^\infty \frac\{1\}\{1+\left(\frac\{1\}\{t\}\right)^4\}\cdot \frac\{1\}\{t^2\}\mathrm\{ d\} t =\int\_0^\infty \frac\{t^2\}\{1+t^4\}\mathrm\{ d\} t =\frac\{1\}\{2\}\int\_0^\infty \frac\{1+t^2\}\{1+t^4\}\mathrm\{ d\} t\\\\ &=\frac\{1\}\{2\}\int\_0^\infty \frac\{\frac\{1\}\{t^2\}+1\}\{\frac\{1\}\{t^2\}+t^2\}\mathrm\{ d\} t =\frac\{1\}\{2\}\int\_0^\infty \frac\{1\}\{\left(t-\frac\{1\}\{t\}\right)^2+2\}\mathrm\{ d\} \left(t-\frac\{1\}\{t\}\right) \stackrel\{s=t-\frac\{1\}\{t\}\}\{=\}\frac\{1\}\{2\}\int\_\{-\infty\}^\{+\infty\} \frac\{1\}\{s^2+2\}\mathrm\{ d\} s\\\\ &=\frac\{1\}\{\sqrt\{2\}\}\int\_0^\infty \frac\{1\}\{1+\left(\frac\{s\}\{\sqrt\{2\}\}\right)^2\}\mathrm\{ d\} \frac\{s\}\{\sqrt\{2\}\} =\frac\{1\}\{\sqrt\{2\}\}\cdot \frac\{\pi\}\{2\} =\frac\{\pi\}\{2\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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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436、 (4)、 求 $\displaystyle \int\_0^\infty \frac\{1+x^2\}\{1+x^4\}\mathrm\{ d\} x$. (河南大学2023年数学分析考研试题) [广义积分 ]
[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\} &\int \frac\{1+x^2\}\{1+x^4\}\mathrm\{ d\} x =\int \frac\{\frac\{1\}\{x^2+1\}\}\{\frac\{1\}\{x^2\}+x^2\}\mathrm\{ d\} x =\int \frac\{\mathrm\{ d\} \left(x-\frac\{1\}\{x\}\right)\}\{\left(x-\frac\{1\}\{x\}\right)^2+2\}\\\\ =&\frac\{1\}\{\sqrt\{2\}\}\int \frac\{\mathrm\{ d\} \left\[\frac\{1\}\{\sqrt\{2\}\}\left(x-\frac\{1\}\{x\}\right)\right\]\} \{\left\[\frac\{1\}\{\sqrt\{2\}\}\left(x-\frac\{1\}\{x\}\right)\right\]^2+1\} =\frac\{1\}\{\sqrt\{2\}\}\arctan \left\[\frac\{1\}\{\sqrt\{2\}\}\left(x-\frac\{1\}\{x\}\right)\right\]+C \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 $\displaystyle =\frac\{\pi\}\{\sqrt\{2\}\}$.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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437、 6、 设 $\displaystyle \lambda > 0$, 讨论 $\displaystyle \int\_0^\infty \frac\{\arctan x\}\{x^\lambda\}\mathrm\{ d\} x$ 的敛散性 (包括绝对收敛与条件收敛). (黑龙江大学2023年数学分析考研试题) [广义积分 ]
[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^\infty \frac\{\arctan x\}\{x^\lambda\}\mathrm\{ d\} x =\int\_0^1 \frac\{\arctan x\}\{x^\lambda\}\mathrm\{ d\} x+\int\_1^\infty \frac\{\arctan x\}\{x^\lambda\}\mathrm\{ d\} x\equiv I\_1+I\_2. \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\} \frac\{\arctan x\}\{x^\lambda\}\sim \frac\{x\}\{x^\lambda\}=\frac\{1\}\{x^\{\lambda-1\}\}\left(x\to 0\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知当且仅当 $\displaystyle \lambda-1 < 1\Leftrightarrow \lambda < 2$ 时, $\displaystyle I\_1$ (绝对) 收敛. 又由
\begin\{aligned\} \frac\{\arctan x\}\{x^\lambda\}\sim\frac\{\pi\}\{2x^\lambda\}\left(x\to +\infty\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知当且仅当 $\displaystyle \lambda > 1$ 时, $\displaystyle I\_2$ (绝对收敛). 综上即知当且仅当 $\displaystyle 1 < \lambda < 2$ 时, 原广义积分绝对收敛.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 08:51:00
