南方科技大学2023年数学分析考研试题参考解答
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1、 (15 分) 求极限 $\displaystyle \lim_{x\to 0}\frac{\mathrm{e}^{x^2+x}-\mathrm{e}^x+2\ln \cos x}{x^3}$. fl: 函数极限
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} \mbox{原式}\overset{\tiny\mbox{Taylor 展开}}{=}&\lim_{x\to 0}\frac{\boxed{\begin{array}{c}1+(x^2+x)+\frac{x^2(x+1)^2}{2}+\frac{x^3(x+1)^3}{3!}+o(x^3)\\ -\left +2\left[(\cos x-1)+o(x^3)\right]\end{array}}}{x^3}\\ =&\lim_{x\to 0}\frac{\left -\left-2\left(-\frac{x^2}{2}\right)+o(x^3)}{x^3}\\ =&1+\frac{1}{3!}-\frac{1}{3!}=1. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
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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2、 (15 分) 设方程 $x^2+y^2+z^2-3xyz=0$ 确定了隐函数 $\displaystyle z=z(x,y), y=y(x,z)$. 再设 $f(x,y,z)=xy^2z^3$.
(1)、 求 $\left.\frac{\partial f}{\partial x}\left(x,y,z(x,y)\right)\right|_{(1,1)}$;
(2)、 求 $\left.\frac{\partial f}{\partial x}\left(x,y(x,z),z\right)\right|_{(1,1)}$.
fl: 多元函数微分学
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $F=x^2+y^2+z^2-3xyz$.
(1)、 把 在 $(1,1,1)$ 处, $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} &\left.\frac{\partial f}{\partial x}\left(x,y,z(x,y)\right)\right|_{(1,1)} =f_x+f_zz_x\\ =&f_x+f_z\left(-\frac{F_x}{F_z}\right) =\frac{f_xF_z-f_zF_x}{F_z}=-2. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
(2)、 把 在 $(1,1,1)$ 处, $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} &\left.\frac{\partial f}{\partial x}\left(x,y(x,z),z\right)\right|_{(1,1)} =f_x+f_yy_x\\ =&f_x+f_y\left(-\frac{F_x}{F_y}\right) =\frac{f_xF_y-f_yF_x}{F_y}=-1. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
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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3、 (15 分) 计算 $\iint_{x^2+y^2\leq 1}|3x+4y|\mathrm{ d} x\mathrm{ d} y$. fl: 重积分
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 作正交变换 $\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} u=\frac{3x+4y}{5}, v=\frac{4x-3y}{5}\Rightarrow u^2+v^2=x^2+y^2, \left|\frac{\partial(u,v)}{\partial (x,y)}\right|=1. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
而 $\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} \mbox{原式}&=5\iint_{u^2+v^2\leq 1} |u|\mathrm{ d} u\mathrm{ d} v =10\iint_{u^2+v^2\leq 1\atop u\gt 0} u\mathrm{ d} u\mathrm{ d} v\\ &=10\int_0^1 u\mathrm{ d} u\int_{-\sqrt{1-u^2}}^{\sqrt{1-u^2}}\mathrm{ d} v =20\int_0^1 u\sqrt{1-u^2}\mathrm{ d} u\stackrel{u=\sin\theta}{=}\cdots =\frac{20}{3}. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
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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4、 (15 分)
(1)、 求 $\sum_{n=1}^\infty n3^{3n}x^{3n}$ 的收敛半径; fl: 幂级数
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle a_n=\left\{\begin{array}{llllllllllll}k 3^{3k},&n=3k,\\ 0,&n=3k+1, 3k+2\end{array}\right.$ 知 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} \varlimsup_{n\to\infty}\sqrt{a_n}=\lim_{k\to\infty}\sqrt{a_{3k}} =\lim_{k\to\infty} \sqrt{k3^{3k}}=3. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
故收敛半径为 $\frac{1}{3}$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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(2)、 求 $f(x)=\frac{x}{\sqrt{2-x}}$ 的麦克劳林级数. fl: 幂级数
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} f(x)=&\frac{x}{\sqrt{2}}\frac{1}{\sqrt{1-\frac{x}{2}}} =\frac{x}{\sqrt{2}}\sum_{n=0}^\infty \frac{\frac{1}{2}\cdot\frac{3}{2}\cdots \frac{2n-1}{2}}{n!}\left(\frac{x}{2}\right)^n\\ =&\sum_{n=0}^\infty \frac{(2n-1)!!}{(2n)!!2^{n+\frac{1}{2}}} x^{n+1}, |x|\lt 2. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
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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5、 (15 分) 求函数 $f(x)=\left\{\begin{array}{llllllllllll}\frac{2}{\pi}x,&0\leq x\leq \frac{\pi}{2},\\ 2-\frac{2}{\pi}x,&\frac{\pi}{2}\lt x\leq\pi\end{array}\right.$ 的正弦级数, 并讨论其敛散性. fl: Fourier级数
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} f(x)\sim \sum_{n=1}^\infty b_n\sin nx, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
