## 张祖锦2023年数学专业真题分类70天之第37天
---
829、 10、 (15 分) 设函数 $\displaystyle f(x)$ 在 $\displaystyle (-\infty,+\infty)$ 上具有一阶连续导数, $\displaystyle L$ 是上半平面 $\displaystyle (y > 0)$ 内的有向分段光滑曲线, 其起点为 $\displaystyle (a,b)$, 终点为 $\displaystyle (c,d)$. 记
\begin\{aligned\} I=\int\_L \frac\{1\}\{y\}[1+y^2f(xy)]\mathrm\{ d\} x+\frac\{x\}\{y^2\}[y^2f(xy)-1]\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\}
(1)、 证明曲线积分 $\displaystyle I$ 与路径 $\displaystyle L$ 无关; (2)、 当 $\displaystyle ab=cd$ 时, 求 $\displaystyle I$ 的值. (重庆大学2023年数学分析考研试题) [曲线曲面积分 ]
[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)、 设
\begin\{aligned\} P=\frac\{1\}\{y\}+yf(xy), Q=xf(xy)-\frac\{x\}\{y^2\}, \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\} P\_y=-\frac\{1\}\{y^2\}+f(xy)+xyf'(xy)=Q\_x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle I$ 与 $\displaystyle L$ 无关. (2)、 设
\begin\{aligned\} C: y=\frac\{ab\}\{x\}=\frac\{cd\}\{x\}, x: a\to c. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则由第 1 步知
\begin\{aligned\} I&=\int\_C \left\[\frac\{1\}\{y\}+yf(ab)\right\]\mathrm\{ d\} x +\left\[xf(ab)-\frac\{x\}\{y^2\}\right\]\mathrm\{ d\} y\\\\ &=f(ab)\int\_C y\mathrm\{ d\} x+x\mathrm\{ d\} y+\int\_C \frac\{1\}\{y\}\mathrm\{ d\} x-\frac\{x\}\{y^2\}\mathrm\{ d\} y\\\\ &=f(ab)\int\_C \mathrm\{ d\} (xy)+\int\_C \mathrm\{ d\} \frac\{x\}\{y\}\\\\ &=f(ab) (cd-ab)+\left(\frac\{c\}\{d\}-\frac\{a\}\{b\}\right) =\frac\{c\}\{d\}-\frac\{a\}\{b\}. \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)/
---
830、 (6)、 设 $\displaystyle \varSigma$ 是 $\displaystyle x^2+y^2+z^2=a^2$ ($a > 0$) 的左半侧, 则 $\displaystyle \iint\_\varSigma(x+y+z)\mathrm\{ d\} S=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (重庆师范大学2023年数学分析考研试题) [曲线曲面积分 ]
[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\}=&\iint\_\{x^2+y^2+z^2=a^2\atop y\leq 0\} (x+y+z)\mathrm\{ d\} S \xlongequal\{\tiny\mbox\{对称性\}\} 2\iint\_\{x^2+y^2+z^2=a^2\atop y\leq 0, z\geq0\} (x+y+z)\mathrm\{ d\} S\\\\ \stackrel\{z=\sqrt\{a^2-x^2-y^2\}\}\{=\}&2\iint\_\{x^2+y^2\leq a^2\atop y\leq 0\}(x+y+\sqrt\{a^2-x^2-y^2\})\frac\{a\}\{\sqrt\{a^2-x^2-y^2\}\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&2a\left\[\iint\_\{x^2+y^2\leq a^2\atop y\leq 0\}\frac\{y\}\{\sqrt\{a^2-x^2-y^2\}\}\mathrm\{ d\} x\mathrm\{ d\} y+\frac\{\pi a^2\}\{2\}\right\]\\\\ =&2a\left\[\int\_\pi^\{2\pi\} \mathrm\{ d\} \theta\int\_0^a \frac\{r\sin\theta\}\{\sqrt\{a^2-r^2\}\}r\mathrm\{ d\} r+\frac\{\pi a^2\}\{2\}\right\]\\\\ =&2a\left\[-2\int\_0^a \frac\{r^2\}\{\sqrt\{a^2-r^2\}\}\mathrm\{ d\} r+\frac\{\pi a^2\}\{2\}\right\] \stackrel\{r=a\sin t\}\{=\}\cdots=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)/
---
831、 9、 (15 分) 确定含参量积分
\begin\{aligned\} g(\alpha)=\int\_0^\infty \frac\{\ln(1+x^3)\}\{x^\alpha\}\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\}
的连续区间. (北京师范大学2023年数学分析考研试题) [含参量积分 ]
[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\} \frac\{\ln(1+x^3)\}\{x^\alpha\}\sim\frac\{x^3\}\{x^\alpha\}=\frac\{1\}\{x^\{\alpha-3\}\}\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 $\displaystyle \alpha < 4$ 时, $\displaystyle \int\_0^1 \frac\{\ln(1+x^3)\}\{x^\alpha\}\mathrm\{ d\} x$ 收敛. 又由
\begin\{aligned\} \alpha\leq 1\Rightarrow&\lim\_\{x\to+\infty\}s \frac\{\frac\{\ln(1+x^3)\}\{x^\alpha\}\}\{\frac\{1\}\{x^\alpha\}\} =\lim\_\{x\to+\infty\}\ln(1+x^3)=+\infty,\\\\ \alpha > 1\Rightarrow&\lim\_\{x\to+\infty\}\frac\{\frac\{\ln(1+x^3)\}\{x^\alpha\}\}\{\frac\{1\}\{x^\frac\{\alpha+1\}\{2\}\}\} =\lim\_\{x\to+\infty\}\frac\{\ln(1+x^3)\}\{x^\frac\{\alpha-1\}\{2\}\}\\\\ &=\lim\_\{t\to+\infty\}\frac\{\ln(1+t^\frac\{6\}\{\alpha-1\})\}\{t\}\xlongequal\{\tiny\mbox\{L'Hospital\}\} \cdots=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 $\displaystyle \alpha > 1$ 时, $\displaystyle \int\_1^\infty \frac\{\ln(1+x^3)\}\{x^\alpha\}\mathrm\{ d\} x$ 收敛. 综上, 当且仅当 $\displaystyle 1 < \alpha < 4$ 时, $\displaystyle g(\alpha)$ 收敛. 对 $\displaystyle \forall\ 0 < \delta < \frac\{3\}\{2\}$, 我们证明 $\displaystyle g(\alpha)$ 在 $\displaystyle 1+\delta\leq \alpha\leq 4-\delta$ 上一致收敛,而 $\displaystyle g$ 连续. 由 $\displaystyle \delta$ 的任意性知 $\displaystyle g$ 在 $\displaystyle (1,4)$ 连续. 事实上, 当 $\displaystyle 1+\delta\leq \alpha\leq 4-\delta$ 时,
