## 张祖锦2023年数学专业真题分类70天之第20天
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438、 8、 试证: $\displaystyle \int\_\{-\infty\}^\{+\infty\}\mathrm\{e\}^\{-x^2\}\mathrm\{ d\} x=\sqrt\{\pi\}$. (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) / 由
\begin\{aligned\} \left\[\int\_\{\mathbb\{R\}\}\mathrm\{e\}^\{-x^2\}\mathrm\{ d\} x\right\]^2=&\iint\_\{\mathbb\{R\}^2\}\mathrm\{e\}^\{-(x^2+y^2)\}\mathrm\{ d\} x\mathrm\{ d\} y =\int\_0^\infty \mathrm\{e\}^\{-r^2\}\cdot 2\pi r\mathrm\{ d\} r=\pi \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \int\_\mathbb\{R\} \mathrm\{e\}^\{-x^2\}\mathrm\{ d\} x=\sqrt\{\pi\}$.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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439、 2、 (15 分) 计算积分 $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-\frac\{1\}\{4s\}\}\cdot s^\{-\frac\{3\}\{2\}\}\cdot \mathrm\{e\}^\{-s\}\mathrm\{ d\} s$. (华中科技大学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^\infty \mathrm\{e\}^\{-\left(s+\frac\{1\}\{4s\}\right)\} s^\{-\frac\{3\}\{2\}\}\mathrm\{ d\} s\\\\ \stackrel\{s^\{-\frac\{1\}\{2\}\}=t\}\{=\}&\int\_\infty^0 \mathrm\{e\}^\{-\left(\frac\{1\}\{t^2\}+\frac\{t^2\}\{4\}\right)\}(-2\mathrm\{ d\} t) =\frac\{2\}\{\mathrm\{e\}\}\int\_0^\infty \mathrm\{e\}^\{-\left(\frac\{t\}\{2\}-\frac\{1\}\{t\}\right)^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\}
设
\begin\{aligned\} u=\frac\{t\}\{2\}-\frac\{1\}\{t\}\Rightarrow \frac\{\mathrm\{ d\} u\}\{\mathrm\{ d\} t\}=\frac\{1\}\{2\}+\frac\{1\}\{t^2\} > 0, t=u+\sqrt\{2+u^2\}\left( > 0!\right) \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\} \mbox\{原式\}=&\frac\{2\}\{\mathrm\{e\}\}\int\_\mathbb\{R\} \mathrm\{e\}^\{-u^2\} \left(1+\frac\{u\}\{\sqrt\{2+u^2\}\}\right)\mathrm\{ d\} u\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&\frac\{4\}\{\mathrm\{e\}\}\int\_0^\infty \mathrm\{e\}^\{-u^2\}\mathrm\{ d\} u=\frac\{4\}\{\mathrm\{e\}\}\frac\{\sqrt\{\pi\}\}\{2\} =\frac\{2\sqrt\{\pi\}\}\{\mathrm\{e\}\}. \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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440、 6、 设广义积分 $\displaystyle \int\_a^\infty f(x)\mathrm\{ d\} x$ 收敛. 证明: $\displaystyle F(x)=\int\_a^x f(t)\mathrm\{ d\} t$ 在 $\displaystyle [a,+\infty)$ 上一致连续. (南昌大学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 $\displaystyle \int\_a^\infty f(x)\mathrm\{ d\} x$ 收敛及 Cauchy 收敛准则知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ X > a,\mathrm\{ s.t.\} \forall\ y > x\geq X, \varepsilon > \left|\int\_x^y f(t)\mathrm\{ d\} t\right| =|F(y)-F(x)|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle F$ 在 $\displaystyle [X,+\infty)$ 上一致连续. 又由 $\displaystyle F$ 在 $\displaystyle [a,X]$ 上连续知其一致连续. 故 $\displaystyle F$ 在 $\displaystyle [a,+\infty)$ 上一致连续.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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441、 6、 判断 $\displaystyle \int\_0^\infty \frac\{\sqrt\{x\}\cos x\}\{x+100\}\mathrm\{ d\} x$ 绝对收敛还是条件收敛. (南京航空航天大学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) / 设 $\displaystyle f(x)=\frac\{\sqrt\{x\}\cos x\}\{x+100\}$, 则 $\displaystyle f'(x)=\frac\{100-x\}\{2\sqrt\{x\}(100+x)^2\} < 0, x > 100$. 故 $\displaystyle f$ 在 $\displaystyle [100,+\infty)$ 上严格递减. 由 Dirichlet 判别法知原广义积分收敛. 又
