Eufisky - The lost book

曲面积分计算

(2012年中科院考研题)设$\rho (x,y,z)$是原点$O$到椭球面$\frac{x^2}2+\frac{y^2}2+z^2=1$的上半部分(即满足$z\geq 0$的部分) $\Sigma$的任一点$(x,y,z)$处的切面的距离,求积分\[\iint_\Sigma \frac z{\rho (x,y,z)}dS.\]
 
所求积分为
 
\[I=\iint_{\Sigma}{\frac{z}{\rho \left( x,y,z \right)}dS}=\frac{1}{2}\iint_{\Sigma}{z\sqrt{x^2+y^2+z^2}dS}.\]
 
记$z=\varphi (x,y), (x,y)\in D$,其中$D$为$x^2+y^2=2$.首先有
 
\begin{align*}dS&=\sqrt{1+\left(\frac{\partial \varphi}{\partial x}\right)^2+\left(\frac{\partial \varphi}{\partial x}\right)^2}dxdy=\sqrt{1+\left(\frac{-x}{2z}\right)^2+\left(\frac{-y}{2z}\right)^2}dxdy\\&=\frac1{2z}\sqrt{x^2+y^2+4z^2}dxdy.\end{align*}
 
因此
 
\begin{align*}I&=\iint_{\Sigma}{\frac{z}{\rho \left( x,y,z \right)}dS}=\frac{1}{2}\iint_{\Sigma}{z\sqrt{x^2+y^2+z^2}dS}\\&=\frac{1}{4}\iint_D\sqrt{x^2+y^2+z^2}\sqrt{x^2+y^2+4z^2}dxdy=\frac{1}{4}\iint_D\sqrt{1+\frac{x^2}2+\frac{y^2}2}\sqrt{4-x^2-y^2}dxdy\\&=\frac14\int_0^{2\pi}d\theta\int_0^{\sqrt{2}}r\sqrt{1+\frac{r^2}2}\sqrt{4-r^2}dr=\frac\pi4\int_0^{\sqrt{2}}\sqrt{1+\frac{u}2}\sqrt{4-u}du\\&=\frac\pi{4\sqrt{2}}\int_0^{\sqrt{2}}\sqrt{8+2u-u^2}du=\frac{\sqrt{2}\pi}{16}\left(\sqrt{10-6\sqrt{2}}+9\arcsin \frac{\sqrt 2-1}3+2\sqrt{2}+9\arcsin \frac13\right).\end{align*}
这是因为
\begin{align*}&\int{\sqrt{8+2u-u^2}du}=u\sqrt{8+2u-u^2}-\int{\frac{u-u^2}{\sqrt{8+2u-u^2}}du}\\&=u\sqrt{8+2u-u^2}-\int{\frac{\left( 8+2u-u^2 \right) -8-u}{\sqrt{8+2u-u^2}}du}\\&=u\sqrt{8+2u-u^2}-\int{\sqrt{8+2u-u^2}du}+\int{\frac{8+u}{\sqrt{8+2u-u^2}}du}\\&=\left( u-1 \right) \sqrt{8+2u-u^2}-\int{\sqrt{8+2u-u^2}du}+\int{\frac{9}{\sqrt{9-\left( u-1 \right) ^2}}du}\\&=\left( u-1 \right) \sqrt{8+2u-u^2}-\int{\sqrt{8+2u-u^2}du}+9\arcsin \frac{u-1}{3}+C,\end{align*}
$$\int{\sqrt{8+2u-u^2}du}=\frac{u-1}{2}\sqrt{8+2u-u^2}+\frac{9}{2}\arcsin \frac{u-1}{3}+C.$$

 

曲线积分的计算

写出圆周的单层位势$$U(a,b)=\int_{x^2+y^2=R^2}\ln \frac 1{\sqrt{(x-a)^2+(y-b)^2}}ds,\quad \text{其中}\,  a^2+b^2\neq R^2$$


解.不妨设$R>0$,否则考察$-R$.令$x=R\cos\theta,y=R\sin\theta$,则$$ds=\sqrt{\left[x'(\theta)\right]^2+\left[y'(\theta)\right]^2}d\theta=Rd\theta.$$因此

\begin{align*}U(a,b)&=R\int_0^{2\pi}\ln \frac 1{\sqrt{(R\cos\theta-a)^2+(R\sin\theta-b)^2}}d\theta\\&=R\int_0^{2\pi}\ln \frac 1{\sqrt{R^2+a^2+b^2-2aR\cos\theta-2bR\sin\theta}}d\theta\\&=-\frac R2\int_0^{2\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\sin(\theta+\varphi)\right)d\theta,\quad \text{其中}\, \tan\varphi=\frac ab\\&=-\frac R2\int_0^{2\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\cos\theta\right)d\theta\\&=-\frac R2\int_0^{\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\cos\theta\right)d\theta-\frac R2\int_\pi^{2\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\cos\theta\right)d\theta\\&=-\frac R2\int_0^{\pi}\ln \left(R^2+a^2+b^2-2R\sqrt{a^2+b^2}\cos\theta\right)d\theta-\frac R2\int_0^\pi\ln \left(R^2+a^2+b^2+2R\sqrt{a^2+b^2}\cos\theta\right)d\theta.\end{align*}

中间几步注意到了对积分变量进行诸如$u=\theta+c$的变换改变积分上下限,而$\sin x$和$\cos x$的最小周期为$2\pi$.因此仍等于$0$到$2\pi$上的积分.

