Math Analysis Homework - Week 11

Class 1

T1

计算曲线围成的面积

\begin{gathered} x=y^2,y=x^2 \\ y=\sin x,y=\cos x,x=0,x=2\pi \\ x=a(\cos t+t\sin t),y=a(\sin t-t\cos t)(0\le t\le 2\pi),x=a \\ r^2=a^2\cos 2\theta (a>0) \end{gathered}

(1)

\begin{gathered} S=\int_0^1 (\sqrt x-x^2)dx \\ =(\dfrac{2}{3} x^{\frac32} -\dfrac{x^3}{3} )\vert_0^1 \\ =\dfrac{1}{3} \end{gathered}

(2)

\begin{gathered} \int_0^{2\pi} \vert \sin x-\cos x \vert dx \\ =\int_0^{2\pi}\vert \sqrt 2\sin(x-\dfrac{\pi}{4} ) \vert dx \\ =2\sqrt 2\int_0^{\pi}\vert \sin{x} \vert \\ =4\sqrt 2 \end{gathered}

(3)

\begin{gathered} S_1 =\dfrac{1}{2}\int_0^{2\pi}(x,y)\times (x+dx,y+dy) \\ =\dfrac{1}{2}\int_0^{2\pi} x(y+dy)-y(x+dx) \\ =\dfrac{1}{2} \int_0^{2\pi}xdy-ydx \\ =\dfrac{a^2}{2} \int_0^{2\pi} ((\cos t+t\sin t)(t\sin t)-(\sin t-t\cos t)(t\cos t))dt \\ =\dfrac{a^2}{2} \int_0^{2\pi} t^2dt \\ =\dfrac{4a^2\pi^3}{3} \\ S_2=\dfrac{1}{2}\times a^2\times 2\pi \\ =\pi a^2 \\ S=S_1+S_2=\dfrac{4a^2\pi^3}{3} +\pi a^2 \end{gathered}

(4)

\begin{gathered} \dfrac{1}{2}\int_0^{2\pi} a^2 \vert \cos(2\theta) \vert d\theta \\ =\dfrac12 2\int_{-\frac\pi4}^{\frac\pi4} a^2 \vert \cos(2\theta) \vert d\theta \\ =\dfrac12a^2 \sin2\theta \vert_{-\frac\pi4}^{\frac\pi4} \\ =a^2 \end{gathered}

T2

求弧长

\begin{gathered} y=\ln \cos x,x\in[0,\dfrac{\pi}{3} ] \\ y=\int_{-\sqrt 3}^x \sqrt{3-t^2}dt \\ \theta=\dfrac{1}{2} (r+\dfrac{1}{r} )(1\le r\le 3) \end{gathered}

(1)

\begin{gathered} \int_0^{\frac\pi3}\sqrt{1+y'^2}dx \\ =\int_0^{\frac\pi3}\sqrt{1+\tan^2x}dx \\ =\int_0^{\frac\pi3}\sec xdx \\ =(\ln \vert \sec x+\tan x \vert) \vert_0^{\frac\pi3} \\ =\ln(2+\sqrt 3) \end{gathered}

(2)

\begin{gathered} \int_{-\sqrt 3}^{\sqrt 3} \sqrt{1+y'^2}dx \\ =\int_{-\sqrt 3}^{\sqrt 3} \sqrt{4-x^2}dx \\ =\int_{-\frac\pi3}^{\frac\pi3}4\cos^2 tdt \\ =\int_{-\frac\pi3}^{\frac\pi3}2(1+\cos 2t)dt \\ =(2x+\sin 2x)\vert_{-\frac\pi3}^{\frac\pi3} \\ =\dfrac{4}{3} \pi+\sqrt 3 \end{gathered}

(3)

\begin{gathered} \int_A^B \sqrt{r'^2+r^2}d\theta \\ =\int_1^3 \sqrt{1+r^2(\dfrac{1}{2} -\dfrac{1}{2r^2} )^2}dr \\ =\int_1^3 \dfrac{r^2+1}{2r} dr \\ =2+\dfrac{\ln 3}{2} \end{gathered}