\begin\{aligned\} \int\_0^\infty \left|\frac\{\sqrt\{x\}\cos x\}\{x+100\}\right|\mathrm\{ d\} x \geq \int\_0^\infty \frac\{\sqrt\{x\}\cos^2x\}\{x+100\}\mathrm\{ d\} x =\int\_0^\infty \frac\{\sqrt\{x\}\}\{2(x+100)\}[1+\cos 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和, 还是发散的, 我们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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442、 7、 讨论反常积分 $\displaystyle \int\_0^\infty \frac\{\sin x\arctan x\}\{x^p\}\mathrm\{ d\} x$ 的敛散性. (上海财经大学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)、 由
\begin\{aligned\} \frac\{\sin x\arctan x\}\{x^p\}\sim\frac\{x\cdot x\}\{x^p\}=\frac\{1\}\{x^\{p-2\}\}\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 $\displaystyle p-2 < 1\Leftrightarrow p < 3$ 时, $\displaystyle \int\_0^1 \cdots$ 绝对收敛; 当 $\displaystyle p-2\geq 1\Leftrightarrow p\geq 3$ 时, $\displaystyle \int\_0^1 \cdots$ 发散. (2)、 (2-1)、 当 $\displaystyle p > 1$ 时, 由
\begin\{aligned\} \int\_1^\infty\left|\frac\{\sin x\arctan x\}\{x^p \}\right|\mathrm\{ d\} x\leq \frac\{\pi\}\{2\}\int\_1^\infty\frac\{1\}\{x^p \}\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 $\displaystyle \int\_1^\infty \cdots$ 绝对收敛. (2-2)、 当 $\displaystyle 0 < p \leq 1$ 时, 由 Dirichlet 判别法知原广义积分收敛. 又由
\begin\{aligned\} \int\_1^\infty \left|\frac\{\sin x \arctan x\}\{x^p \}\right|\mathrm\{ d\} x \geq\frac\{\pi\}\{4\}\int\_1^\infty \frac\{\sin^2x\}\{x^p \}\mathrm\{ d\} x =\frac\{\pi\}\{8\}\int\_0^\infty\frac\{1-\cos 2x\}\{x^p \}\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 $\displaystyle \int\_1^\infty \cdots$ 条件收敛. (2-3)、 当 $\displaystyle p \leq 0$ 时, 由
\begin\{aligned\} \int\_\{2n\pi+\frac\{\pi\}\{4\}\}^\{2n\pi+\frac\{3\pi\}\{4\}\} \frac\{\sin x\arctan x\}\{x^p \}\mathrm\{ d\} x \geq \frac\{\frac\{1\}\{\sqrt\{2\}\}\frac\{\pi\}\{4\}\}\{1^p \}\cdot\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\}
及 Cauchy 收敛准则知 $\displaystyle \int\_1^\infty \cdots$ 发散. 综上, 当 $\displaystyle 1 < p < 3$ 时, 原广义积分绝对收敛; 当 $\displaystyle 0 < p < 1$ 时, 原广义积分条件收敛; 当 $\displaystyle p\leq 0$ 或 $\displaystyle p\geq 3$ 时, 原广义积分发散.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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443、 13、 解答如下问题: (1)、 证明反常积分 $\displaystyle \int\_0^\infty \frac\{\arctan x\}\{x^p\}\mathrm\{ d\} x$ ($1 < p < 2$) 收敛; (上海大学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^p\}\mathrm\{ d\} x =\int\_0^1 \frac\{\arctan x\}\{x^p\}\mathrm\{ d\} x+\int\_1^\infty \frac\{\arctan x\}\{x^p\}\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^p\}\sim \frac\{x\}\{x^p\}=\frac\{1\}\{x^\{p-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 $\displaystyle p-1 < 1\Leftrightarrow p < 2$ 时, $\displaystyle I\_1$ (绝对) 收敛. 又由
