Math Analysis Homework - Week 9

Class 1

T1

\begin{gathered} \begin{cases} f(x)\in C[a,b], \\ \forall g\in C[a,b],\int_a^b f(x)g(x)dx=0 \end{cases} \\ \implies f(x)=0 \end{gathered}

反证,假设存在 $f(x_0)\ne 0$ ,不妨设是 $f(x_0)>0$ ,则存在 $(x_0-\delta,x_0+\delta)$ 使得 $\forall x\in (x_0-\delta,x_0+\delta),f(x)>\dfrac{f(x_0)}2$ .

取

\begin{gathered} g(x)=\begin{cases} 1,\vert x-x_0 \vert <\dfrac{\delta}{2} \\ \vert x-x_0-\dfrac{\delta}{2} \vert,\vert x-x_0 \vert \in [\dfrac{\delta}{2},\delta) \\ 0,\text{otherwise} \end{cases} \\ \implies \int f(x)g(x)dx \ge \dfrac{f(x_0)}{2} \delta>0 \end{gathered}

T2

\begin{gathered} \begin{cases} a,b>0,f(x)\in C[-a,b] \\ f(x)>0,\int_{-a}^bxf(x)dx=0 \end{cases} \\ \implies \int_{-a}^b x^2f(x)dx\le ab\int_{-a}^b f(x)dx \end{gathered}

Sol1:

\begin{gathered} \int_{-a}^0x^2f(x)dx\le \int_{-a}^0 f(x)(-x)adx=\int_{0}^b f(x)xadx\le \int_0^b f(x)abdx \\ \text{similarily, }\int_{0}^b x^2f(x)dx\le \int_{-a}^0 f(x)abdx \\ \implies \int_{-a}^b x^2f(x)dx \\ \le \int_{-a}^0 f(x)abdx+\int_0^b f(x)abdx \\ =ab\int_{-a}^b f(x)dx \\ \end{gathered}

Sol2:

\begin{gathered} \int_{-a}^b (x+a)(b-x)f(x)dx\ge 0 \\ \implies \int_{-a}^b (-x^2+(b-a)x+ab)f(x)dx\ge 0 \end{gathered}

中间是 $0$ ,就弄完了.

[think] 后面这个做法看起来已经称为一种套路了.就是对于 $\int x^kf(x)$ 的形式能凑出来的最优情形.

T3

\begin{gathered} f(x)\in C[0,1],f(x)>0 \\ \implies \int_0^1 f(x)dx\int_0^1 \dfrac{1}{f(x)} dx\ge 1 \end{gathered}

\begin{gathered} 1=\int_0^1 \sqrt{f(x)}\cdot \dfrac{1}{\sqrt{f(x)}} dx\le (\int_0^1 \sqrt{f(x)}^2dx)^\frac12 (\int_0^1 \dfrac{1}{\sqrt{f(x)}} dx)^\frac12 \\ \implies \int_0^1 f(x)dx\int_0^1 \dfrac{1}{f(x)} dx\ge 1 \end{gathered}

T4

\begin{gathered} \begin{cases} f(x) \in C[0,\pi] \\ \int_0^\pi f(\theta)\cos \theta d\theta = \int_0^\pi f(\theta)\sin \theta d\theta =0 \\ \end{cases} \\ \implies \exists x_1,x_2\in (0,\pi),f(x_i)=0 \end{gathered}

若 $f(x)$ 无零点,则 $f(x)\sin (x)$ 无变号零点, $\int_0^\pi f(x)\sin x dx\ne 0$ , $f$ 至少一个变号零点.

若 $f(x)$ 恰有一个零点 $x_1$ .

考虑令 $\int_0^\pi f(x)(\sin x\cos \varphi+\cos x\sin \varphi)dx=0=\int_0^\pi f(x)\sin(x+\varphi)dx$ ,于是令 $\phi=-x_1$ ,则 $\int f(x)\sin(x+\varphi)dx$ 不变号,不为 $0$ ,矛盾,得证.

