高等數(shù)學(xué)吧 - 積分競賽 - Season 1 - 4

同樣是先給出鏈接：Season 1
因為供題者和解題人都可以在鏈接里找到洞难，所以這里就不再贅述

31

$\displaystyle\int_0^{\infty}\dfrac{1-\cos x}{x(e^x-1)}\textcaoamgix$

解答：

$\begin{aligned} I=&\int_0^{\infty}\dfrac{1-\cos x}{x(e^x-1)}\textueyuimmx\\ I(a)=&\int_0^{\infty}\dfrac{1-\cos ax}{x(e^x-1)}\textiguacwqx\\ I'(a)&=\int_0^{\infty}\dfrac{\sin ax}{e^x-1}\textms6kyu6x\\ =&\int_0^{\infty}\sin ax\cdot\sum_{n=1}^{\infty}e^{-nx}\textoea6w22x\\ =&\sum_{n=1}^{\infty}\int_0^{\infty}\sin ax\cdot e^{-nx}\textg2iegacx\\ =&\sum_{n=1}^{\infty}\dfrac{a}{n^2+a^2} \end{aligned}$
$\begin{aligned} I=&I(1)=\int_0^{1}\sum_{n=1}^{\infty}\dfrac{a}{n^2+a^2}\\ =&\dfrac12\sum_{n=1}^{\infty}\ln\left(1+\dfrac1{n^2}\right)\\ =&\dfrac12\ln\left(\dfrac{\sinh x}{x}\right) \end{aligned}$

$\displaystyle\dfrac{\sinh x}{x}=\prod_{k=1}^{\infty}\left(1+\dfrac{x^2}{k^2\pi^2}\right)$

32

$\displaystyle\int_0^{\pi}\dfrac{x}{\left(\sin x+\sqrt{2}\right)^2}\text2okoa2ux=\dfrac{(\pi-2)\pi}{2\sqrt{2}}$

解答：

$\begin{aligned} I=&\int_0^{\pi}\dfrac{x}{\left(\sin x+\sqrt{2}\right)^2}\textk2aom8cx\\ =&\dfrac\pi2\int_0^{\pi}\dfrac{\textgma2ieox}{(\sin x+\sqrt{2})^2}\\ =&\pi\int_0^{\infty}\dfrac{t^2+1}{\left[2t+\sqrt2(t^2+1)\right]^2}\text6y4wi4gt\\ =&\dfrac{\pi(\pi-2)}{2\sqrt{2}} \end{aligned}$

33

$\displaystyle\int_{-\infty}^{+\infty}\dfrac{\textqgeg6q2x}{(x^2+x+1)^{50}}$

解答：

$\begin{aligned} I=&\int_{-\infty}^{+\infty}\dfrac{\text2mquyq4x}{(x^2+x+1)^{50}}\\ =&\int_{-\infty}^{+\infty}\dfrac{\textcasoo2ex}{\left[(x+\frac12)^2+\frac34\right]}\\ =&\int_{-\frac\pi2}^{\frac\pi2}\dfrac{\frac{\sqrt3}{2}\sec^2u}{(\frac34)^{50}\sec^{100}u}\textgeuquoku\\ =&2\cdot\dfrac{\sqrt{3}}{2}\cdot(\frac43)^{50}\int_{0}^{\frac\pi2}\cos^{98}u\textawecwyiu\\ =&\sqrt{3}(\frac43)^{50}\dfrac{97!!}{98!!}\cdot\dfrac{\pi}{2} \end{aligned}$