\begin\{aligned\} \frac\{\arctan x\}\{x^p\}\sim\frac\{\pi\}\{2x^p\}\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 $\displaystyle p > 1$ 时, $\displaystyle I\_2$ (绝对收敛). 综上即知当且仅当 $\displaystyle 1 < p < 2$ 时, 原广义积分绝对收敛. 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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444、 (4)、 $\displaystyle \int\_0^\infty\frac\{\mathrm\{ d\} x\}\{(1+x^2)(1+x^3)\}$. (四川大学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\}\stackrel\{\frac\{1\}\{x\}=t\}\{=\}&\int\_0^\infty \frac\{\frac\{1\}\{t^2\}\mathrm\{ d\} t\}\{\left(1+\frac\{1\}\{t^2\}\right)\left(1+\frac\{1\}\{t^3\}\right)\} =\int\_0^\infty \frac\{t^3\mathrm\{ d\} t\}\{(1+t^2)(1+t^3)\}\equiv I\\\\ =&\frac\{1\}\{2\}\left(\mbox\{原式\}+I\right)=\frac\{1\}\{2\}\int\_0^\infty \frac\{\mathrm\{ d\} x\}\{1+x^2\}=\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](
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445、 7、 (15 分) 设 $\displaystyle p > 0$, 讨论广义积分 $\displaystyle \int\_1^\infty \frac\{\sin \left(x+\frac\{1\}\{x\}\right)\}\{x^p\}\mathrm\{ d\} x$ 的收敛性与绝对收敛性. (苏州大学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) / 设题中广义积分为 $\displaystyle I$. (1)、 当 $\displaystyle p \leq 0$ 时, 由
\begin\{aligned\} &\quad \ 2n\pi+\frac\{\pi\}\{4\}\leq x\leq 2n\pi+\frac\{\pi\}\{2\} \Rightarrow \frac\{1\}\{2n\pi+\frac\{\pi\}\{2\}\} < \frac\{1\}\{x\} < \frac\{1\}\{2n\pi+\frac\{\pi\}\{4\}\}\\\\ &\Rightarrow 2n\pi+\frac\{\pi\}\{4\} +\frac\{1\}\{2n\pi+\frac\{\pi\}\{2\}\} < x+\frac\{1\}\{x\} < 2n\pi+\frac\{\pi\}\{2\}\frac\{1\}\{2n\pi+\frac\{\pi\}\{4\}\}\\\\ &\Rightarrow \sin \left(x+\frac\{1\}\{x\}\right)\geq \frac\{1\}\{\sqrt\{2\}\}, \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\_\{2n\pi+\frac\{\pi\}\{4\}\}^\{2n\pi+\frac\{\pi\}\{2\}\} \frac\{\sin \left(x+\frac\{1\}\{x\}\right)\}\{x^p\}\mathrm\{ d\} x &\geq \frac\{\left(2n\pi+\frac\{\pi\}\{4\}\right)^\{-p\}\}\{\sqrt\{2\}\}\cdot \frac\{\pi\}\{4\}\geq \frac\{\pi\}\{4\sqrt\{2\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Cauchy 收敛准则知 $\displaystyle I$ 发散. (2)、 当 $\displaystyle 0 < p \leq 1$ 时,
\begin\{aligned\} \int\_1^\infty \frac\{\sin \left(x+\frac\{1\}\{x\}\right)\}\{x^p\} \mathrm\{ d\} x=\int\_1^\infty \frac\{\left(1-\frac\{1\}\{x^2\}\right)\sin \left(x+\frac\{1\}\{x\}\right)\}\{x^p \left(1-\frac\{1\}\{x^p\}\right)\}\mathrm\{ d\} x.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
注意到
\begin\{aligned\} \left|\int\_1^A\sin \left(x+\frac\{1\}\{x\}\right) \left(1-\frac\{1\}\{x^2\}\right)\mathrm\{ d\} x\right| &=\left|\int\_1^A \sin\left(x+\frac\{1\}\{x\}\right)\mathrm\{ d\} \left(x+\frac\{1\}\{x\}\right)\right|\\\\ &=\left|\cos 2-\cos\left(A+\frac\{1\}\{A\}\right)\right|\\\\ &\leq 2,\\\\ \frac\{1\}\{x^p\left(1-\frac\{1\}\{x^2\}\right)\} &=\frac\{1\}\{x^p-x^\{p-2\}\}\searrow 0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
我们可由 Dirichlet 判别法推得 $\displaystyle I$ 收敛. 进一步,
\begin\{aligned\} \left|\frac\{\sin \left(x+\frac\{1\}\{x\}\right)\}\{x^p\}\right| \geq \frac\{\sin^2\left(x+\frac\{1\}\{x\}\right)\}\{x^p\} =\frac\{1\}\{2\}\left\[\frac\{1\}\{x^p\}-\frac\{\cos 2\left(x+\frac\{1\}\{x\}\right)\}\{x^p\}\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
同 $\displaystyle (I)$ 的论述知 $\displaystyle \int\_1^\infty \frac\{\cos 2\left(x+\frac\{1\}\{x\}\right)\}\{x^p\}\mathrm\{ d\} x$ 收敛. 由 $\displaystyle \int\_1^\infty \frac\{1\}\{x^p\}\mathrm\{ d\} x=+\infty$ 知
\begin\{aligned\} \int\_1^\infty \frac\{\left|\sin \left(x+\frac\{1\}\{x\}\right)\right|\}\{x^p\}\mathrm\{ d\} x &\geq \int\_1^\infty \frac\{\sin^2\left(x+\frac\{1\}\{x\}\right)\}\{x^p\}\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\}