\begin\{aligned\} 0 < x < 1\Rightarrow&\frac\{\ln(1+x^3)\}\{x^\alpha\}\leq \frac\{x^3\}\{x^\{4-\delta\}\}=\frac\{1\}\{x^\{1-\delta\}\},\\\\ x > 1\Rightarrow&\frac\{\ln(1+x^3)\}\{x^\alpha\}\leq\frac\{\ln(1+x^3)\}\{x^\{1+\delta\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 Weierstrass 判别法即知 $\displaystyle \int\_0^\infty \frac\{\ln(1+x^3)\}\{x^\alpha\}\mathrm\{ d\} x$ 关于 $\displaystyle 1+\delta\leq \alpha\leq 4-\delta$ 一致收敛. 综上即知 $\displaystyle g$ 的连续区间为 $\displaystyle (1,4)$.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)/
---
832、 12、 设 $\displaystyle 0 < p < 2, I(y)=\int\_1^\infty \frac\{\cos xy\}\{x^p\}\mathrm\{ d\} x$. (1)、 讨论 $\displaystyle I(y)$ 在 $\displaystyle (0,+\infty)$ 上的一致收敛性; (2)、 讨论 $\displaystyle I(y)$ 在 $\displaystyle (0,+\infty)$ 上的连续性. (东南大学2023年数学分析考研试题) [含参量积分 ]
[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)、 (1-1)、 当 $\displaystyle 1 < p < 2$ 时, 由 $\displaystyle \left|\frac\{\cos xy\}\{x^p\}\right|\leq\frac\{1\}\{x^p\}$ 及 Weierstrass 判别法知 $\displaystyle I(y)$ 关于 $\displaystyle y > 0$ 一致收敛. (1-2)、 当 $\displaystyle 0 < p\leq 1$ 时, 由
\begin\{aligned\} \sup\_\{y > 0\}\left|\int\_A^B \frac\{\cos xy\}\{x^p\}\mathrm\{ d\} x\right| \geq \lim\_\{y\to0^+\}\left|\int\_A^B \frac\{\cos xy\}\{x^p\}\mathrm\{ d\} x\right| =\left|\int\_A^B \frac\{1\}\{x^p\}\mathrm\{ d\} x\right|\xrightarrow\{B\to+\infty\}+\infty \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 $\displaystyle I(y)$ 关于 $\displaystyle y > 0$ 不一致收敛. (2)、 对 $\displaystyle \forall\ \delta > 0$, 由 $\displaystyle \sup\_\{y\geq \delta\}\left|\int\_1^A\cos xy\mathrm\{ d\} x\right|\leq\sup\_\{y\geq \delta\}\frac\{2\}\{y\}=\frac\{2\}\{\delta\}$ 及 Dirichlet 判别法知 $\displaystyle I(y)$ 关于 $\displaystyle y\geq \delta$ 一致收敛, 而 $\displaystyle I(y)$ 在 $\displaystyle [\delta,+\infty)$ 上连续. 由 $\displaystyle \delta$ 的任意性知 $\displaystyle I(y)$ 在 $\displaystyle (0,+\infty)$ 上连续.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)/
---
833、 (2)、 设 $\displaystyle x,t > 0$, 则含参量积分
\begin\{aligned\} I(t)=\int\_0^\infty \frac\{\mathrm\{e\}^\{-x\}-\mathrm\{e\}^\{-xt\}\}\{x\}\mathrm\{ d\} x=\underline\{\ \ \ \ \ \ \ \ \ \ \}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(复旦大学2023年分析(第6,7,8,9,10题没做)考研试题) [含参量积分 ]
[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(t)=&\int\_0^\infty \int\_1^t \mathrm\{e\}^\{-xs\}\mathrm\{ d\} s\mathrm\{ d\} x =\int\_1^t \int\_0^\infty \mathrm\{e\}^\{-xs\}\mathrm\{ d\} x\mathrm\{ d\} s\\\\ =&\int\_1^t \frac\{1\}\{s\}\mathrm\{ d\} s=\ln t. \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)/
---
834、 (2)、 设 $\displaystyle \alpha > 0$. (2-1)、 证明 $\displaystyle \int\_0^\infty \frac\{\mathrm\{ d\} x\}\{(1+x^2)(1+x^2y^2)\}$ 关于 $\displaystyle y$ 在 $\displaystyle [0,\alpha]$ 上一致收敛; (2-2)、 求 $\displaystyle \int\_0^\infty \frac\{\arctan (\alpha x)\}\{x(1+x^2)\}\mathrm\{ d\} x$. (河海大学2023年数学分析考研试题) [含参量积分 ]
[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 $\displaystyle \sup\_\{y\in\mathbb\{R\}\}\frac\{1\}\{(1+x^2)(1+x^2y^2)\}\leq\frac\{1\}\{1+x^2\}$ 及 Weierstrass 判别法知题中含参量广义积分关于 $\displaystyle y$ 在 $\displaystyle \mathbb\{R\}$ 上一致收敛. 结论得证. (2-2)、 由 Abel 判别法知 $\displaystyle f(\alpha)=\int\_0^\infty \frac\{\arctan (\alpha x)\}\{x(1+x^2)\}\mathrm\{ d\} x$ 收敛. 由第 1 步知
\begin\{aligned\} f'(\alpha)=&\int\_0^\infty \frac\{\mathrm\{ d\} x\}\{(1+x^2)(1+\alpha^2x^2)\}\\\\ =&\frac\{1\}\{1-\alpha^2\}\int\_0^\infty\left(\frac\{1\}\{1+x^2\}-\frac\{1\}\{1+\alpha^2x^2\}\right)\mathrm\{ d\} x\\\\ &=\frac\{1\}\{1-\alpha^2\}\left(\frac\{\pi\}\{2\}-\frac\{\pi\}\{2\}\alpha\right) =\frac\{\pi\}\{2(1+\alpha)\}. \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\} f(\alpha)=f(0)+\int\_0^\alpha f'(t)\mathrm\{ d\} t=\frac\{\pi\}\{2\}\ln(1+\alpha). \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)/