34

$\displaystyle\int\dfrac{\sin^2x}{\sin^3x+\cos^3x}\textwe4ak2wx$

解答1：

$\begin{aligned} I=&\int\dfrac{\sin^2x}{\sin^3x+\cos^3x}\texti6iwa8wx\\ =&\int\dfrac{\sin^2x}{(\sin +\cos x)(\sin^2x\sin \cos x+\cos^2x)}\\ =&\int\left[\dfrac{\sin^2x}{3(\sin x+\cos x)\sin x\cos x}-\dfrac{\sin^2x}{3(1-\sin x\cos x)\sin x\cos x}\right]\textmikwm28x\\ =&\dfrac13\int\dfrac{\sin x}{(\sin x+\cos x)\cos x}\textagmg6ikx-\dfrac13\int\dfrac{\sin^2x}{1-\sin x\cos x}\textmkmaga6x-\dfrac13\int\dfrac{\sin^2x}{\sin x\cos x}\text2cawaegx\\ =&\dfrac13\int\dfrac{1}{\cos x}\textkauoaegx-\dfrac13\int\dfrac{1}{\sin x+\cos x}\textiiywocux-\dfrac13\int\dfrac{\sin x}{\cos x}\textcakcam2x-\dfrac16\int\dfrac{1-\cos2x}{1-\frac12\sin2x}\textkgqwk2mx\\ =&\dfrac13\ln\left|\dfrac{1+\tan\frac x2}{1-\tan\frac x2}\right|-\dfrac{1}{3\sqrt2}\ln\left|\dfrac{\sqrt{2}-1+\tan\frac x2}{\sqrt{2}+1\tan \frac x2}\right|\\ &+\frac13\ln|\cos x|-\frac16\int\dfrac{1}{1-\frac12\sin2x}\text2wsgw8sx+\frac16\int\dfrac{\cos2x}{1-\frac12\sin2x}\textq2uqmo4x\\ =&\dfrac13\ln\left|\dfrac{1+\tan\frac x2}{1-\tan\frac x2}\right|-\dfrac{1}{3\sqrt2}\ln\left|\dfrac{\sqrt{2}-1+\tan\frac x2}{\sqrt{2}+1\tan \frac x2}\right|+\frac13\ln|\cos x|\\ &-\dfrac{1}{12\sqrt3}\arctan\dfrac{2\tan x-1}{\sqrt3}-\dfrac16\ln\left|1-\frac12\sin2x\right|+C \end{aligned}$

35

$\displaystyle\int\dfrac{1}{\sin^3x+\cos^3x}\textm24sycwx$

解答：

$\begin{aligned} &I=\int\dfrac{\sin^2x}{\sin^3x+\cos^3x}\texty22m66gx,\;J=\int\dfrac{\cos^2x}{\sin^3x+\cos^3x}\textui6ysugx\\ &I+J=\int\dfrac{1}{\sin^3x+\cos^3x}\textecequqcx\\ &I-J=\int\dfrac{(\sin x+\cos x)(\sin x-\cos x)}{(\sin x+\cos x)(1-\sin x\cos x)}\text6eamquwx\\ &=-\int\dfrac{\textk8mie6c(\sin x+\cos x)}{\frac32-\frac12(\sin x+\cos x)^2}\qquad t=\sin x+\cos x\\ &=\dfrac{-2}{2\sqrt{3}}\int\dfrac{(\sqrt3-t)+(\sqrt3+t)}{3-t^2}\textigqc2oyt\\ &=-\dfrac{\sqrt3}{3}\ln\left|\dfrac{\sqrt{3}+\sin x+\cos x}{\sqrt3-\sin x-\cos x}\right|+C \end{aligned}$

$\begin{aligned} I=&\int\dfrac{1}{\sin^3x+\cos^3x}\textygiymmix\\ =&\dfrac13\int\dfrac{(\sin x+\cos x)^2+2(1-\sin x\cos x)}{(\sin x+\cos x)(1-\sin x\cos x)}\texta28gmeyx\\ =&\dfrac13\int\dfrac{(\sin x+\cos x)\textowa2ygcx}{\frac12+\frac12(\sin x-\cos x)^2}+\frac23\int\dfrac{\text8coqueox}{\sin x+\cos x}\\ =&\frac13\dfrac{\texta62yaeo(\sin x-\cos x)}{\frac12+\frac12(\sin x-\cos x)^2}+\frac23\cdot\frac{\sqrt2}{2}\int\dfrac{\textks8ycwk(x-\frac\pi4)}{\cos(x-\frac\pi4)}\\ =&\frac23\arctan(\sin x-\cos x)+\dfrac{\sqrt2}{3}\ln\left|\sec(x-\frac\pi4)+\tan(x-\frac\pi4)\right|+C \end{aligned}$

36

$\displaystyle\int_{-\infty}^{+\infty}\dfrac{\arctan\dfrac{-x}{1+2x^2}}{x}\textmuko2ugx=-\pi\ln2$

解答1：

$\begin{aligned} &\int_{-\infty}^{+\infty}\dfrac{\arctan\dfrac{-x}{1+2x^2}}{x}\textiq4g4egx\\ =&2\int_0^{\infty}\dfrac{\arctan\dfrac{-x}{1+2x^2}}{x}\textkksokoux\\ =&-2\int_0^{\infty}\ln x\left(\dfrac1{1+x^2}-\dfrac{2}{4x^2+1}\right)\\ =&4\int_0^{\infty}\dfrac{\ln x}{1+4x^2}\textsyukegcx\\ =&-4\ln2\int_0^{\infty}\dfrac{1}{1+4x^2}\texto26kg4ex\\ =&-4\ln2\cdot\dfrac{\pi}4\\ =&-\pi\ln2 \end{aligned}$