(3)、 当 $\displaystyle p > 1$ 时, 由 $\displaystyle \left|\frac\{\sin \left(x+\frac\{1\}\{x\}\right)\}\{x^p\}\right|\leq\frac\{1\}\{x^p\}$ 及比较判别法知 $\displaystyle I$ 绝对收敛. 综上, 当 $\displaystyle p\leq 0$ 时, $\displaystyle I$ 发散; 当 $\displaystyle 0 < p\leq 1$ 时, $\displaystyle I$ 条件收敛; 当 $\displaystyle p > 1$ 时, $\displaystyle I$ 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](
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446、 10、 (1)、 证明: $\displaystyle \int\_0^\infty \frac\{\sin^3x\}\{x\}\mathrm\{ d\} x$ 收敛. (天津大学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\} &\sin^3x=\sin x(1-\cos^2x) =\sin x-\sin x\frac\{1+\cos 2x\}\{2\}\\\\ =&\frac\{\sin x\}\{2\}-\frac\{1\}\{4\}[\sin 3x+\sin (-x)] =\frac\{3\sin x-\sin 3x\}\{4\} \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\} \mbox\{原式\}=&\int\_0^\infty \frac\{\sin x\}\{x\}\mathrm\{ d\} x-\frac\{1\}\{4\}\int\_0^\infty \frac\{\sin 3x\}\{x\}\mathrm\{ d\} x\\\\ \stackrel\{3x=t\}\{=\}&\frac\{3\}\{4\}\int\_0^\infty \frac\{\sin x\}\{x\}\mathrm\{ d\} x -\frac\{1\}\{4\}\int\_0^\infty \frac\{\sin t\}\{t\}\mathrm\{ d\} t=\frac\{1\}\{2\}\int\_0^\infty \frac\{\sin x\}\{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\}
而由 Dirichlet 判别法知 $\displaystyle \int\_0^\infty \frac\{\sin x\}\{x\}\mathrm\{ d\} x$ 收敛, 而原式也收敛.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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447、 2、 讨论积分 $\displaystyle \int\_0^\infty \left\[\left(1-\frac\{\sin x\}\{x\}\right)^\{-\frac\{1\}\{p\}\}-1\right\]\mathrm\{ d\} x$ 的敛散性 (绝对敛散性). (同济大学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) / $\displaystyle \mbox\{原式\}=\int\_0^1+\int\_1^\infty \cdots\equiv I\_1+I\_2$. 由
\begin\{aligned\} \left(1-\frac\{\sin x\}\{x\}\right)^\{-\frac\{1\}\{p\}\}\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \left\[\frac\{1\}\{3!\}x^2+o(x^2)\right\]^\{-\frac\{1\}\{p\}\} \sim (3!)^\frac\{1\}\{p\} x^\{-\frac\{2\}\{p\}\}, x\to 0 \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 $\displaystyle \frac\{2\}\{p\} < 1\Leftrightarrow p > 2$ 时, $\displaystyle I\_1$ 收敛且为绝对收敛. 再者,
\begin\{aligned\} x > 1\Rightarrow& \left|\frac\{\sin x\}\{x\}\right| < 1 \Rightarrow \left(1-\frac\{\sin x\}\{x\}\right)^\{-\frac\{1\}\{p\}\}-1&\xlongequal[\tiny\mbox\{展开\}]\{\tiny\mbox\{Taylor\}\} \left\[1+\frac\{1\}\{p\}\frac\{\sin x\}\{x\}+O\left(\frac\{1\}\{x^2\}\right)\right\]\\\\ &=\frac\{1\}\{p\}\frac\{\sin x\}\{x\}+O\left(\frac\{1\}\{x^2\}\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
蕴含 $\displaystyle I\_2$ 是一个条件收敛广义积分与绝对收敛广义积分的和, 是条件收敛的. 综上即知原广义积分收敛 (且为条件收敛) 当且仅当 $\displaystyle p > 2$.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](
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---