---
835、 7、 证明含参变量积分 $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-\alpha x^2\}\mathrm\{ d\} x$ 在 $\displaystyle 0 < \alpha\_0 < \alpha < \infty$ 上一致收敛, 并问其在 $\displaystyle 0 < \alpha < \infty$ 是否一致收敛? (华东师范大学2023年数学分析考研试题) [含参量积分 ]
[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 $\displaystyle \sup\_\{\alpha > \alpha\_0\}\mathrm\{e\}^\{-\alpha x^2\}=\mathrm\{e\}^\{-\alpha\_0 x^2\}$ 及 Weierstrass 判别法即知 $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-\alpha x^2\}\mathrm\{ d\} x$ 在 $\displaystyle 0 < \alpha\_0 < \alpha < \infty$ 上一致收敛. 又由
\begin\{aligned\} \sup\_\{\alpha > 0\} \int\_\{n-1\}^n \mathrm\{e\}^\{-\alpha x^2\}\mathrm\{ d\} x \geq \int\_\{n-1\}^n \mathrm\{e\}^\{-\frac\{x^2\}\{n^2\}\}\mathrm\{ d\} x \geq \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\}
及 Cauchy 收敛准则知 $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-\alpha x^2\}\mathrm\{ d\} x$ 在 $\displaystyle 0 < \alpha < \infty$ 上不一致收敛.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)/
---
836、 9、 证明 $\displaystyle g(s)=\int\_0^\infty x^\{s-1\}\mathrm\{e\}^\{-2x\}\mathrm\{ d\} x$ 对 $\displaystyle s\in (0,+\infty)$ 可微. (华南理工大学2023年数学分析考研试题) [含参量积分 ]
[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 $\displaystyle x^\{s-1\}\mathrm\{e\}^\{-2x\}\sim x^\{s-1\}, x\to 0$ 知 $\displaystyle \int\_0^1 x^\{s-1\}\mathrm\{e\}^\{-2x\}\mathrm\{ d\} x$ 收敛. 又由
\begin\{aligned\} \lim\_\{x\to+\infty\}\frac\{x^\{s-1\}\mathrm\{e\}^\{-2x\}\}\{\mathrm\{e\}^\{-x\}\}=\lim\_\{x\to+\infty\}\frac\{x^\{s-1\}\}\{\mathrm\{e\}^x\}=0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及比较判别法知 $\displaystyle \int\_1^\infty x^\{s-1\}\mathrm\{e\}^\{-2x\}\mathrm\{ d\} x$ 收敛. 写出
\begin\{aligned\} \int\_0^\infty \frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} s\}\left\[x^\{s-1\}\mathrm\{e\}^\{-2x\}\right\]\mathrm\{ d\} x =\int\_0^\infty x^\{s-1\}\ln x \mathrm\{e\}\{-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 $\displaystyle \forall\ 0 < a < b < \infty$, 由
\begin\{aligned\} &0 < x\leq 1\Rightarrow |x^\{s-1\}\ln x\mathrm\{e\}^\{-x\}|\leq x^\{a-1\} |\ln x|,\\\\ &\lim\_\{x\to 0^+\}\frac\{x^\{a-1\}\ln x\}\{x^\{\frac\{a\}\{2\}-1\}\} =\lim\_\{x\to 0^+\}\frac\{\ln x\}\{x^\{-\frac\{a\}\{2\}\}\}\xlongequal\{\tiny\mbox\{L'Hospital\}\} \cdots 0\Rightarrow \int\_0^1 x^\{a-\}|\ln 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\}
及 Weierstrass 判别法知 $\displaystyle \int\_0^1 x^\{s-1\}\ln x\mathrm\{e\}^\{-x\}\mathrm\{ d\} x$ 关于 $\displaystyle s\in [a,b]$ 一致收敛. 又由
\begin\{aligned\} &x\geq 1\Rightarrow |x^\{s-1\}\ln x\mathrm\{e\}^\{-x\} \leq x^\{b-1\}(x-1)\mathrm\{e\}^\{-x\}=x^b \mathrm\{e\}^\{-x\},\\\\ &\lim\_\{x\to+\infty\}\frac\{x^b \mathrm\{e\}^\{-x\}\}\{\mathrm\{e\}^\{-\frac\{x\}\{2\}\}\}=\lim\_\{x\to+\infty\}\frac\{x^b\}\{\mathrm\{e\}^\frac\{x\}\{2\}\}=0\Rightarrow \int\_1^\infty x^b \mathrm\{e\}^\{-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\}
及 Weierstrass 判别法知 $\displaystyle \int\_1^\infty x^\{s-1\}\ln x\mathrm\{e\}^\{-x\}\mathrm\{ d\} x$ 关于 $\displaystyle s\in [a,b]$ 一致收敛. 故
\begin\{aligned\} g'(s)=\int\_0^\infty x^\{s-1\}\ln x\mathrm\{e\}^\{-x\}, s\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\}
由 $\displaystyle a,b$ 的任意性知 $\displaystyle g$ 在 $\displaystyle (0,+\infty)$ 上可导.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)/
---
837、 8、 (15 分) 设 $\displaystyle I(\alpha)=\int\_0^\infty \frac\{\sin \alpha x\}\{x\}\mathrm\{ d\} x, 0\leq \alpha\leq b < \infty$. 证明: (1)、 若 $\displaystyle a > 0$, 则 $\displaystyle I(\alpha)$ 关于 $\displaystyle \alpha$ 在 $\displaystyle (a,b)$ 上一致收敛; (2)、 试问 $\displaystyle I(\alpha)$ 关于 $\displaystyle \alpha$ 在 $\displaystyle (0,b)$ 上是否一致收敛, 并说明理由. (华中师范大学2023年数学分析考研试题) [含参量积分 ]