T5

\begin{gathered} \begin{cases} f\in C[a,b],f(x)\ge 0 \\ M=\max f([a,b]) \end{cases}\\ \implies \lim_{n \to \infty} (\int_a^b f^n(x)dx)^\frac1n=M \end{gathered}

\begin{gathered} \int_a^b f^n(x)\le (b-a)M^n \\ \implies \lim_{n \to \infty} (\int_a^b f^n(x)dx)^\frac1n\le \lim_{n \to \infty} ((b-a)M^n)^\frac1n=M \\ \text{let } f(x_0)=M \\ f(x)\in C[a,b] \implies \forall p<1,\exists \delta,\ s.t.\ \\ \forall x \in (x_0-\delta,x_0+\delta)\cap [a,b],f(x)>pM \\ \implies \int_a^b f^n(x)dx>2\delta(pM)^n \\ \implies \forall p\in(0,1), \lim_{n \to \infty} (\int_a^b f^n(x)dx)^\frac1n>\lim_{n \to \infty} (2\delta(pM)^n)^\frac1n=pM \\ \implies \lim_{n \to \infty} (\int_a^b f^n(x)dx)^\frac1n\ge M \\ \implies \lim_{n \to \infty} (\int_a^b f^n(x)dx)^\frac1n= M \\ \end{gathered}

T6

\begin{gathered} \begin{cases} f \in C[a,b],f(x)\ge 0 \\ f(x) \text{ is strictly increasing} \\ \forall p,\exists x_p\in [a,b] \\ f^p(x_p)=\dfrac{1}{b-a} \int_a^b f^p(t)dt \end{cases} \\ \implies \lim_{p \to +\infty} x_p =b \end{gathered}

$f$ 严格单增且连续则 $f^{-1}$ 存在且连续,故只需证

\begin{gathered} f(b)=\lim_{p \to +\infty} f(x_p)=\lim_{p \to +\infty} (\dfrac{1}{b-a} \int_a^b f^p(t)dt)^\frac1p \\ \end{gathered}

而

\begin{gathered} \lim_{p \to +\infty} (\dfrac{1}{b-a} \int_a^b f^p(t)dt)^\frac1p =\lim_{p \to +\infty} (\dfrac{1}{b-a} )^\frac1p (\int_a^b f^p(t)dt)^\frac1p \\ \xlongequal{by T5}\lim_{p \to +\infty} (\dfrac{1}{b-a} )^\frac1p M^\frac1p \\ =M=f(b) \end{gathered}

T7

\begin{gathered} \begin{cases} f(x)\in C[0,1]\cap D(0,1) \\ f(1)=2\int_0^\frac12 xf(x)dx \end{cases} \\ \implies \exists \xi\in (0,1) \\ f(\xi)+\xi f'(\xi)=0 \end{gathered}

\begin{gathered} \text{let } g(x)=xf(x) \\ g(1)=2\int_0^\frac12 g(x)dx=2\times \dfrac{1}{2} g(\xi),\xi\in (0,\dfrac{1}{2} ) \\ g(1)=g(\xi) \xRightarrow{\text{Rolle's Theorem} }\exist \xi',g'(\xi')=0=f(\xi)+\xi f'(\xi) \end{gathered}

T8

\begin{gathered} \lim_{n \to \infty} \int_0^{\frac\pi2}\sin^n xdx \end{gathered}

\begin{gathered} =\lim_{n \to \infty} \int_0^{\frac\pi2-\delta_n}\sin^n xdx+\int_{\frac\pi2-\delta_n}^{\frac\pi2}\sin^n xdx \\ =\lim_{n \to \infty} (\dfrac{\pi}{2} -\delta)\sin^n \xi+\delta_n\sin^n\xi_2 \\ \le \lim_{n \to \infty} \dfrac{\pi}{2}\cos^n(\delta_n)+\delta_n \\ \text{let }\delta_n=\text{max } v \\ s.t.\\ \cos^n(v)<\dfrac{1}{n} \\ \implies \lim_{n \to \infty} \delta_n=\lim_{n \to \infty} \arccos((\dfrac{1}{n} )^\frac1n)=\arccos(1)=0 \\ \implies Ans=\lim_{n \to \infty} \dfrac{\pi}{2} \cos^n(\delta_n)+\delta_n=0 \end{gathered}

T9

\begin{gathered} \begin{cases} f(x) \text{ is integrable on any limited range} \\ \lim_{x \to +\infty} f(x)=l \end{cases} \\ \implies \lim_{x \to +\infty} \dfrac{1}{x} \int_0^x f(t)dt=l \end{gathered}

\begin{gathered} \exists X,x>X \implies \vert f(x)-l \vert <\epsilon \\ \lim_{x \to +\infty} \dfrac{1}{x} \int_0^x f(t)dt \\ =\lim_{x \to +\infty} \dfrac{1}{x} (\int_0^X f(t)dt+\int_X^x f(t)dt) \\ =\lim_{x \to +\infty} \dfrac{1}{x} \int_X^xf(t)dt \\ \in(\lim_{x \to +\infty} \dfrac{1}{x} (l-\epsilon)(x-X),\lim_{x \to +\infty} (l+\epsilon)(x+X )) \\ =(l-\epsilon,l+\epsilon) \end{gathered}

因为对任意 $\epsilon>0$ 均成立,于是得证.