448、 (5)、 $\displaystyle \int\_0^\infty\frac\{\sin^4 x\}\{x^2\}\mathrm\{ d\} x$. (西安交通大学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\{\sin^2x(1-\cos^2x)\}\{x^2\}\mathrm\{ d\} x =\int\_0^\infty\frac\{\sin^2x\}\{x^2\}\mathrm\{ d\} x-\frac\{1\}\{4\}\int\_0^\infty\frac\{\sin^22x\}\{x^2\}\mathrm\{ d\} x\\\\ \stackrel\{2x=t\}\{=\}&\int\_0^\infty\frac\{\sin^2x\}\{x^2\}\mathrm\{ d\} x-\frac\{1\}\{4\}\int\_0^\infty\frac\{\sin^2t \}\{\frac\{t^2\}\{4\}\}\mathrm\{ d\}\frac\{t\}\{2\} =\frac\{1\}\{2\}\int\_0^\infty\frac\{\sin^2x\}\{x^2\}\mathrm\{ d\} x\\\\ =&-\frac\{1\}\{2\}\int\_0^\infty\sin^2 x\mathrm\{ d\} \frac\{1\}\{x\} \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \frac\{1\}\{2\}\int\_0^\infty\frac\{\sin 2x\}\{x\}\mathrm\{ d\} x\\\\ \stackrel\{2x=t\}\{=\}&\frac\{1\}\{2\}\int\_0^\infty\frac\{\sin t\}\{t\}\mathrm\{ d\} t =\frac\{1\}\{2\}\cdot\frac\{\pi\}\{2\}=\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)/
---
449、 8、 (15 分) 判断反常积分 $\displaystyle \int\_0^\infty \frac\{\sin\left(x+\frac\{1\}\{x\}\right)\}\{x^p\}\mathrm\{ d\} x$ 的敛散性. (西南财经大学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) / 设题中反常积分为
\begin\{aligned\} I=\int\_0^1 \frac\{\sin\left(x+\frac\{1\}\{x\}\right)\}\{x^p\}\mathrm\{ d\} x+\int\_1^\infty \frac\{\sin\left(x+\frac\{1\}\{x\}\right)\}\{x^p\}\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\}
(1)、 当 $\displaystyle p\leq 0$ 时 $\displaystyle I\_2$ 发散. 在作变换 $\displaystyle x=\frac\{1\}\{t\}$ 后知当 $\displaystyle \alpha\geq 2$ 时, $\displaystyle I\_1$ 发散. (2)、 以下考虑 $\displaystyle 0 < \alpha < 2$. 重新写出
\begin\{aligned\} I=&\int\_0^1 \frac\{\left(1-\frac\{1\}\{x^2\}\right)\sin\left(x+\frac\{1\}\{x\}\right)\}\{x^p\left(1-\frac\{1\}\{x^2\}\right)\}\mathrm\{ d\} x +\int\_1^\infty \frac\{\left(1-\frac\{1\}\{x^2\}\right)\sin\left(x+\frac\{1\}\{x\}\right)\}\{x^p\left(1-\frac\{1\}\{x^2\}\right)\}\mathrm\{ d\} x \equiv I\_3+I\_4. \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\} &\left|\int\_1^A \left(1-\frac\{1\}\{x^2\}\right)\sin\left(x+\frac\{1\}\{x\}\right)\mathrm\{ d\} x\right| =\left|\left.-\cos \left(x+\frac\{1\}\{x\}\right)\right|\_1^A\right|\\\\ =&\left|\cos 2-\cos \left(A+\frac\{1\}\{A\}\right)\right|\leq 2, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle \frac\{1\}\{x^p \left(1-\frac\{1\}\{x^2\}\right)\} =\frac\{1\}\{x^p-\frac\{1\}\{x^\{2-p\}\}\}\searrow 0$, 及 Dirichlet 判别法知 $\displaystyle I\_4$ 收敛. 作变换 $\displaystyle x=\frac\{1\}\{t\}$ 后即知 $\displaystyle I\_3$ 也收敛. 故 $\displaystyle I$ 在 $\displaystyle (0,2)$ 内收敛. (3)、 写出
\begin\{aligned\} \left|\frac\{\sin\left(x+\frac\{1\}\{x\}\right)\}\{x^p\}\right| \geq \frac\{\sin^2\left(x+\frac\{1\}\{x\}\right)\}\{x^p\} =\frac\{1\}\{2x^p\}-\frac\{1\}\{2x^a\} \cos 2\left(x+\frac\{1\}\{x\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
类似第 2 步知当 $\displaystyle 0 < p < 2$ 时, $\displaystyle \int\_0^\infty \frac\{1\}\{2x^a\} \cos 2\left(x+\frac\{1\}\{x\}\right)\mathrm\{ d\} x$ 收敛. 但 $\displaystyle 0 < p\leq 1\Rightarrow \int\_1^\infty\frac\{1\}\{2x^p\}\mathrm\{ d\} x=+\infty$, 当 $\displaystyle 1 < p < 2$ 时, $\displaystyle \int\_0^1 \frac\{1\}\{2x^p\}\mathrm\{ d\} x=+\infty$. 故 $\displaystyle \int\_0^\infty \left|\frac\{\sin\left(x+\frac\{1\}\{x\}\right)\}\{x^p\}\right|\mathrm\{ d\} x=+\infty$. 综上, 当且仅当 $\displaystyle 0 < p < 2$ 时, $\displaystyle I$ 收敛, 且为条件收敛.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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---
450、 (6)、 对于广义积分 $\displaystyle \int\_1^\infty \frac\{\mathrm\{ d\} x\}\{x^\lambda \ln x\}$, 下列说法正确的是 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$.