再由比较常见的Poisson积分公式有$$\int_0^\pi\ln (a\pm b\cos x)dx=\pi\ln\frac{a+\sqrt{a^2-b^2}}{2},\quad a\geq b\geq 0$$此公式证明只需把积分看成是关于$b$的函数,对$b$求导即可.

由此得$$\int_0^\pi\ln \left(a^2\pm 2ab\cos x+b^2\right)dx=\begin{cases}2\pi\ln a,&a\geq b\geq 0\\2\pi\ln b,&b\geq a\geq 0\end{cases}$$因此所求积分为

$$U(a,b)=\begin{cases}-2\pi |R|\ln |R|,& a^2+b^2<R^2\\-\pi |R|\ln \left(a^2+b^2\right),& a^2+b^2>R^2\end{cases}$$


在惯性系内一不受外力作用的刚性飞行器绕固定点转动的动态可用Euler方程描述\begin{align*}J_1\dot\omega_1&=(J_2-J_3)\omega_2\omega_3,\\J_2\dot\omega_2&=(J_3-J_1)\omega_3\omega_1,\\J_3\dot\omega_3&=(J_1-J_2)\omega_1\omega_2.\end{align*}其中$\omega_1,\omega_2,\omega_3$为刚体转动角速度的投影, $J_1,J_2,J_3$为惯性主轴的转动惯量且$J_1,J_2,J_3$均大于$0$.

(1)研究

含奇点的第二型曲面积分计算

谢惠民下册上的一道题:求

$$I=\iint_{\Sigma}{\left( x^3+\frac{1}{x} \right) dydz+\left( y^2-xz \right) dzdx+\left( z^3+\frac{z}{x^2} \right) dxdy},$$其中$\Sigma$是球面$x^2+y^2+z^2=2z$,取外侧.

解.注意到球面上的圆$x=0,y^2+z^2=2z$是上述积分的奇点,我们考察两半球$\Sigma_1:(x-\varepsilon)^2+y^2+(z-1)^2=1,x\geq\varepsilon$和$\Sigma_2:(x+\varepsilon)^2+y^2+(z-1)^2=1,x\leq -\varepsilon$, 其中$\varepsilon$为足够小的正数.并记$\Gamma_1$为圆盘$x=\varepsilon,y^2+(z-1)^2=1$,而$\Gamma_2$为圆盘$x=-\varepsilon,y^2+(z-1)^2=1$.

利用球的极坐标方程

\[x=\varepsilon+r\sin \varphi \cos \theta ,y=r\sin \varphi \sin \theta ,z=r\cos \varphi +1,\quad 0\leq \varphi\leq \pi,-\pi/2\leq\theta\leq \pi/2,0\leq r\leq 1\]

以及

\[x=-\varepsilon+r\sin \varphi \cos \theta ,y=r\sin \varphi \sin \theta ,z=r\cos \varphi +1,\quad 0\leq \varphi\leq \pi,\pi/2\leq\theta\leq 3\pi/2,0\leq r\leq 1\]