T10

\begin{gathered} f(x) \text{ is integrable on } [A,B] \\ \implies \forall a<b \in (A,B) \\ \lim_{h \to 0} \int_a^b \vert f(x+h)-f(x) \vert dx=0 \end{gathered}

\begin{gathered} \forall T=\{ a=t_0<\ldots< t_n=b \} \\ \lim_{\vert\vert T \vert\vert \to 0} \sum _{i = 1} ^{n} w_i(t_i-t_{i-1})=0 \\(w_i=\sup_{x,y \in [t_{i-1},t_i]} f(x)-f(y)) \end{gathered}

对一个 $h$ ,按长度 $h$ 分割得到 $\forall i<n,t_i-t_{i-1}=h$ 的 $T$ ,则 $\forall x\in [t_{i-1},t_i]$ , $\vert f(x+h)-f(x)\vert\le w_i+w_{i+1}$ ,于是

\begin{gathered} \lim_{h \to 0} \int_a^b \vert f(x_h)-f(x) \vert dx\le \lim_{h \to 0} \sum_{i=1}^{n-1}(t_i-t_{i-1})(w_i+w_{i+1})+(t_n-t_{n-1})w_n \\ \le \lim_{h \to 0} 2 \sum _{i = 1} ^{n}w_i(t_i-t_{i-1}) \\ =\lim_{\vert\vert T \vert\vert \to 0} 2w_i(t_i-t_{i-1}) \\ =0 \end{gathered}

T11

\begin{gathered} \lim_{x \to +\infty} \dfrac{\int_0^x (\arctan t)^2dt}{\sqrt{1+x^2}} \end{gathered}

\begin{gathered} \xlongequal{\text{L'Hospital} } \lim_{x \to +\infty} \dfrac{\sqrt{1+x^2}\arctan^2 x}{x} \\ =\lim_{x \to +\infty} \sqrt{ 1+\dfrac{1}{x^2} } \arctan^2 x \\ =\dfrac{\pi^2}{4} \end{gathered}

T12

\begin{gathered} \begin{cases} b>0,f(x) \in C[0,b] \\ f(x) \text{ is strictly increasing} \end{cases} \\ \implies 2\int_0^b xf(x)dx\ge b\int_0^b f(x)dx \end{gathered}

\begin{gathered} \int_0^b f(x)(2x-b)dx \\ =-\int_0^\frac{b}2f(x)(b-2x)dx+\int_\frac{b}2^bf(x)(2x-b)dx \\ =\int_0^\frac{b}2 f(x)(b-2x)(f(b-x)-f(x))dx\ge 0 \end{gathered}

T13

\begin{gathered} f(x)\in D[0,1],f(0)=0,f'(x)\in[0,1] \\ \implies \int_0^1 f^3(x)dx\le (\int_0^1 f(x)dx)^2 \end{gathered}

\begin{gathered} \text{let } F(x)=\int_0^x f^3(t)dt-(\int_0^x f(t)dt)^2 \\ F'(x)=f^3(x)-2(\int_0^x f(t)dt)f(x) \\ \text{let } F_1(x)=f^2(x)-2\int_0^x f(t)dt \\ F_1'(x)=2f(x)f'(x)-2f(x)\le 0 \\ \implies F_1'(x)\le F_1'(0)=0 \\ \implies F'(x)<0 \\ \implies F(x)<F(0)=0 \\ \text{Q.E.D} \end{gathered}

T14

\begin{gathered} \begin{cases} f(x) \in C[0,+\infty) \\ x>0 \implies \int_0^x f(t)dt=\dfrac{1}{2} xf(x) \end{cases} \\ \implies f(x)=cx,x>0 \end{gathered}

$x=0$ 时,显然同样 $\int_0^x f(t)dt=\dfrac{1}{2}xf(x)$ ,这个式子对 $[0,+\infty)$ 成立.

两边同时求导, $f(x)=\dfrac{1}{2} f(x)+\dfrac{1}{2} xf'(x)$ ,于是

\begin{gathered} f(x)=xf'(x) \\ \dfrac{f(x)-xf'(x)}{x^2} =0 \\ \implies (\dfrac{f(x)}{x} )'=0 \\ \implies f(x)=cx \end{gathered}

T15

\begin{gathered} f(x) \in C[0,+\infty),f(x)>0 \\ \implies \phi(x)=\dfrac{\int_0^x tf(t)dt}{\int_0^x f(t)dt} \text{ is strictly increasing} \end{gathered}

\begin{gathered} \int_0^x tf(t)dt<\int_0^x xf(t)dt \\ \implies \phi'(x)=\dfrac{xf(x)\int_0^x f(t)dt-f(x)\int_0^x tf(t)dt}{(\int_0^x f(t)dt)^2}>0 \\ \text{Q.E.D} \end{gathered}

T16

\begin{gathered} \int_0^1 (2x-1)e^{x^2-1}dx \end{gathered}

等价于积 $\int_0^1 e^{x^2}dx$ ,感觉做不了.