A. $\displaystyle \lambda > 1$ 时收敛, $\displaystyle \lambda\leq 1$ 时收敛
B. $\displaystyle \lambda > 1, \lambda\leq 1$ 时都发散
C. $\displaystyle \lambda > 1$ 时收敛, $\displaystyle \lambda\leq 1$ 时发散
C. $\displaystyle \lambda > 1, \lambda\leq 1$ 时收敛 (长安大学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) / $\displaystyle C$. 当 $\displaystyle \lambda > 1$ 时, 由 $\displaystyle \lim\_\{x\to+\infty\}\frac\{\frac\{1\}\{x^\lambda\ln x\}\}\{\frac\{1\}\{x^\lambda\}\} =\lim\_\{x\to+\infty\}\frac\{1\}\{\ln x\}=0$ 及比较判别法知原广义积分收敛. 当 $\displaystyle \lambda=1$ 时,
\begin\{aligned\} \mbox\{原式\}=\int\_1^\infty \frac\{\mathrm\{ d\} x\}\{x\ln x\}=\int\_0^\infty \frac\{\mathrm\{ d\} t\}\{t\}=+\infty; \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
当 $\displaystyle \lambda < 1$ 时, 由 $\displaystyle \lim\_\{x\to+\infty\}\frac\{\frac\{1\}\{x^\lambda\ln x\}\}\{\frac\{1\}\{x^\frac\{1+\lambda\}\{2\}\}\} =\lim\_\{x\to+\infty\}\frac\{x^\frac\{1-\lambda\}\{2\}\}\{\ln x\}=+\infty$ 及比较判别法知原广义积分发散.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)/
---
451、 3、 (10 分) 讨论 $\displaystyle \int\_1^\infty \frac\{\cos^2x\}\{x\}\mathrm\{ d\} x$ 的敛散性. (中国海洋大学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) / $\displaystyle \mbox\{原式\}=\frac\{1\}\{2\}\int\_1^\infty \frac\{1+\cos 2x\}\{x\}\mathrm\{ d\} x =\frac\{1\}\{2\}\int\_1^\infty \frac\{\mathrm\{ d\} x\}\{x\}+\frac\{1\}\{2\}\int\_1^\infty \frac\{\cos 2x\}\{x\}\mathrm\{ d\} x$, 作为一个发散广义积分与收敛广义积分 (由 Dirichlet 判别法) 的和, 是发散的.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)/
---
452、 3、 (15 分) 求积分 $\displaystyle \int\_0^\infty (x+1)\mathrm\{e\}^\{-x^2\}\mathrm\{ d\} x$. (中国科学技术大学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\} \left\[\int\_0^\infty \mathrm\{e\}^\{-x^2\}\mathrm\{ d\} x\right\]^2&=\int\_0^\infty \mathrm\{e\}^\{-x^2\}\mathrm\{ d\} x\int\_0^\infty \mathrm\{e\}^\{-y^2\}\mathrm\{ d\} y\\\\ &=\iint\_\{(0,\infty)^2\}\mathrm\{e\}^\{-(x^2+y^2)\}\mathrm\{ d\} x\mathrm\{ d\} y =\int\_0^\infty \mathrm\{e\}^\{-r^2\}r\mathrm\{ d\} r\int\_0^\frac\{\pi\}\{2\}\mathrm\{ d\} \theta =\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\}
知 $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-x^2\}\mathrm\{ d\} x=\frac\{\sqrt\{\pi\}\}\{2\}$. 故
\begin\{aligned\} \mbox\{原式\}=\left.-\frac\{\mathrm\{e\}^\{-x^2\}\}\{2\}\right|\_0^\infty+\frac\{\sqrt\{\pi\}\}\{2\}=\frac\{1+\sqrt\{\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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
453、 10、 计算 $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-x^2\}\cos (\alpha x)\mathrm\{ d\} x$. (中国科学院大学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(\alpha)=\int\_0^\infty \mathrm\{e\}^\{-x^2\}\cos ()\mathrm\{ d\} x=\frac\{\sqrt\{\pi\}\}\{2\}\mathrm\{e\}^\{-\frac\{\alpha^2\}\{4\}\}. \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 $\displaystyle I(0)=\int\_0^\infty \mathrm\{e\}^\{-x^2\}\mathrm\{ d\} x=\frac\{\sqrt\{\pi\}\}\{2\}$,
\begin\{aligned\} I'(\alpha)&=\int\_0^\infty -x\mathrm\{e\}^\{-x^2\} \sin (\alpha x)\mathrm\{ d\} x =\frac\{1\}\{2\}\int\_0^\infty \sin (\alpha x)\mathrm\{ d\} \mathrm\{e\}^\{-x^2\}\\\\ &=-\frac\{1\}\{2\}\int\_0^\infty \alpha\cos(\alpha x)\mathrm\{e\}^\{-x^2\}\mathrm\{ d\} x =-\frac\{\alpha\}\{2\}I(\alpha) \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\} &\frac\{\mathrm\{ d\} I\}\{I\}=-\frac\{\alpha\}\{2\}\mathrm\{ d\} \alpha \Rightarrow \ln I=-\frac\{\alpha^2\}\{4\}+C\_1\\\\ \Rightarrow& I(\alpha)=C\mathrm\{e\}^\{-\frac\{\alpha^2\}\{4\}\} \stackrel\{I(0)=\frac\{\sqrt\{\pi\}\}\{2\}\}\{\Rightarrow\} I(\alpha)=\frac\{\sqrt\{\pi\}\}\{2\}\mathrm\{e\}^\{-\frac\{\alpha^2\}\{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](