由Gauss公式可知

\begin{align*}I_{11}&=\iiint_{D_1}{\left( 3x^2+2y+3z^2 \right) dxdydz}\\&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{d\theta \int_0^{\pi}{d\varphi \int_0^1{\left[ 3\left( \varepsilon +r\sin \varphi \cos \theta \right) ^2+2r\sin \varphi \sin \theta +3\left( r\cos \varphi +1 \right) ^2 \right] r^2\sin \varphi dr}}}\\&=2\left( \varepsilon ^2+1 \right) \pi +\frac{3\pi}{2}\varepsilon +\frac{4\pi}{5}.\end{align*}
\begin{align*}I_{12}&=\iint_{\Gamma _1}{\left( x^3+\frac{1}{x} \right) dydz+\left( y^2-xz \right) dzdx+\left( z^3+\frac{z}{x^2} \right) dxdy}\\&=\iint_{\Gamma _1}{\left( \varepsilon ^3+\frac{1}{\varepsilon} \right) dydz}=\left( \varepsilon ^3+\frac{1}{\varepsilon} \right) \pi .\end{align*}
\begin{align*}I_{21}&=\iiint_{D_2}{\left( 3x^2+2y+3z^2 \right) dxdydz}\\&=\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}{d\theta \int_0^{\pi}{d\varphi \int_0^1{\left[ 3\left( -\varepsilon +r\sin \varphi \cos \theta \right) ^2+2r\sin \varphi \sin \theta +3\left( r\cos \varphi +1 \right) ^2 \right] r^2\sin \varphi dr}}}\\&=2\left( \varepsilon ^2+1 \right) \pi +\frac{3\pi}{2}\varepsilon +\frac{4\pi}{5}\end{align*}
\begin{align*}I_{22}&=\iint_{\Gamma _2}{\left( x^3+\frac{1}{x} \right) dydz+\left( y^2-xz \right) dzdx+\left( z^3+\frac{z}{x^2} \right) dxdy}\\&=\iint_{\Gamma _2}{\left( -\varepsilon ^3-\frac{1}{\varepsilon} \right) dydz}=-\left( \varepsilon ^3+\frac{1}{\varepsilon} \right) \pi.\end{align*}
因此$$I‘=I_{11}+I_{21}-I_{12}-I_{22}=4\left( \varepsilon ^2+1 \right) \pi +3\pi \varepsilon +\frac{8\pi}{5}\rightarrow \frac{28}{5}\pi,$$即$I=\frac{28}{5}\pi$.
设$J$为关于$x\left( t \right) $和$t$的连续函数,满足
$$\frac{\partial J}{\partial t}=\frac{1}{4}\left( \frac{\partial J}{\partial x} \right) ^2-x^2-\frac{1}{2}x^4,\qquad \text{其中}J\left[ x\left( 1 \right) ,1 \right] =0$$
求$J\left[ x\left( t \right) ,t \right]$.
 
关于 I will not change, no matter how U change … 
笙歌姐,这句话何解?
 
文科生:“不论你怎么移情别恋,我是不会变心的”理科生:“电流不随电压的变化而变化。”

I am here,because U are here.

$$IR\cdot \frac{\varepsilon S}{4\pi kd}\cdot \lim_{n\rightarrow \infty}\frac{\prod_{k=1}^n{k^k}}{n^{n^2/2+n/2+1/12}e^{-n^2/4}}\cdot k\ln W$$
Glaisher-Kinkelin constant

与双重对数函数有关的积分

这里

求$$\int_0^{\frac{\pi}{2}}{\frac{x^2}{1+\cos ^2x}dx}.$$


对于$|b|<a$,注意到

\begin{align*}\frac{a^2-b^2}{a^2-2ab\cos x+b^2}&=\frac{a}{a-e^{ix}b}+\frac{be^{-ix}}{a-e^{-ix}b}\\&=\sum_{n=0}^\infty \left(\frac{b}{a}\right)^ne^{inx}+\frac{be^{-ix}}{a}\sum_{n=0}^\infty\left(\frac{b}{a}\right)^ne^{-inx}\\&=1+\sum_{n=1}^\infty \left(\frac{b}{a}\right)^ne^{inx}+ \sum_{n=1}^{\infty}\left(\frac{b}{a}\right)^{n}e^{-inx}\\&=1+2\sum_{n=1}^\infty \left(\frac{b}{a}\right)^n\cos(n x),\end{align*}
因此
\begin{equation*}1+2\sum_{n=1}^\infty \left(\frac{b}{a}\right)^n\cos(n x)=\frac{a^2-b^2}{a^2+b^2-2ab\cos x},\qquad\qquad\mbox{对于}\, |b|<a.\end{equation*}

令$a=\frac{2+\sqrt{2}}{2}$和$b=\frac{-2+\sqrt{2}}{2}$,我们有

\begin{align*}1+2\sum_{n=1}^\infty \left(2\sqrt{2}-3\right)^n\cos(n x)=\frac{2\sqrt{2}}{3+\cos x}.\end{align*}
由$\displaystyle\int_0^{\pi}{x^2\cos \left( nx \right) dx}=2\pi \frac{\cos \left( \pi n \right)}{n^2}$可知
\begin{align*}I&=\int_0^{\frac{\pi}{2}}{\frac{x^2}{1+\cos ^2x}dx}=\int_0^{\frac{\pi}{2}}{\frac{x^2}{1+\frac{\cos 2x+1}{2}}dx}\\&=\int_0^{\frac{\pi}{2}}{\frac{2x^2}{3+\cos 2x}dx}=\frac{1}{4}\int_0^{\pi}{\frac{x^2}{3+\cos x}dx}\\&=\frac{1}{8\sqrt{2}}\int_0^{\pi}{x^2\left[ 1+2\sum_{n=1}^{\infty}{\left( 2\sqrt{2}-3 \right) ^n\cos \left( nx \right)} \right] dx}\\&=\frac{1}{8\sqrt{2}}\left[ \frac{\pi ^3}{3}+2\sum_{n=1}^{\infty}{\left( 2\sqrt{2}-3 \right) ^n\int_0^{\pi}{x^2\cos \left( nx \right) dx}} \right]\\&=\frac{\sqrt{2}}{48}\pi ^3+\frac{\sqrt{2}}{4}\pi \sum_{n=1}^{\infty}{\frac{\left( 2\sqrt{2}-3 \right) ^n\cos \left( \pi n \right)}{n^2}}\\&=\frac{\sqrt{2}}{48}\pi ^3+\frac{\sqrt{2}}{4}\pi \sum_{n=1}^{\infty}{\frac{\left( 3-2\sqrt{2} \right) ^n}{n^2}}\\&=\frac{\sqrt{2}}{48}\pi ^3+\frac{\sqrt{2}}{4}\pi \mathrm{Li}_2\left( 3-2\sqrt{2} \right).\end{align*}