T3

过点 $P(1,0)$ 作抛物线 $y=\sqrt{x-2}$ 的切线，求该切线与抛物线及 $x$ 轴所围成的平面图形绕 $x$ 轴、 $y$ 轴旋转而成的旋转体体积.

\begin{gathered} x=y^2+2 \end{gathered}

切线: $y=\dfrac{1}{2} x-\dfrac{1}{2} \iff x=2y+1$

绕y:

\begin{gathered} \pi\int_0^1 ((y^2+2)^2-(2y+1)^2) dy \\ =\dfrac{6\pi}{5} \end{gathered}

绕x:

\begin{gathered} \pi\int_2^3 ((\dfrac{1}{2} x-\dfrac{1}{2} )^2-(\sqrt{x-2})^2)dx+\dfrac{1}{3} 1\times \pi\times \dfrac{1}{2^2} \\ =\dfrac{1}{6} \pi \end{gathered}

T4

求心形线的一段 $r=a(1+\cos\theta) \left(0 \leqslant \theta \leqslant \frac{\pi}{2}\right)$ 与 $\theta = \frac{\pi}{2}$ 和极轴所围成图形绕极轴旋转一周所得立体的体积.

\begin{gathered} \int_a^b y^2dx \\ =\pi\int_{\frac\pi2}^0 r^2\sin^2 t(r'\cos t-r\sin t)dt \\ =-\pi\int_0^{\frac\pi2} r^2\sin^2 t(r'\cos t-r\sin t)dt \\ =-\pi (\int_0^{\frac\pi2} (r^2r'dt)(\sin^2t\cos t)-\int_0^{\frac\pi2} r^3\sin^3dt) \\ =-\pi (\dfrac{r^3}{3}\sin^2t\cos t\vert_0^{\frac\pi2}-\int_0^{\frac\pi2} \dfrac{r^3}{3} (2\sin t\cos^2 t-\sin^3)dt-\int_0^{\frac\pi2} r^3\sin^3 dt) \\ =\int_0^{\frac\pi2} \dfrac{2\pi}{3}r^3 \sin tdt \end{gathered}

所以

\begin{gathered} V=\int_0^{\frac \pi2} \dfrac{2\pi}{3} a^3(1+\cos \theta)^3\sin \theta d\theta \\ =\int_0^{\frac \pi2} \dfrac{2\pi}{3} a^3(1+\cos \theta)^3d(1+\cos\theta) \\ =\dfrac{5}{2} \pi a^3 \end{gathered}

T5

证明：图形 $0 \leqslant y \leqslant y(x), a \leqslant x \leqslant b$ 绕 $y$ 轴旋转所得的旋转体体积为 $V_y = 2\pi \int_a^b xy(x)\mathrm{d}x.$ 并由此计算： (1) 由 $y=x(x-1)^2$ , $y=0$ 所围图形绕 $y$ 轴旋转所得的旋转体体积; (2) 由 $y=\sin x \; (0 \leqslant x \leqslant \pi)$ , $y=0$ 所围图形绕 $y$ 轴旋转所得的旋转体体积.

微元法,取一个内径为 $x$ ,外径为 $x+dx$ ,高为 $y$ 的圆环柱体,体积为 $\pi y((x+dx)^2-x^2)=2\pi yxdx+y\pi(dx)^2\approx 2\pi yxdx$ ,累加即得.