其中 $\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} n\gt 1\Rightarrow b_n&=\frac{2}{\pi}\int_0^\pi f(x)\sin nx\mathrm{ d} x=\left\{\begin{array}{llllllllllll}0,&n=2k,\\ \frac{8(-1)^{k-1}}{(2k-1)^2\pi^2},&n=2k-1.\end{array}\right. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
由 $f$ 分段光滑知 $\displaystyle f(x)=\frac{8}{\pi^2}\sum_{k=1}^\infty \frac{(-1)^{k-1}}{(2k-1)^2}\sin (2k-1)x$. 所以 $f$ 的 Fourier 级数是收敛的. 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)/
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6、 (15 分) 证明:曲面 $\displaystyle \sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{a}\ (a\gt 0)$ 上任一点的切平面在三个坐标轴上的截距之和为 $\displaystyle a$. fl: 多元函数微分学
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 曲面上点 $(x,y,z)$ 处的切平面方程为 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} \frac{X-x}{\sqrt{x}}+\frac{Y-y}{\sqrt{y}}+\frac{Z-z}{\sqrt{z}}=0\Leftrightarrow \frac{X}{\sqrt{x}}+\frac{Y}{\sqrt{y}}+\frac{Z}{\sqrt{z}}=\sqrt{a}, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
其在 $x,y,z$ 轴上的截距分别为 $\displaystyle \sqrt{ax},\ \sqrt{ay},\ \sqrt{az}$, 它们的和为 $\sqrt{ax}+\sqrt{ay}+\sqrt{az}=a$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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7、 (15 分) $F(t)=\int_0^{t^2}\mathrm{ d} x\int_{x-t}^{x+t}\cos (x^2+y^2-t^2)\mathrm{ d} y$. 求 $F'(t)$. fl: 微分
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $g(x,t)=\int_{x-t}^{x+t}\cos(x^2+y^2-t^2)\mathrm{ d} y$, 则 $\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} g_t(x,t)=&\int_{x-t}^{x+t}\left[-\sin(x^2+y^2-t^2)\right]\cdot(-2t)\mathrm{ d} y\\ &+\cos\left -\cos\left(-1)\\ =&2t\int_{x-t}^{x+t} \sin(x^2+y^2-t^2)\mathrm{ d} y +\cos (2x^2+2xt)+\cos(2x^2-2xt). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
而 $F(t)=\int_0^{t^2} g(x,t)\mathrm{ d} x$ 蕴含 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} F'(t)=&\int_0^{t^2}g_t(x,t)\mathrm{ d} x+g(t^2,t)\cdot 2t\\ =&\int_0^{t^2}\left[\begin{array}{c}2t\int_{x-t}^{x+t} \sin(x^2+y^2-t^2)\mathrm{ d} y\\ +\cos (2x^2+2xt)+\cos(2x^2-2xt)\end{array}\right]\mathrm{ d} x\\ &+2t\int_{t^2-t}^{t^2+t} \cos(t^4+y^2-t^2)\mathrm{ d} y. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
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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8、 (15 分) 求 $\oint_C \frac{(x-y)\mathrm{ d} x+(x+4y)\mathrm{ d} y}{x^2+4y^2}$, 其中 $C:x^2+y^2=1$, 取逆时针. fl: 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)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $P=\frac{x-y}{x^2+4y^2}, Q=\frac{x+4y}{x^2+4y^2}$, 则 $P_y=Q_x$, 而由 Green 公式知 $\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} I=&\int_{x^2+4y^2=\varepsilon^2}\frac{(x-y)\mathrm{ d} x+(x+4y)\mathrm{ d} y}{x^2+4y^2}\\ =&\frac{1}{\varepsilon^2} \int_{x^2+4y^2=\varepsilon^2} (x-y)\mathrm{ d} x+(x+4y)\mathrm{ d} y\\ =&\frac{1}{\varepsilon^2} \int_{x^2+4y^2\leq \varepsilon^2}2\mathrm{ d} x\mathrm{ d} y =\frac{2}{\varepsilon^2}\cdot \left(\pi \varepsilon\frac{\varepsilon}{2}\right) =\pi. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
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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9、 (15 分) 设 $f(x)$ 在 $\displaystyle $ 上 Riemann 可积, 且在 $\displaystyle x=1$ 处左连续. 证明:$\displaystyle \tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} \lim_{n\to\infty}\int_0^1 nx^{n-1}f(x)\mathrm{ d} x=f(1). \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
fl: 积分与极限
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 写出 $\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} &\left|\int_0^1 nx^{n-1}f(x)\mathrm{ d} x-f(1)\right| =\left|\int_0^1 nx^{n-1}\mathrm{ d} x\right|\\ \leq& \int_0^{1-\delta}+\int_{1-\delta}^1\cdots \leq 2\max_{}|f|\cdot (1-\delta)^n+\max_{}|f-f(1)| \equiv I_1+I_2. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
对 $\forall\ \varepsilon\gt 0$, 先取 $\delta\gt 0$ 使得 $\displaystyle I_2 \lt\varepsilon/2$; 再对该 $\delta$, 存在 $N$, 使得 $n\gt N\Rightarrow I_2 \lt\varepsilon/2\Rightarrow \left|\int_0^1 nx^{n-1}f(x)\mathrm{ d} x-f(1)\right|\lt \varepsilon$. 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)/
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10、 (15 分) 设 $f(x)$ 在 $\displaystyle [0,+\infty)$ 上连续且非负, $\displaystyle \int_0^\infty f(x)\mathrm{ d} x$ 收敛.