[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\} \sup\_\{a < \alpha < b\}\left|\int\_0^A \sin \alpha x\mathrm\{ d\} x\right|\leq \sup\_\{a < \alpha < b\}\frac\{2\}\{\alpha\}=\frac\{2\}\{a\}, \frac\{1\}\{x\}\searrow 0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Dirichlet 判别法知 $\displaystyle I(\alpha)$ 关于 $\displaystyle \alpha\in (a,b)$ 一致收敛. (2)、 由
\begin\{aligned\} &\sup\_\{0 < \alpha < b\}\left|\int\_n^\{2n\}\frac\{\sin(\alpha x)\}\{x\}\mathrm\{ d\} x\right| \stackrel\{\alpha x=t\}\{=\}\sup\_\{0 < \alpha < b\}\left|\int\_\{n\alpha\}^\{2n\alpha\}\frac\{\sin t\}\{t\}\mathrm\{ d\} t\right|\\\\ \geq& \left|\int\_\{n\cdot\frac\{\pi\}\{2n\}\}^\{2n\cdot\frac\{\pi\}\{2n\}\}\frac\{\sin t\}\{t\}\mathrm\{ d\} t\right|=\int\_\frac\{\pi\}\{2\}^\pi \frac\{\sin t\}\{t\}\mathrm\{ d\} t > 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Cauchy 收敛准则知 $\displaystyle I(\alpha)$ 关于 $\displaystyle \alpha$ 在 $\displaystyle (0,b)$ 上不一致收敛.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)/
---
838、 (8)、 已知 $\displaystyle b > a > 0$, 求积分 $\displaystyle \int\_0^1 \frac\{x^b-x^a\}\{\ln x\}\sin \ln \frac\{1\}\{x\}\mathrm\{ d\} x$. (暨南大学2023年数学分析考研试题) [含参量积分 ]
[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\_a^b x^t\mathrm\{ d\} t\cdot \sin\left(\ln\frac\{1\}\{x\}\right)\mathrm\{ d\} x =\int\_a^b \int\_0^1 x^t \sin \left(\ln \frac\{1\}\{x\}\right)\mathrm\{ d\} x\mathrm\{ d\} t\\\\ &\stackrel\{\ln\frac\{1\}\{x\}=s\}\{=\}\int\_a^b \int\_\infty^0 \mathrm\{e\}^\{-st\} \sin s\cdot (-\mathrm\{e\}^\{-s\})\mathrm\{ d\} s\mathrm\{ d\} t =\int\_a^b \frac\{1\}\{1+(t+1)^2\}\mathrm\{ d\} t\\\\ &=\arctan (1+b)-\arctan (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)/
---
839、 11、 对 $\displaystyle a > 0$, 证明: $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-xy\}\sin x\mathrm\{ d\} x$ 在 $\displaystyle (0,+\infty)$ 上收敛, 在 $\displaystyle [a,+\infty)$ 上不一致收敛, 在 $\displaystyle (0,+\infty)$ 上非一致收敛. (南昌大学2023年数学分析考研试题) [含参量积分 ]
[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 $\displaystyle \left|\int\_0^A \sin x\mathrm\{ d\} x\right|\leq 2$, $\displaystyle \mathrm\{e\}^\{-xy\}$ 关于 $\displaystyle x$ 递减, 且 $\displaystyle y > 0$ 蕴含 $\displaystyle \lim\_\{x\to+\infty\}\mathrm\{e\}^\{-xy\}=0$. 据 Dirichlet 判别法即知 $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-xy\}\sin x\mathrm\{ d\} x$ 关于 $\displaystyle y\in (0,+\infty)$ 收敛. (2)、 对 $\displaystyle \forall\ a > 0$, $\displaystyle \mathrm\{e\}^\{-xy\}$ 关于 $\displaystyle x$ 递减, 且
\begin\{aligned\} \sup\_\{y\geq \delta\} \mathrm\{e\}^\{-xy\}=\mathrm\{e\}^\{-\delta x\}\xrightarrow\{x\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\}
由 Dirichlet 判别法即知 $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-xy\}\sin x\mathrm\{ d\} x$ 关于 $\displaystyle y\in [a,\infty)$ 一致收敛. (3)、 往用反证法证明 $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-xy\}\sin x\mathrm\{ d\} x$ 在 $\displaystyle (0,+\infty)$ 上非一致收敛. 若不然,
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ X > 0,\mathrm\{ s.t.\} \forall\ x\_2 > x\_1 > X, \forall\ y > 0, \left|\int\_\{x\_1\}^\{x\_2\} \mathrm\{e\}^\{-xy\}\sin x\mathrm\{ d\} x\right| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle y\to 0^+$ 得 $\displaystyle \left|\int\_\{x\_1\}^\{x\_2\} \sin x\mathrm\{ d\} x\right|\leq\varepsilon$. 但这与 $\displaystyle \int\_\{2n\pi\}^\{2n\pi+\pi\}\sin x\mathrm\{ d\} x=2$ 矛盾. 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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
840、 (3)、 (20 分) 计算积分 $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-px\}\frac\{\sin bx-\sin ax\}\{x\}\mathrm\{ d\} x$ $\displaystyle (p > 0, b > a > 0$). (山东大学2023年数学分析考研试题) [含参量积分 ]