(1)

\begin{gathered} V=2\pi\int_0^1 x^2(x-1)^2 dx \\ =2\pi \int_0^1 (x^4-2x^3+x^2)dx \\ =\dfrac{\pi}{15} \end{gathered}

(2)

\begin{gathered} V=2\pi \int_0^\pi x\sin xdx \\ =2\pi^2 \end{gathered}

Class 2

T1

计算下列反常积分：

\begin{gathered} (5) \int_0^{+\infty} \frac{1+x^2}{1+x^4} \mathrm{d}x \end{gathered}

\begin{gathered} =\int_0^{+\infty} \dfrac{1+\dfrac{1}{x^2} }{x^2+\dfrac{1}{x^2} } dx \\ \text{let }t=x-\dfrac{1}{x} \\ Ans=\int_{-\infty}^{+\infty} \dfrac{1}{t^2+2}dt \\ \text{let } t=\sqrt 2\tan t \\ Ans=\int_{-\frac{\pi}2}^{\frac{\pi}2} \dfrac{\sqrt 2\sec^2 t}{2\sec^2 t} dt \\ =\dfrac{\sqrt 2\pi}{2} \end{gathered}

T2

计算下列反常积分：

\begin{gathered} (6) \int_0^{+\infty} \frac{x \mathrm{e}^{-x}}{\left(1+\mathrm{e}^{-x}\right)^2} \mathrm{d}x \end{gathered}

\begin{gathered} \int \dfrac{e^{-x}}{(1+e^{-x})^2} dx \\ =\int -\dfrac{dt}{(1+t)^2} \\ =\dfrac{1}{1+t}+C \\ =\dfrac{1}{1+e^{-x}}+C \\ \text{let } C=-1 \\ \implies Ans=\dfrac{-xe^{-x}}{1+e^{-x}}\vert_0^{+\infty} -\int_0^{+\infty} \dfrac{-e^{-x}}{1+e^{-x}}dx \\ = 0+\ln 2 \\ =\ln 2 \end{gathered}

T3

判断下列无穷积分的敛散性：

\begin{gathered} (4) \int_0^{+\infty} \frac{x}{\mathrm{e}^x + \mathrm{e}^{-x}} \mathrm{d}x \end{gathered}

\begin{gathered} \le \int_0^{\infty} \dfrac{x}{e^x} \\ \le \int_0^\infty \dfrac{x}{1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6} } \\ \le \int_0^1 \dfrac{x}{e^x}+\int_1^\infty \dfrac{6}{x^2} \\ \le 1+\int_1^{\infty} \dfrac{6}{x^2} \end{gathered}

第二项收敛.所以收敛.

T4

判断下列无穷积分的敛散性：

\begin{gathered} (5) \int_1^{+\infty} \left[ \ln\left(1+\frac{1}{x^2}\right) - \frac{1}{1+x^2} \right] \mathrm{d}x \end{gathered}

\begin{gathered} \ln(1+\dfrac{1}{x^2} )-\dfrac{1}{1+x^2} \\ =\dfrac{1}{x^2}-\dfrac{1}{2x^4}+o(\dfrac{1}{x^5} ) -\dfrac{1}{x^2} +\dfrac{1}{x^4} \\ =\dfrac{1}{2x^4} +o(\dfrac{1}{x^5} ) \\ \implies \lim_{x \to \infty} \dfrac{\left[ \ln\left(1+\frac{1}{x^2}\right) - \frac{1}{1+x^2} \right]}{\dfrac{1}{2x^4} } =1 \\ \implies \text{Convergent} \end{gathered}

T5

判断下列无穷积分的敛散性：

\begin{gathered} (6) \int_1^{+\infty} x\left(1-\cos\frac{1}{x^2}\right)^p \mathrm{d}x, \quad p \in \mathbb{R} \end{gathered}

\begin{gathered} \cos \dfrac{1}{x^2} =1-\dfrac{1}{2x^4}+o(x^5) \\ \implies \lim_{x \to \infty} \dfrac{x(1-\cos \dfrac{1}{x^2} )^p}{x^{1-4p}} =C\in (0,\infty) \\ \int_1^{+\infty} x(1-\cos \dfrac{1}{x^2} )^p\begin{cases} \text{ is convergent} ,p>\dfrac{1}{2} \\ \text{ isn't convergent} ,p\le \dfrac{1}{2} \end{cases} \end{gathered}