T17

\begin{gathered} \lim_{n \to \infty} \sum _{k = 1} ^{n} \sqrt{ \dfrac{(n+k)(n+k+1)}{n^4} } \end{gathered}

\begin{gathered} =\lim_{n \to \infty} \sum _{k = 1} ^{n} \dfrac{1}{n} \sqrt{ (\dfrac{k}{n} +1)(\dfrac{k+1}{n} +1) } \\ \end{gathered}

令 $f(x)=x,T=\{ t_i=1+\dfrac in \},\xi_i=\sqrt{(\dfrac{k}{n} +1)(\dfrac{k+1}{n} +1)}$ ,于是原式就是

\begin{gathered} \int_1^2 xdx=\dfrac{3}{2} \end{gathered}

T18

\begin{gathered} \forall x\in[-1,+\infty),f(x)=\int_{-1}^x \dfrac{e^{\frac1t}}{t^2(1+e^\frac1t)^2} dt \\ \implies f(x)=? \end{gathered}

\begin{gathered} \text{let } u=e^{\frac1t} \\ f(x)=-\int_{\frac1e}^{\frac1{e^x}} \dfrac{1}{(1+u)^2} du \\ =\dfrac{1}{e^\frac1x+1} -\dfrac{e}{e+1},x<0 \\ f(0)=1-\dfrac{e}{1+e} =\dfrac{1}{1+e} \\ x>0 \implies f(x)=f(0)+f(x)-f(0^+)=\dfrac{1}{e^\frac1x+1} -\dfrac{e}{1+e} +1 \end{gathered}

Class 2

T1

\begin{gathered} \int_0^1 \vert 1-2x \vert dx \end{gathered}

\begin{gathered} =2\int_0^{\frac12}(1-2x)dx \\ =2(x-x^2\vert^{\frac12}_0)dx \\ =\dfrac{1}{2} \end{gathered}

T2

\begin{gathered} \int_{-\frac\pi2}^{\frac\pi2} \sqrt{ \cos x-\cos^3 x } dx \end{gathered}

\begin{gathered} =-2\int^{\frac\pi2}_{0} \sqrt{ \cos x } d\cos x \\ =-\dfrac{4}{3} \cos^{\frac32} x \vert_0^{\frac\pi2} \\ =\dfrac{4}{3} \end{gathered}

T3

\begin{gathered} \int_{\frac1e}^e \vert \ln x \vert dx \end{gathered}

\begin{gathered} =-\int_{\frac1e}^1 \ln xdx+\int_1^e \ln xdx \\ =-(x\ln x-x)\vert_{\frac1e}^1+(x\ln x-x)\vert_1^e \\ =1-\dfrac{2}{e}+1 \\ =2-\dfrac{2}{e} \end{gathered}

T4

\begin{gathered} \int_0^\pi e^x \cos^2 xdx \end{gathered}

\begin{gathered} =e^x\cos^2 x \vert_0^\pi+\int_0^\pi e^x 2\cos x\sin xdx \\ =e^\pi-1+\int_0^\pi e^x \sin 2x dx \\ \end{gathered}

\begin{gathered} \int_0^\pi e^x \sin 2xdx \\ =\dfrac{1}{2}\int_0^{2\pi }e^{\frac{x}{2} }\sin xdx \\ =(e^{\frac x2}\sin x\vert _0^{2\pi})-\int_0^{2\pi}e^{\frac x2}\cos xdx \\ =(e^{\frac x2}\sin x\vert _0^{2\pi})-(2e^{\frac x2}\cos x\vert_0^{2\pi})-2\int_0^{2\pi}e^{\frac x2}\sin xdx \\ \implies \int_0^\pi e^x\sin 2xdx=-\dfrac{2}{5} e^\pi+\dfrac{2}{5} \implies Ans=\dfrac{3}{5} e^\pi-\dfrac{3}{5} \end{gathered}

T5

\begin{gathered} \int_1^e (x\ln x)^2 dx \\ =(\dfrac{1}{3} x^3\ln^2 x-\dfrac{2}{9} x^3\ln x+\dfrac{2}{27} x^3)\vert_1^e \\ =\dfrac{5e^3-2}{27} \end{gathered}