[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\}^\{-px\}\mathrm\{ d\} x\int\_a^b \cos xt\mathrm\{ d\} t =\int\_a^b \mathrm\{ d\} t\int\_0^\infty \mathrm\{e\}^\{-px\}\cos xt\mathrm\{ d\} x\\\\ &\stackrel\{xt=s\}\{=\}\int\_a^b \frac\{1\}\{t\}\mathrm\{ d\} t\int\_0^\infty \mathrm\{e\}^\{-\frac\{p\}\{t\}s\}\cos s\mathrm\{ d\} s \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \int\_a^b \frac\{1\}\{t\}\cdot\frac\{\frac\{p\}\{t\}\}\{1+\left(\frac\{p\}\{t\}\right)^2\}\mathrm\{ d\} t\\\\ &=p\int\_a^b \frac\{\mathrm\{ d\} t\}\{t^2+p^2\}=\arctan\frac\{b\}\{p\}-\arctan \frac\{a\}\{p\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
此处可以交换次积分次序是因为 $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-px\}\cos xt\mathrm\{ d\} x$ 关于 $\displaystyle t\in [a,b]$ 是一致收敛的. 这可由 $\displaystyle |\mathrm\{e\}^\{-px\}\cos xt|\leq \mathrm\{e\}^\{-p x\}$ 及 Weierstrass 判别法得到.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)/
---
841、 (2)、 (15 分) 设函数 $\displaystyle f(x)$ 在闭区间 $\displaystyle [0,1]$ 上连续且恒正, 讨论
\begin\{aligned\} F(y)=\int\_0^1\frac\{yf(x)\}\{x^2+y^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\}
的连续性, 并证明你的结论. (山东大学2023年数学分析考研试题) [含参量积分 ]
[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)、 设 $\displaystyle m=\min\_\{[0,1]\}f > 0$, 则
\begin\{aligned\} F(y)\geq m\int\_0^1 \frac\{y\}\{x^2+y^2\}\mathrm\{ d\} x\stackrel\{x=yt\}\{=\} m\int\_0^y \frac\{\mathrm\{ d\} t\}\{1+t^2\} =m\arctan \frac\{1\}\{t\}. \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\} \varliminf\_\{y\to 0^+\}F(y)\geq m\cdot\frac\{\pi\}\{2\} > 0=F(0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这表明 $\displaystyle F$ 在 $\displaystyle y=0$ 处不连续. (2)、 对 $\displaystyle \forall\ 0 < \delta < A < \infty$, 由 $\displaystyle \frac\{yf(x)\}\{x^2+y^2\}$ 在
\begin\{aligned\} \left\\{(x,y); x\in [0,1], y\in [\delta,A]\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
上连续知 $\displaystyle F$ 在 $\displaystyle [\delta,A]$ 上连续. 由 $\displaystyle \delta,A$ 的任意性知 $\displaystyle F$ 在 $\displaystyle \forall\ y > 0$ 处连续. (3)、 由 $\displaystyle F$ 是奇函数知 $\displaystyle F$ 在除了 $\displaystyle y=0$ 外都连续.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)/
---
842、 4、 (10 分) 求含参量积分 $\displaystyle I(a)=\int\_0^\infty \mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}\cos (ax)\mathrm\{ d\} x$. (苏州大学2023年数学分析考研试题) [含参量积分 ]
[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 $\displaystyle \left|\mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}\cos (ax)\right|\leq \mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}$ 及 Weierstrass 判别法知 $\displaystyle \int\_0^\infty \mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}\cos (ax)\mathrm\{ d\} x$ 关于 $\displaystyle a\in\mathbb\{R\}$ 一致收敛, 从而 $\displaystyle I(a)$ 连续, 且
\begin\{aligned\} I(0)=&\int\_0^\infty \mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}\mathrm\{ d\} x =\sqrt\{\iint\_\{(0,\infty)^2\}\mathrm\{e\}^\{-\frac\{x^2+y^2\}\{2\}\}\mathrm\{ d\} x\mathrm\{ d\} y\} =\sqrt\{\int\_0^\infty \mathrm\{e\}^\{-\frac\{r^2\}\{2\}\}\cdot \frac\{2\pi r\}\{4\}\mathrm\{ d\} r\} =\sqrt\{\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\}
又由
\begin\{aligned\} \int\_0^\infty \frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} a\}\left\[\mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}\cos(ax)\right\]\mathrm\{ d\} x =-x\mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}\sin (ax), |-x\mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}\sin (ax)|\leq |x|\mathrm\{e\}^\{-\frac\{x^2\}\{2\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Weierstrass 判别法知 $\displaystyle \int\_0^\infty \frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} a\}\left\[\mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}\cos(ax)\right\]\mathrm\{ d\} x$ 关于 $\displaystyle a\in\mathbb\{R\}$ 一致收敛, 而 $\displaystyle I(a)$ 可导, 且