T6

判断下列无穷积分的敛散性（含绝对收敛性与条件收敛性）：

\begin{gathered} (3) \int_1^{+\infty} \sin\left(\frac{\sin x}{x}\right) \mathrm{d}x \end{gathered}

\begin{gathered} \int_a^b \sin(\dfrac{\sin x}{x} )dx \\ =\sum_{i=A}^B \int_{k\pi}^{(k+1)\pi}\sin(\dfrac{\sin x}{x} )dx \\ +\int_a^{A\pi}\sin(\dfrac{\sin x}{x} )dx+\int_{B\pi}^b \sin(\dfrac{\sin x}{x} )dx \end{gathered}

其中第一项因为正负交替且递减,求和式绝对值小于第一项 $\int_{k\pi}^{(k+1)\pi}\sin (\dfrac{\sin x}x)$ .于是这三项都为 $\int_c^d \sin \dfrac{\sin x}x<2\pi \dfrac{1}{x}$ ,从而小于 $\dfrac{6\pi}{a}$ ,应用柯西收敛准则得知手收敛.

而

\begin{gathered} \int_1^{+\infty}{\left \vert \sin \dfrac{\sin x}{x} dx \right \vert} \\ > \int_1^{+\infty} {\left\vert \dfrac{\sin x}{x}dx\right \vert} -{\left\vert\int_1^{+\infty}\dfrac16(\dfrac{\sin x}{x} )^3dx \right \vert} \end{gathered}

第一项发散,第二项收敛,故发散.

条件收敛.

T7

判断下列无穷积分的敛散性（含绝对收敛性与条件收敛性）：

\begin{gathered} (4) \int_1^{+\infty} \frac{\cos(x^p)}{x} \mathrm{d}x, \quad p \in \mathbb{R} \end{gathered}

$p>0$ 时.

\begin{gathered} =\int_1^{+\infty}\dfrac{\cos x}{x^{\frac1p}}x^{\frac1p-1} dx \\ =\int_1^{+\infty}\dfrac{\cos x}{x} dx \\ \end{gathered}

$\cos x$ 积分有界, $\dfrac{1}{x}$ 递减且收敛到 $0$ ,原式收敛.

绝对值:

\begin{gathered} \vert \dfrac{\cos x}{x} \vert \ge \dfrac{\sin (x+\frac\pi2)}{x+\frac\pi2} \end{gathered}

而后一项积分发散.

故原式条件收敛.

$p\le 0$ 时, $\lim_{x\to +\infty}\cos x^p=1$

于是取 $X$ 足够大使 $\cos x^p>\dfrac{1}{2}$ , $\int_X^{+\infty}\dfrac{\cos(x^p)}{x} dx>\dfrac{1}{2}\int_X^{+\infty}\dfrac1x=\infty$ ,发散.

T8

设 $P_m(x)$ 和 $P_n(x)$ 分别为 $m$ 和 $n$ 次多项式，并且当 $x \geqslant a$ 时， $P_n(x) > 0$ . 试研究

\begin{gathered} \int_a^{+\infty} \frac{P_m(x)}{P_n(x)} \sin x \mathrm{d}x \end{gathered}

的绝对收敛性和条件收敛性.

不妨设 $x\to +\infty$ 时 $P_m(x)>0$

若 $m\ge n$ , $\lim_{x\to \infty}\dfrac{P_m(x)}{P_n(x)}=A>0$

\begin{gathered} \exists X,x>X \implies \dfrac{P_m(x)}{P_n(x)} >B(B<A) \\ \implies {\left \vert \int_{2n\pi+\frac\pi4}^{2n\pi+\frac\pi2} \dfrac{P_m(x)}{P_n(x)} \sin xdx \right \vert} \\ >\dfrac{\sqrt 2B\pi}{8} \end{gathered}

发散.

若 $m<n-1$ ,

\begin{gathered} \int {\left \vert \dfrac{P_m(x)}{P_n(x)} \sin x \right \vert} \le \int {\left \vert \dfrac{1}{x^r} \right \vert}<+\infty, (r>1) \end{gathered}

绝对收敛

若 $m=n-1$ ,

\begin{gathered} \dfrac{P_m(x)}{P_n(x)}=\dfrac{1}{x} +o(\dfrac{1}{x}) \end{gathered}

其中第一部分最后条件收敛,第二部分最后收敛,故整体条件收敛.