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---
454、 6、 (10 分) 求反常积分 $\displaystyle \int\_0^\infty \frac\{\arctan x\}\{x^p\}\mathrm\{ d\} x$ 的敛散性. (中国矿业大学(北京)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^p\}\mathrm\{ d\} x =\int\_0^1 \frac\{\arctan x\}\{x^p\}\mathrm\{ d\} x+\int\_1^\infty \frac\{\arctan x\}\{x^p\}\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^p\}\sim \frac\{x\}\{x^p\}=\frac\{1\}\{x^\{p-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 $\displaystyle p-1 < 1\Leftrightarrow p < 2$ 时, $\displaystyle I\_1$ (绝对) 收敛. 又由
\begin\{aligned\} \frac\{\arctan x\}\{x^p\}\sim\frac\{\pi\}\{2x^p\}\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 $\displaystyle p > 1$ 时, $\displaystyle I\_2$ (绝对收敛). 综上即知当且仅当 $\displaystyle 1 < p < 2$ 时, 原广义积分收敛, 且为绝对收敛.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](
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---
455、 (8)、 $\displaystyle \int\_2^\infty \frac\{1\}\{x\ln^\{2023\}x\}\mathrm\{ d\} x=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (重庆师范大学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\}\stackrel\{\ln x=t\}\{=\}\int\_\{\ln 2\}^\infty \frac\{\mathrm\{ d\} t\}\{t^\{2023\}\} =\left.\frac\{1\}\{2022\}t^\{-2022\}\right|\_\{t=\ln 2\}^\{t=+\infty\} =\frac\{1\}\{2022\ln^\{2022\}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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---
456、 6、 (15 分) 设 $\displaystyle a\_\{2n-1\}=\frac\{1\}\{n\}, a\_\{2n\}=\int\_n^\{n+1\}\frac\{\mathrm\{ d\} x\}\{x\}$. 证明: $\displaystyle \sum\_\{n=1\}^\infty (-1)^na\_n$ 条件收敛. (安徽大学2023年高等代数考研试题) [数项级数 ]
[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)、 设级数 $\displaystyle \sum\_\{n=1\}^\infty (-1)^n a\_n$ 的前 $\displaystyle n$ 项和为 $\displaystyle S\_n$, 则由题意知
\begin\{aligned\} S\_\{2n\}&=-a\_1+a\_2+\cdots-a\_\{2n-1\}+a\_\{2n\}\\\\ &=-\left(1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n\}\right)\\\\ &\quad +\left\[(\ln 2-\ln 1)+\cdots+\left(\ln (n+1)-\ln n\right)\right\]\\\\ &=\ln (n+1)-\left(1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 由
\begin\{aligned\} S\_\{2(n+1)\}-S\_\{2n\} =&\ln (n+2)-\left(1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n+1\}\right)\\\\ & -\ln (n+1)+\left(1+\frac\{1\}\{2\}+\cdots+\frac\{1\}\{n\}\right)\\\\ =&\ln \left(1+\frac\{1\}\{n+1\}\right)-\frac\{1\}\{n+1\} < 0,\\\\ S\_\{2n\}=&\ln (n+1)-\sum\_\{k=1\}^n \frac\{1\}\{k\} > \ln(n+1)-1-\sum\_\{k=2\}^n \int\_\{k-1\}^k\frac\{1\}\{x\}\mathrm\{ d\} x\\\\ =&\ln (n+1)-1-\int\_1^n\frac\{1\}\{x\}\mathrm\{ d\} x =\ln \left(1+\frac\{1\}\{n\}\right)-1 > -1 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及单调有界定理知 $\displaystyle \lim\_\{n\to\infty\}S\_\{2n\}$ 存在, 设为 $\displaystyle S$. 则
\begin\{aligned\} \lim\_\{n\to\infty\}S\_\{2n+1\}=\lim\_\{n\to\infty\}\left(S\_\{2n\}+\frac\{1\}\{n+1\}\right)=S. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \lim\_\{n\to\infty\}S\_n=S$. 这表明 $\displaystyle \sum\_\{n=1\}^\infty (-1)^n a\_n=S$. (3)、 由
\begin\{aligned\} \sum\_\{k=1\}^\{2n\}|a\_k| &\geq \sum\_\{k=1\}^n|a\_\{2k\}| =\sum\_\{k=1\}^n \frac\{1\}\{k\} > \sum\_\{k=1\}^n \int\_k^\{k+1\}\frac\{1\}\{x\}\mathrm\{ d\} x =\ln (n+1)\to +\infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \sum\_\{n=1\}^\infty |a\_n|$ 发散. 而 $\displaystyle \sum\_\{n=1\}^\infty (-1)^na\_n$ 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)/
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457、 4、 (15 分) 设数列 $\displaystyle \left\\{a\_n\right\\}$ 为单调递减的非负数列. 证明: 级数 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 与 $\displaystyle \sum\_\{n=1\}^\infty 2^na\_\{2^n\}$ 具有相同的敛散性. (北京科技大学2023年数学分析考研试题) [数项级数 ]