\begin\{aligned\} I'(a)=&-\int\_0^\infty x\mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}\sin (ax)\mathrm\{ d\} x =\int\_0^\infty \sin(ax)\mathrm\{ d\} \mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}\\\\ \xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\}& -a\int\_0^\infty \mathrm\{e\}^\{-\frac\{x^2\}\{2\}\}\cos(ax)\mathrm\{ d\} x =-aI(a). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} \left\[\mathrm\{e\}^\frac\{a^2\}\{2\}I(a)\right\]'=0\Rightarrow I(a)=I(0)\mathrm\{e\}^\{-\frac\{a^2\}\{2\}\} =\sqrt\{\frac\{\pi\}\{2\}\}\mathrm\{e\}^\{-\frac\{a^2\}\{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)/
---
843、 13、 证明: 含参量积分 $\displaystyle \int\_0^\infty \frac\{x\sin(tx)\}\{1+x^2\}\mathrm\{ d\} x$ 关于 $\displaystyle t\in [\delta,+\infty)$ 一致收敛, 其中 $\displaystyle \delta > 0$, 但关于 $\displaystyle t\in (0,+\infty)$ 不一致收敛. (太原理工大学2023年数学分析考研试题) [含参量积分 ]
[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 $\displaystyle f(x)=\frac\{x\}\{1+x^2\}$, 则由 $\displaystyle f'(x)=\frac\{1-x^2\}\{(1+x^2)^2\}$ 知当 $\displaystyle x\geq 1$ 时, $\displaystyle f(x)\searrow$, 且 $\displaystyle \lim\_\{x\to+\infty\}f(x)=0$. 再者,
\begin\{aligned\} \sup\_\{t\geq \delta\}\left|\int\_0^A \sin(tx)\mathrm\{ d\} x\right| =\sup\_\{t\geq \delta\}\left|\int\_0^\{tA\}\sin \tau\frac\{\mathrm\{ d\}\tau\}\{\tau\}\right| \leq \frac\{2\}\{\delta\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 Dirichlet 判别法知 $\displaystyle \int\_0^\infty \frac\{x\sin(tx)\}\{1+x^2\}\mathrm\{ d\} x$ 关于 $\displaystyle t\in [\delta,+\infty)$ 一致收敛. 又由
\begin\{aligned\} &\sup\_\{t > 0\}\int\_n^\{2n\}\frac\{x\sin(tx)\}\{1+x^2\}\mathrm\{ d\} x \geq \int\_n^\{2n\}\frac\{x\sin\frac\{\pi\}\{4n\}x\}\{1+x^2\}\mathrm\{ d\} x\\\\ \geq& \frac\{1\}\{\sqrt\{2\}\}\int\_n^\{2n\}\frac\{x\}\{1+x^2\}\mathrm\{ d\} x =\frac\{1\}\{2\sqrt\{2\}\}\ln\frac\{4n^2+1\}\{n^2+1\} \xrightarrow\{n\to\infty\}\frac\{\ln 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\}
及 Cauchy 收敛准则知 $\displaystyle \int\_0^\infty \frac\{x\sin(tx)\}\{1+x^2\}\mathrm\{ d\} x$ 关于 $\displaystyle t\in (0,+\infty)$ 不一致收敛.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)/
---
844、 3、 设 $\displaystyle I(p)=\int\_0^\infty \mathrm\{e\}^\{-px\}\frac\{\sin x\}\{x\}\mathrm\{ d\} x$, $\displaystyle 0 < a < b$. (1)、 证明: $\displaystyle I(p)$ 在 $\displaystyle [a,b]$ 上一致收敛; (2)、 求 $\displaystyle I(p)$ 及 $\displaystyle I(0)$. (同济大学2023年数学分析考研试题) [含参量积分 ]
[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 $\displaystyle \sup\_\{p\in [a,b]\}\left|\mathrm\{e\}^\{-px\}\frac\{\sin x\}\{x\}\right|\leq \sup\_\{p\in [a,b]\}\mathrm\{e\}^\{-px\} =\mathrm\{e\}^\{-pa\}$ 及 Weierstrass 判别法知 $\displaystyle I(p)$ 在 $\displaystyle [a,b]$ 上一致收敛. (2)、 $\displaystyle I(0)=\int\_0^\infty \frac\{\sin x\}\{x\}\mathrm\{ d\} x=\frac\{\pi\}\{2\}$. 由
\begin\{aligned\} \int\_0^\infty \frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} p\}\left(\mathrm\{e\}^\{-px\}\frac\{\sin x\}\{x\}\right)\mathrm\{ d\} x =-\int\_0^\infty \mathrm\{e\}^\{-px\}\sin x\mathrm\{ d\} x\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} -\frac\{1\}\{1+p^2\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
关于 $\displaystyle p\in [a,b]$ 一致收敛知
\begin\{aligned\} I(p)=I(0)+\int\_0^p I'(\tau)\mathrm\{ d\} \tau=\frac\{\pi\}\{2\}-\arctan p=\arctan \frac\{1\}\{p\}. \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)/
---
845、 3、 解答题. (1)、 求 $\displaystyle F(\alpha)=\int\_0^\infty \frac\{\ln \left(1+x^\frac\{5\}\{2\}\right)\}\{x^\alpha\}\mathrm\{ d\} x$ 的连续区间. (武汉大学2023年数学分析考研试题) [含参量积分 ]
[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\} \frac\{\ln(1+x^\frac\{5\}\{2\})\}\{x^\alpha\}\sim \frac\{x^\frac\{5\}\{2\}\}\{x^\alpha\}=\frac\{1\}\{x^\{\alpha-\frac\{5\}\{2\}\}\}, 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 $\displaystyle \alpha-\frac\{5\}\{2\} < 1\Leftrightarrow \alpha < \frac\{7\}\{2\}$ 时, $\displaystyle \int\_0^1 \frac\{\ln \left(1+x^\frac\{5\}\{2\}\right)\}\{x^\alpha\}\mathrm\{ d\} x$ 收敛. 又由
\begin\{aligned\} &\lim\_\{x\to+\infty\}\frac\{\frac\{\ln(1+x^\frac\{5\}\{2\})\}\{x^\alpha\}\}\{\frac\{1\}\{x^\frac\{\alpha+1\}\{2\}\}\} =\lim\_\{x\to+\infty\}\frac\{\ln(1+x^\frac\{5\}\{2\})\}\{x^\frac\{\alpha-1\}\{2\}\} =\left\\{\begin\{array\}\{llllllllllll\}0,&\alpha > 1,\\\\ +\infty,&\alpha\leq 1\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\}