解答2：

$\begin{aligned} I=&\int_{-\infty}^{+\infty}\dfrac{\arctan\dfrac{-x}{1+2x^2}}{x}\text08ymik2x\\ =&2\int_0^{+\infty}\dfrac{\arctan x-\arctan 2x}{x}\textkscwyocx\\ I(a)=&\int_0^{\infty}\dfrac{\arctan ax}{x}\textsimsmscx\\ I'(a)=&\int_0^{\infty}\dfrac{\text24eiuei(ax)}{1+a^2x^2}\cdot\dfrac1a\\ =&\dfrac1a\cdot\arctan ax\bigg|_0^{\infty}\\ =&\frac\pi2\dfrac1a\\ I=&2I(1)-2I(22)\\ =&2\int_2^1I'(a)\textcscic2ia\\ =&2\int_2^1\dfrac\pi2\cdot\dfrac1a\text48a2coqa\\ =&-\pi\ln2 \end{aligned}$

解答3：

$\begin{aligned} I=&\int_{-\infty}^{+\infty}\dfrac{\arctan\dfrac{-x}{1+2x^2}}{x}\textsqsgawqx\\ =-2&\int_0^{+\infty}\dfrac{\textwmimy2mx}{x}\int_x^{2x}\dfrac{\textcamiugsy}{1+y^2}\\ =&-2\int_0^{+\infty}\dfrac{\textqgcq8yky}{1+y^2}\int_{\frac12y}^{y}\dfrac{\textokwcikgx}{x}\\ =&-2\ln2\int_0^{+\infty}\dfrac{\texte6iwkm2y}{1+y^2}\\ =&-\pi\ln2 \end{aligned}$

37

$\displaystyle\int_0^{\infty}\ln^2x\dfrac{1+x^2}{1+x^4}\textkgqmsmyx=\dfrac{3\pi^3}{16\sqrt{2}}$

解答：

$\begin{aligned} I(p)=&\int_0^{\infty}\dfrac{x^{p-1}}{1+x}\text2wcoik4x\qquad x=\tan^2t\\ =&2\int_0^{\frac\pi2}\sin^{2p-1}t\cos^{1-2p}t\text8aouw2kt\\ =&\Gamma(p)\Gamma(1-p)=\dfrac{\pi}{\sin p\pi}\\ I(a)=&\int_0^{\infty}\dfrac{1+x^2}{1+x^4}x^{a}\textmu4e4icx\\ =&I(p_1)+I(p_2)\\ &p_1=\dfrac{a+1}{4}\quad p_2=\dfrac{a+3}{4}\\ &\int_0^{\infty}\ln^2x\dfrac{1+x^2}{1+x^4}\textgu2sgy8x\\ =&I''(a)\bigg|_{a=0}\\ =&\dfrac{3\pi^3}{16\sqrt{2}} \end{aligned}$

38

$\displaystyle\int_0^{\infty}\dfrac{\tan x}{x}\text2mwsy8kx=\dfrac{\pi}{2}$

這也先坑著吧，主值積分是啥我也不清楚

39

$\displaystyle\int_0^{\infty}\dfrac{\sin x}{\cosh x-\cos x}\dfrac{x^k}{k!}\textiiseya2x$

解答：

$\begin{aligned} &注意到\\ &\dfrac{\sin x}{\cosh x-\cos x}=2\sum_{n=1}^{\infty}e^{nx}\sin nx\\ &\int_0^{\infty}x^{m-1}e^{-ax}\sin bx\textkikeimox\\ &=\dfrac{(m-1)!}{(a^2+b^2)^{\frac m2}}\sin\left(m\arctan \dfrac 蛾狗{a}\right)\\ &\int_0^{\infty}\dfrac{\sin x}{\cosh x-\cos x}\dfrac{x^k}{k!}\text6kugcmix\\ =&2\int_0^{\infty}\dfrac{x^k}{k!}\sum_{n=1}^{\infty}e^{-nx}\sin nx\textkyekgsox\\ =&2\sum_{n=1}^{\infty}\dfrac{\sin\dfrac{k+1}{4}\pi}{(2n^2)^{\frac{k+1}2}}\\ =&\sum_{n=1}^{\infty}\dfrac{\sin\dfrac{k+1}{4}\pi}{(\sqrt{2})^{k-1}n^{k+1}}\\ =&\dfrac{\zeta(k+1)\sin\dfrac{k+1}{4}\pi}{(\sqrt{2})^{k-1}} \end{aligned}$