(1)、 证明: $\forall\ a\gt 0$, $\displaystyle \int_0^\infty \ln \left\mathrm{ d} x$ 收敛;
(2)、 若 $\displaystyle f(x)\gt 0, \forall\ x\gt 0; \left\{a_n\right\}$ 是单调递增且趋于 $\displaystyle +\infty$ 的正数列, 求 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} \lim_{n\to\infty}a_n\int_0^\infty \ln \left\mathrm{ d} x, \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
并验证结果准确性.
fl: 广义积分
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
(1)、 由对数不等式 $\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} \frac{x}{1+x}\lt \ln(1+x)\lt x, x\gt -1, x\neq 0 \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
知 $\ln \left\leq\frac{f(x)}{a}$, 而由比较判别法知 $\int_0^\infty \ln \left\mathrm{ d} x$ 收敛.
(2)、 设 $\displaystyle g(x,t)=\left\{\begin{array}{llllllllllll}f(x),&x\gt 0, t=0,\\ \frac{\ln \left}{t},&x\gt 0, t\gt 0.\end{array}\right.$ 则 $\displaystyle g$ 在 $\displaystyle \times $ 上连续, 而一致连续. 由含参量正常积分的性质知 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} \lim_{t\to 0^+}\int_0^A g(x,t)\mathrm{ d} x=\int_0^A g(x,0)\mathrm{ d} x=\int_0^A f(x)\mathrm{ d} x. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
对 $\forall\ 0\lt t\lt 1$, $\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} &\left|\frac{1}{t}\int_0^\infty \ln \left\mathrm{ d} x-\int_0^\infty f(x)\mathrm{ d} x\right|\\ \leq& \int_0^A \left|\frac{\ln \left}{t}-f(x)\right|\mathrm{ d} x +\int_A^\infty \frac{\ln \left}{t}\mathrm{ d} x+\int_A^\infty f(x)\mathrm{ d} x\\ \leq&\int_0^A \left\mathrm{ d} x+2\int_A^\infty f(x)\mathrm{ d} x. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
令 $t\to 0^+$ 得 $\displaystyle \tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} \varlimsup_{t\to 0^+}\left|\frac{1}{t}\int_0^\infty \ln \left\mathrm{ d} x-\int_0^\infty f(x)\mathrm{ d} x\right| \leq 2\int_A^\infty f(x)\mathrm{ d} x. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
再令 $A\to\infty$ 得 $\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} 0\leq& \varlimsup_{t\to 0^+}\left|\frac{1}{t}\int_0^\infty \ln \left\mathrm{ d} x-\int_0^\infty f(x)\mathrm{ d} x\right|\\ \leq& \varlimsup_{t\to 0^+}\left|\frac{1}{t}\int_0^\infty \ln \left\mathrm{ d} x-\int_0^\infty f(x)\mathrm{ d} x\right| \leq 0. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
故 $\lim_{t\to 0^+}\frac{1}{t}\int_0^\infty \ln \left\mathrm{ d} x=\int_0^\infty f(x)\mathrm{ d} x$. 由归结原理知 $\tiny\boxed{@跟锦数学微信公众号}$
$$\begin{aligned} \mbox{原式}=\lim_{t\to 0^+}\frac{1}{t}\int_0^\infty \ln \left\mathrm{ d} x=\int_0^\infty f(x)\mathrm{ d} x. \tiny\boxed{\begin{array}{c}\mbox{跟锦数学微信公众号}\\\mbox{www.zhangzujin.cn}\end{array}}\end{aligned}$$
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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