知当且仅当 $\displaystyle \alpha > 1$ 时, $\displaystyle \int\_1^\infty \frac\{\ln \left(1+x^\frac\{5\}\{2\}\right)\}\{x^\alpha\}\mathrm\{ d\} x$ 收敛. 故 $\displaystyle F(\alpha)$ 的定义域为 $\displaystyle \left(1,\frac\{7\}\{2\}\right)$. 对 $\displaystyle \forall\ \delta\in (0,1)$, 由
\begin\{aligned\} 0 < x < 1\Rightarrow& \sup\_\{1+\delta\leq \alpha\leq \frac\{7\}\{2\}-\delta\}\frac\{\ln \left(1+x^\frac\{5\}\{2\}\right)\}\{x^\alpha\}=\frac\{\ln (1+x^\frac\{5\}\{2\})\}\{x^\{\frac\{7\}\{2\}-\delta\}\} \sim \frac\{1\}\{x^\{1-\delta\}\},\\\\ x > 1\Rightarrow&\sup\_\{1+\delta\leq \alpha\leq \frac\{7\}\{2\}-\delta\}\frac\{\ln \left(1+x^\frac\{5\}\{2\}\right)\}\{x^\alpha\} =\frac\{\ln (1+x^\frac\{5\}\{2\})\}\{x^\{1+\delta\}\}\ll \frac\{1\}\{x^\{1+\frac\{\delta\}\{2\}\}\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Weierstrass 判别法知 $\displaystyle \int\_0^\infty \frac\{\ln \left(1+x^\frac\{5\}\{2\}\right)\}\{x^\alpha\}\mathrm\{ d\} x$ 关于 $\displaystyle \alpha\in \left\[1+\delta,\frac\{7\}\{2\}-\delta\right\]$ 一致收敛, 而连续. 由 $\displaystyle \delta\in(0,1)$ 的任意性知 $\displaystyle F(\alpha)$ 在 $\displaystyle \left(1,\frac\{7\}\{2\}\right)$ 上连续.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)/
---
846、 (3)、 证明: $\displaystyle I(x)=\int\_0^\infty \frac\{\sin (xy)\}\{y\}\mathrm\{ d\} y$ 在 $\displaystyle (0,+\infty)$ 上内闭一致收敛. (武汉理工大学2023年数学分析考研试题) [含参量积分 ]
[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 $\displaystyle \int\_0^\infty \frac\{\sin s\}\{s\}\mathrm\{ d\} s$ 收敛及 Cauchy 收敛准则知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ A > 0,\mathrm\{ s.t.\}\forall\ A\_2 > A\_1 > A, \left|\int\_\{A\_1\}^\{A\_2\}\frac\{\sin s\}\{s\}\mathrm\{ d\} s\right| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle X=\frac\{A\}\{\delta\}$, 则 $\displaystyle \forall\ X\_2 > X\_1 > X$, $\displaystyle xX\_2 > xX\_1\geq \delta X=A$,
\begin\{aligned\} \sup\_\{x\geq \delta\} \left|\int\_\{X\_1\}^\{X\_2\} \frac\{\sin (xy)\}\{y\}\mathrm\{ d\} y\right| \stackrel\{xy=s\}\{=\}\sup\_\{x\geq \delta\} \left|\int\_\{xX\_1\}^\{xX\_2\}\frac\{\sin s\}\{s\}\mathrm\{ d\} s\right|\leq \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 Cauchy 收敛准则即知 $\displaystyle \int\_0^\infty \frac\{\sin xy\}\{y\}\mathrm\{ d\} y$ 在 $\displaystyle [\delta,+\infty)$ 上一致收敛.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)/
---
847、 7、 (15 分) 证明: $\displaystyle f(x)=\int\_0^\infty \frac\{\sin xy\}\{y\}\mathrm\{ d\} y$ 在 $\displaystyle (0,+\infty)$ 上不一致收敛, 但在 $\displaystyle (0,+\infty)$ 上连续. (西北大学2023年数学分析考研试题) [含参量积分 ]
[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 $\displaystyle \int\_0^\infty \frac\{\sin s\}\{s\}\mathrm\{ d\} s$ 收敛及 Cauchy 收敛准则知
\begin\{aligned\} \forall\ \varepsilon > 0,\exists\ A > 0,\mathrm\{ s.t.\}\forall\ A\_2 > A\_1 > A, \left|\int\_\{A\_1\}^\{A\_2\}\frac\{\sin s\}\{s\}\mathrm\{ d\} s\right| < \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取 $\displaystyle X=\frac\{A\}\{\delta\}$, 则 $\displaystyle \forall\ X\_2 > X\_1 > X$, $\displaystyle xX\_2 > xX\_1\geq \delta X=A$,
\begin\{aligned\} \sup\_\{x\geq \delta\} \left|\int\_\{X\_1\}^\{X\_2\} \frac\{\sin (xy)\}\{y\}\mathrm\{ d\} y\right| \stackrel\{xy=s\}\{=\}\sup\_\{x\geq \delta\} \left|\int\_\{xX\_1\}^\{xX\_2\}\frac\{\sin s\}\{s\}\mathrm\{ d\} s\right|\leq \varepsilon. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 Cauchy 收敛准则即知 $\displaystyle \int\_0^\infty \frac\{\sin xy\}\{y\}\mathrm\{ d\} y$ 在 $\displaystyle [\delta,+\infty)$ 上一致收敛, 从而 $\displaystyle f(x)$ 在 $\displaystyle [\delta,+\infty)$ 上连续. 由 $\displaystyle \delta$ 的任意性知 $\displaystyle f(y)$ 在 $\displaystyle (0,+\infty)$ 上连续. (2)、 由