[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
\begin\{aligned\} \sum\_\{n=1\}^\infty a\_n&=a\_1+(a\_2+a\_3)+(a\_4+\cdots+a\_7) +(a\_8+\cdots+a\_\{15\})+\cdots\\\\ &\leq a\_1+2a\_2+2^2a\_4+a^3a\_8+\cdots =\sum\_\{n=0\}^\infty 2^na\_\{2^n\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 发散 $\displaystyle \Rightarrow \sum\_\{n=0\}^\infty 2^na\_\{2^n\}$ 发散, $\displaystyle \sum\_\{n=0\}^\infty 2^na\_\{2^n\}$ 收敛 $\displaystyle \Rightarrow \sum\_\{n=1\}^\infty a\_n$ 收敛. (2)、 又由
\begin\{aligned\} \sum\_\{n=1\}^\infty a\_n&=a\_1+a\_2+(a\_3+a\_4)+(a\_5+\cdots+a\_8)+(a\_9+\cdots+a\_\{16\})+\cdots\\\\ &\geq a\_1+a\_2+2a\_\{2^2\}+2^2a\_\{2^3\} +2^3a\_\{2^4\}+\cdots\\\\ &=\frac\{a\_1\}\{2\}+\frac\{1\}\{2\}\left(a\_1+2a\_2+2^2a\_\{2^2\}+2^3a\_\{2^3\} +2^4a\_\{2^4\}+\cdots\right)\\\\ &=\frac\{a\_1\}\{2\}+\frac\{1\}\{2\}\sum\_\{n=0\}^\infty 2^na\_\{2^n\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 收敛 $\displaystyle \Rightarrow \sum\_\{n=0\}^\infty 2^na\_\{2^n\}$ 收敛, $\displaystyle \sum\_\{n=0\}^\infty 2^na\_\{2^n\}$ 发散 $\displaystyle \Rightarrow \sum\_\{n=1\}^\infty a\_n$ 发散.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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458、 2、 判断级数 $\displaystyle \sum\_\{n=2\}^\infty \frac\{\ln (\mathrm\{e\}^n+n^2)\}\{n^2\ln^2n\}$ 的敛散性. (北京邮电大学2023年数学分析考研试题) [数项级数 ]
[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\} &\lim\_\{n\to\infty\}\frac\{\frac\{\ln (\mathrm\{e\}^n+n^2)\}\{n^2\ln^2n\}\}\{\frac\{1\}\{n\ln^2n\}\} =\lim\_\{n\to\infty\}\frac\{\ln(\mathrm\{e\}^n+n^2)\}\{n\}\\\\ \xlongequal[\tiny\mbox\{原理\}]\{\tiny\mbox\{归结\}\}& \lim\_\{x\to+\infty\}\frac\{\ln(\mathrm\{e\}^x+x^2)\}\{x\} \xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{x\to+\infty\}\frac\{\mathrm\{e\}^x+2x\}\{\mathrm\{e\}^x+x^2\}=1,\\\\ &\sum\_\{n=2\}^\infty \frac\{1\}\{n\ln^2n\}\leq \frac\{1\}\{2\ln^22\}+\sum\_\{n=3\}^\infty \int\_\{n-1\}^n\frac\{\mathrm\{ d\} x\}\{x\ln^2x\} =\frac\{1\}\{2\ln^22\}+\frac\{1\}\{\ln 2\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原级数收敛, 且为绝对收敛.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)/
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459、 (7)、 设 $\displaystyle \left\\{a\_n\right\\}, \left\\{b\_n\right\\}$ 均不为零, 且 $\displaystyle \lim\_\{n\to\infty\}\frac\{a\_n\}\{b\_n\}=1$. 若级数 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 收敛, 级数 $\displaystyle \sum\_\{n=1\}^\infty b\_n$ 是否收敛, 为什么? (大连理工大学2023年数学分析考研试题) [数项级数 ]
[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
\begin\{aligned\} a\_n=\frac\{(-1)^n\}\{\sqrt\{n\}\}, b\_n=\frac\{(-1)^n\}\{\sqrt\{n\}\}+\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\}
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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460、 (3)、 若 $\displaystyle p > 1$, 证明: 数项级数 $\displaystyle \sum\_\{n=2\}^\infty \frac\{\sin \left(2\pi \sqrt\{n^2+1\}\right)\}\{\ln^p n\}$ 收敛. (电子科技大学2023年数学分析考研试题) [数项级数 ]
[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 $\displaystyle a\_n=\frac\{\sin \left(2\pi \sqrt\{n^2+1\}\right)\}\{\ln^p n\}$, 则
\begin\{aligned\} a\_n=&\frac\{\sin \left(2\pi \left(\sqrt\{n^2+1\}-2n\pi\right)\right)\}\{\ln^p n\} =\frac\{1\}\{\ln^pn\} \sin\frac\{2\pi\}\{\sqrt\{n^2+1\}+n\}\\\\ \sim&\frac\{1\}\{\ln^pn\}\cdot\frac\{2\pi\}\{2n\}=\frac\{\pi\}\{n\ln^pn\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由积分判别法即知 $\displaystyle \sum\_\{n=1\}^\infty a\_n$ 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 08:51:26