T6

\begin{gathered} \int_0^1 (1-x^2)^n dx \end{gathered}

\begin{gathered} x=\sin t \\ \int_0^1 (1-x^2)^n dx \\ =\int_0^{\frac{\pi}{2}} \cos^{2n+1}tdt \end{gathered}

\begin{gathered} I_m=\int_0^{\frac{\pi}{2}}\cos^m tdt \\ =\sin t\cos^{m-1}t\vert_0^{\frac{\pi}{2}}+(m-1)\int_0^{\frac{\pi}{2}}\sin^2 t\cos^{m-2}tdt \\ =+(m-1)\int_0^{\frac{\pi}{2}}\cos^{m-2}tdt-(m-1)\int_0^{\frac{\pi}{2}}\cos^m tdt \\ =+(m-1)I_{m-2}-(m-1)I_m \\ \implies I_m=\dfrac{m-1}m I_{m-2} \\ I_1=1 \\ Ans=I_{2n+1}=\dfrac{(2n)!!}{(2n+1)!!} \end{gathered}

T7

\begin{gathered} \int_0^{\frac\pi4}\cos^7 (2x)dx \end{gathered}

\begin{gathered} =\dfrac{1}{2}\int_0^{\frac\pi2}\cos^7 xdx \\ =\dfrac{1}{2} I_7 \\ =\dfrac{8}{35} \end{gathered}

T8

\begin{gathered} \int_0^{\frac\pi4}\ln(1+\tan x)dx \end{gathered}

\begin{gathered} I \\ =\int_0^{\frac\pi4}\ln(1+\tan x)dx \\ =\int_0^{\frac\pi4}\ln(1+\tan(\dfrac{\pi}4-x))dx \\ =\int_0^{\frac\pi4}\ln(1+\dfrac{1-\tan x}{1+\tan x} )dx \\ =\int_0^{\frac\pi4}(\ln2-\ln (1+\tan x))dx\\ =\dfrac{\pi}{4} \ln 2-I \\ \implies I=\dfrac{\pi}{8} \ln 2 \end{gathered}

T9

\begin{gathered} \int_0^{\frac\pi2}\dfrac{1}{1+\tan^\alpha x} dx(\alpha>0) \end{gathered}

\begin{gathered} I=\int_0^{\frac\pi2}\dfrac{1}{1+\tan^\alpha x} dx \\ =\int_0^{\frac\pi2}\dfrac{\tan \alpha x}{1+\tan \alpha x} dx \\ =\int_0^\frac\pi21dx-\int_0^\frac\pi2 \dfrac{dx}{1+\tan \alpha x} \\ =\dfrac{\pi}{2} -I \\ \implies I=\dfrac{\pi}{4} \end{gathered}

T10

\begin{gathered} \int_0^{n\pi} x \vert \sin x \vert dx,n\in N \end{gathered}

\begin{gathered} I_k=\int_{k\pi}^{(k+1)\pi} x \vert \sin x \vert dx \\ =\int_{(k-1)\pi}^{k\pi} (x+\pi) \vert \sin (x+\pi) \vert dx \\ =\int_{(k-1)\pi}^{k\pi} x \vert \sin x \vert dx +\pi\int_{(k-1)\pi}^{k\pi}\vert \sin x \vert \\ =I_{k-1} +2\pi \\ I_0=\int_0^\pi x\sin xdx \\ =(-x\cos x+\sin x) \vert_0^\pi \\ =\pi \\ \implies Ans=\sum _{i = 0} ^{n-1} I_i \\ =n^2\pi \end{gathered}

T11

\begin{gathered} f(x)\in C(R), \\ \text{calculate } \dfrac{d}{dx} \int_0^x tf(x^2-t^2)dt \end{gathered}

\begin{gathered} F(x)=\int_0^x tf(x^2-t^2)dt \\ \text{let } s=x^2-t^2,ds=-2tdt \\ F(x)=\int_{x^2}^0 \dfrac{-1ds}{2} f(s) \\ =\int_0^{x^2} \dfrac{1}{2} f(s)ds \\ F'(x)=2x \dfrac{1}{2} f(x^2)=xf(x^2) \end{gathered}

T12

\begin{gathered} f(x)\in C(R), \\ f(x)=x+2\int_0^1 f(t)dt \\ \text{calculate } f(x) \end{gathered}

\begin{gathered} f'(x)=1 \implies f(x)=x+C \\ x+C=x+2\int_0^1 (t+C)dt \\ =x+2(\dfrac{t^2}{2} +Ct)\vert_0^1 \\ =x+1+2C \\ \implies C=-1,f(x)=x-1 \end{gathered}