\begin\{aligned\} \sup\_\{x > 0\}\left|\int\_\frac\{n\pi\}\{4\}^\frac\{3n\pi\}\{4\}\frac\{\sin(xy)\}\{y\}\mathrm\{ d\} y\right| \geq \int\_\frac\{n\pi\}\{4\}^\frac\{3n\pi\}\{4\}\frac\{\sin \frac\{1\}\{n\}y\}\{y\}\mathrm\{ d\} y \geq \frac\{1\}\{\sqrt\{2\}\}\int\_\frac\{n\pi\}\{4\}^\frac\{3n\pi\}\{4\}\frac\{\mathrm\{ d\} y\}\{y\} =\frac\{1\}\{\sqrt\{2\}\}\ln 3 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 Cauchy 收敛准则即知 $\displaystyle \int\_0^\infty \frac\{\sin xy\}\{y\}\mathrm\{ d\} y$ 在 $\displaystyle (0,+\infty)$ 上不一致收敛.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)/
---
848、 6、 (8 分) 证明 $\displaystyle \int\_0^\infty \frac\{\sin xy\}\{1+y^2\}\mathrm\{ d\} y$ 关于 $\displaystyle x\in (-\infty,+\infty)$ 一致收敛. (长安大学2023年数学分析考研试题) [含参量积分 ]
[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 $\displaystyle \left|\frac\{\sin xy\}\{1+y^2\}\right|\leq\frac\{1\}\{1+y^2\}$ 及 Weierstrass 判别法即知 $\displaystyle \int\_0^\infty \frac\{\sin xy\}\{1+y^2\}\mathrm\{ d\} y$ 关于 $\displaystyle x\in (-\infty,+\infty)$ 一致收敛.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)/
---
849、 11、 (20 分) $\displaystyle \forall\ x\in\mathbb\{R\}$, 证明: $\displaystyle \int\_0^\infty\mathrm\{e\}^\{-t\}\frac\{\sin tx\}\{t\}\mathrm\{ d\} t=\arctan x$. (郑州大学2023年数学分析考研试题) [含参量积分 ]
[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) / 左右两端都是关于 $\displaystyle x$ 的奇函数, 而不妨设 $\displaystyle x > 0$. 此时,
\begin\{aligned\} \mbox\{左端\}&=\int\_0^\infty \mathrm\{e\}^\{-t\}\mathrm\{ d\} t\int\_0^x \cos st\mathrm\{ d\} s =\int\_0^x \mathrm\{ d\} s\int\_0^\infty \mathrm\{e\}^\{-t\}\cos st\mathrm\{ d\} s\\\\ &\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \int\_0^x \frac\{\mathrm\{ d\} s\}\{1+s^2\}=\arctan x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
这里可以交换积分次序是因为由 $\displaystyle |\mathrm\{e\}^\{-t\}\cos st|\leq \mathrm\{e\}^\{-t\}$ 及 Weierstrass 判别法知 $\displaystyle \int\_0^\infty \mathrm\{e\}\{-t\}\cos st\mathrm\{ d\} t$ 关于 $\displaystyle s\in[0, x]$ 一致收敛.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)/
---
850、 7、 (15 分) 考虑如下含参量的反常积分
\begin\{aligned\} I(\alpha)=\int\_1^\infty \mathrm\{e\}^\{-\alpha x\}\sin 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\}
证明以上反常积分在 $\displaystyle \alpha\in (0,1]$ 上不是一致收敛的. (中国科学技术大学2023年数学分析考研试题) [含参量积分 ]
[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\} &\sup\_\{0 < \alpha\leq 1\}\left|\int\_\{2k\pi+\frac\{\pi\}\{4\}\}^\{2k\pi+\frac\{3\pi\}\{4\}\} \mathrm\{e\}^\{-\alpha x\}\sin x\mathrm\{ d\} x\right| \geq \frac\{1\}\{\sqrt\{2\}\}\sup\_\{0 < \alpha\leq 1\}\left|\int\_\{2k\pi+\frac\{\pi\}\{4\}\}^\{2k\pi+\frac\{3\pi\}\{4\}\} \mathrm\{e\}^\{-\alpha x\}\mathrm\{ d\} x\right|\\\\ \geq&\frac\{1\}\{\sqrt\{2\}\}\lim\_\{\alpha\to 0^+\}\left|\int\_\{2k\pi+\frac\{\pi\}\{4\}\}^\{2k\pi+\frac\{3\pi\}\{4\}\} \mathrm\{e\}^\{-\alpha x\}\mathrm\{ d\} x\right| =\frac\{1\}\{\sqrt\{2\}\}\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 收敛准则知题中反常积分在 $\displaystyle \alpha\in (0,1]$ 上不是一致收敛的.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)/
---
851、 2、 讨论含参量积分 $\displaystyle I(s)=\int\_0^\infty x^\{s-1\}\mathrm\{e\}^\{-x\}\mathrm\{ d\} x, s\in\mathbb\{R\}$ 的收敛性. (中国矿业大学(徐州)2023年数学分析考研试题) [含参量积分 ]
[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 $\displaystyle x^\{s-1\}\mathrm\{e\}^\{-s\}\sim \frac\{1\}\{x^\{1-s\}\}, x\to 0$ 知当当且仅当 $\displaystyle 1-s < 1\Leftrightarrow s > 0$ 时, $\displaystyle \int\_0^1 x^\{s-1\}\mathrm\{e\}^\{-x\}\mathrm\{ d\} x$ 收敛. 当 $\displaystyle s > 0$ 时, 由
\begin\{aligned\} &x > 1\Rightarrow x^s\mathrm\{e\}^\{-x\}\leq x^\{
+1\}\mathrm\{e\}^\{-s\},\\\\ &\lim\_\{x\to+\infty\}\frac\{x^\{+1\}\mathrm\{e\}^\{-s\}\}\{\frac\{1\}\{x^2\}\} =\lim\_\{x\to+\infty\}\frac\{x^\{+3\}\}\{\mathrm\{e\}^s\}\xlongequal\{\tiny\mbox\{L'Hospital\}\} \cdots =0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及比较判别法知 $\displaystyle \int\_1^\infty x^\{s-1\}\mathrm\{e\}^\{-x\}\mathrm\{ d\} x$ 收敛. 综上即知当且仅当 $\displaystyle s > 0$ 时, $\displaystyle I(s)$ 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 09:14:32