上面那個注意到并不是那么簡單注意到的倒淫，見一下:

$\begin{aligned} &\sum_{n=1}^{\infty}e^{-nt}\sin nx=\frac12\dfrac{\sin x}{\cosh t-\cos x}\\ &1+2\sum_{n=1}^{\infty}e^{-nt}\cos nx=\dfrac{\sinh t}{\cosh t-\cos x}\\ &注意到\\ &\dfrac{p\sin x}{1-2p\cos x+p^2}=\sum_{k=1}^{\infty}p^k\sin kx\quad |p|<1\\ &令\;p=e^{-t}翔曲，其中n=1,2,3,\cdots,t>0\\ &于是有： \\ &\dfrac{e^{-t}\sin x}{1-2e^{-t}\cos x+e^{-2t}}=\sum_{k=1}^{\infty}e^{-kt}\sin kx\\ &化簡得：\\ &\sum_{n=1}^{\infty}e^{-nt}\sin nx=\frac12\dfrac{\sin x}{\cosh t-\cos x}\quad ,t>0\\ &同樣有：\\ &\dfrac{1-p\cos x}{1-2p\cos x+p^2}=\sum_{k=1}^{\infty}p^k\cos kx\\ &\dfrac{1-p^2}{1-2p\cos x+p^2}=1+2\sum_{k=1}^{\infty}p^k\cos kx\\ &同樣可得： \\ &\dfrac12\dfrac{e^t-\cos x}{\cosh t-\cos x}=\sum_{k=1}^{\infty}e^{-kt}\cos kx\qquad t>0\\ &\dfrac{\sinh t}{\cosh t-\cos x}=1+2\sum_{k=1}^{\infty}p^k\cos kx\qquad t>0 \end{aligned}$

這邊的注意到可參見我的復(fù)變函數(shù)習(xí)題：

40

$\displaystyle\int_0^{\infty}\dfrac{\textaqeeawyx}{\left[x^4+(1+2\sqrt{2})x^2+1\right]\left[x^{100}-x^{98}+\cdots+1\right]}=\dfrac{\pi}{2(1+\sqrt{2})}$

解答：

$\begin{aligned} I=&\int_0^{\infty}\dfrac{\text2geicgax}{\left[x^4+(1+2\sqrt{2})x^2+1\right]\left[x^{100}-x^{98}+\cdots+1\right]}\\ =&\int_0^{\infty}\dfrac{\frac1{x^2}\textqeq86c2x}{\left[\frac1{x^4}+(1+2\sqrt2)\frac1{x^2}+1\right][\frac1{x^{100}}-\frac1{x^{98}}+\cdots+1]}\\ =&\int_0^{\infty}\dfrac{x^{102}\text62cm868x}{\left[x^4+(1+2\sqrt{2})x^2+1\right]\left[x^{100}-x^{98}+\cdots+1\right]}\\ 2I=&\int_0^{\infty}\dfrac{1+x^{102}\textsiyo6ywx}{\left[x^4+(1+2\sqrt{2})x^2+1\right]\left[x^{100}-x^{98}+\cdots+1\right]}\\ =&\int_0^{\infty}\dfrac{(x^2+1)(x^{100}-x^{98}+\cdots+1)\text2ae482yx}{\left[x^4+(1+2\sqrt{2})x^2+1\right]\left[x^{100}-x^{98}+\cdots+1\right]}\\ =&\int_0^{\infty}\dfrac{x^2+1}{x^4+(1+2\sqrt{2})x^2+1}\textycmio6sx\\ =&\int_0^{\infty}\dfrac{1+\frac1{x^2}}{x^2+\frac1{x^2}+(1+2\sqrt{2})}\textoew2ougx\\ =&\int_0^{\infty}\dfrac{\textqwokugs(x-\frac1x)}{(x-\frac1x)^2+3+2\sqrt2}\\ =&\dfrac{1}{1+\sqrt2}\arctan\left(\dfrac{t-\frac{1}{t}}{1+\sqrt2}\right)\bigg|_0^{\infty}\\ =&\dfrac{\pi}{1+\sqrt2}\\ I=&\dfrac{\pi}{2(1+\sqrt2)} \end{aligned}$

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高等數(shù)學(xué)吧 - 積分競賽 - Season 1 - 4

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