T9

证明无穷积分的对数判别法：设 $f(x) \in C[1, +\infty)$ 且恒正，若 $\lim_{x\to+\infty} \frac{\ln f(x)}{\ln x} = -\lambda$ ，则当 $\lambda > 1$ 时无穷积分 $\int_1^{+\infty} f(x) \mathrm{d}x$ 收敛.

\begin{gathered} \lim_{x \to +\infty} \dfrac{\ln f(x)}{\ln x} =-\lambda \\ \implies \exists 1<a<\lambda,X \ s.t.\ x>X \implies \dfrac{\ln f(x)}{\ln x} <-a, \\ \implies f(x)<\dfrac{1}{x^a} \end{gathered}

应用比较判别法, $\int_1^{\infty}\dfrac{1}{x^a}$ 收敛,得证.

T10

设在 $[a, +\infty)$ 上满足： $g(x) \leqslant f(x) \leqslant h(x)$ ，且 $\int_a^{+\infty} g(x) \mathrm{d}x$ 与 $\int_a^{+\infty} h(x) \mathrm{d}x$ 收敛，请问 $\int_a^{+\infty} f(x) \mathrm{d}x$ 是否收敛？

\begin{gathered} f(x)-g(x)\le h(x)-g(x)<+\infty \\ f(x)=g(x)+(f(x)-g(x))<+\infty \end{gathered}

收敛.

T11

设函数 $f(x)$ 在 $[a, +\infty)$ 上单调减少且趋于 0. 证明：无穷积分 $\int_a^{+\infty} f(x) \mathrm{d}x$ 与 $\int_a^{+\infty} f(x)\sin^2 x \mathrm{d}x$ 同敛散.

\begin{gathered} I_1=\int_a^{+\infty}f(x)dx \\ I_2=\int_a^{+\infty}f(x)\sin^2 xdx\le I_1 \end{gathered}

故 $I_1$ 收敛推出 $I_2$ 收敛.

若 $I_1$ 发散,因为 $f$ 递减,则

\begin{gathered} F(x)=\int_a^x f(t)dt \\ F(x)<C+4\sum_{i=a} \int_{2i\pi+\frac{\pi}4}^{2i\pi+\frac{3\pi}4}f(t)dt=C+G(x) \end{gathered}

而

\begin{gathered} \int_a^x f(t)dt\sin^2 tdt \\ = H(x) \\ >\sum_{i=a} \int_{2i\pi+\frac\pi4}^{2i\pi+\frac{3\pi}4}f(t)\sin^2tdt \\ >\sum_{i=a} \int_{2i\pi+\frac\pi4}^{2i\pi+\frac{3\pi}4}f(t)\dfrac{1}{2}dt \\ =\dfrac{1}{2} G(x) \end{gathered}

于是 $F(x)<kH(x)$ 推出 $I_2$ 发散.

T12

设函数 $f(x)$ 在 $[a, +\infty)$ 上连续可微，且无穷积分 $\int_a^{+\infty} f(x) \mathrm{d}x$ 与 $\int_a^{+\infty} f'(x) \mathrm{d}x$ 都收敛. 证明： $\lim_{x\to+\infty} f(x) = 0$ .

\begin{gathered} \exists X,\forall c,d>X, {\left \vert \int_c^d f'(x)dx \right \vert} =\vert f(d)-f(c) \vert <\epsilon \end{gathered}

于是 $f(x)$ 收敛,设收敛到 $a$ ,若 $a\ne 0$ ,存在 $0<b<\vert a\vert$

则

\begin{gathered} \exists X,\forall x>X,\vert f(x) \vert >b,\forall x_1,x_2>X,f(x_1)f(x_2)>0 \\ \vert \int_{a}^\infty f(x)dx \vert =\int_{a}^\infty \vert f(x) \vert dx>\int_a^{\infty}bdx=+\infty \end{gathered}

矛盾.于是得证.