求$$\int_0^{\pi}{\frac{x^2}{1+\sin ^2x}dx}.$$


令$t=x-\frac\pi2$,我们有

\begin{align*}J&=\int_0^{\pi}{\frac{x^2}{1+\sin ^2x}dx}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\frac{\left( t+\frac{\pi}{2} \right) ^2}{1+\cos ^2t}dt}\\&=2\int_0^{\frac{\pi}{2}}{\frac{t^2}{1+\cos ^2t}dt}+\frac{\pi ^2}{2}\int_0^{\frac{\pi}{2}}{\frac{1}{1+\cos ^2t}dt}\\&=2\int_0^{\frac{\pi}{2}}{\frac{t^2}{1+\cos ^2t}dt}+\frac{\pi ^2}{2}\int_0^{\frac{\pi}{2}}{\frac{1}{\sin ^2t+2\cos ^2t}dt}\\&=2\int_0^{\frac{\pi}{2}}{\frac{t^2}{1+\cos ^2t}dt}+\frac{\pi ^2}{2}\int_0^{\frac{\pi}{2}}{\frac{1}{\tan ^2t+2}d\left( \tan t \right)}\\&=\frac{\sqrt{2}}{24}\pi ^3+\frac{\sqrt{2}}{2}\pi \mathrm{Li}_2\left( 3-2\sqrt{2} \right) +\frac{\sqrt{2}}{8}\pi ^3\\&=\frac{\sqrt{2}}{2}\pi \mathrm{Li}_2\left( 3-2\sqrt{2} \right) +\frac{\sqrt{2}}{6}\pi ^3.\end{align*}

求$$\int_0^1{\int_0^1{\int_0^1{\int_0^1{\frac{\left( 1-x^2y^2z^2t^2 \right) dxdydzdt}{\sqrt{\left( 1-x^2 \right) \left( 1-y^2 \right) \left( 1-z^2 \right) \left( 1-t^2 \right) \left( 1+x^2y^2z^2t^2 \right)}}}}}}.$$

解.原积分等于$$I=\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\frac{1-\cos ^2\alpha \cos ^2\beta \cos ^2\theta \cos ^2\gamma}{\sqrt{1+\cos ^2\alpha \cos ^2\beta \cos ^2\theta \cos ^2\gamma}}d\alpha d\beta d\theta d\gamma}}}}.$$

由于$$\frac{1}{\sqrt{1+x}}=\sum_{n=0}^{\infty}{\binom{-1/2}{n}x^n}=1+\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-1 \right) !!}{2^nn!}x^n},$$

因此

\begin{align*}\frac{1-x^2}{\sqrt{1+x^2}}&=\left( 1-x^2 \right) \left( 1+\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-1 \right) !!}{2^nn!}x^{2n}} \right) \\&=1+\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-1 \right) !!}{2^nn!}x^{2n}}-x^2-\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-1 \right) !!}{2^nn!}x^{2n+2}}\\&=1-x^2+\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-1 \right) !!}{2^nn!}x^{2n}}-\sum_{n=2}^{\infty}{\frac{\left( -1 \right) ^{n-1}\left( 2n-3 \right) !!}{2^{n-1}\left( n-1 \right) !}x^{2n}}\\&=1-x^2-\frac{x^2}{2}+\sum_{n=2}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-3 \right) !!}{2^nn!}\left( 2n-1+2n \right) x^{2n}}\\&=1-\frac{3x^2}{2}+\sum_{n=2}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-3 \right) !!}{2^nn!}\left( 4n-1 \right) x^{2n}}.\end{align*}

由于$$\int_0^{\frac{\pi}{2}}{\cos ^{2n}xdx}=\frac{\sqrt{\pi}\Gamma \left( n+\frac{1}{2} \right)}{2\Gamma \left( n+1 \right)}=\frac{\sqrt{\pi}}{2n!}\cdot \frac{\left( 2n-1 \right) !!}{2^n}\sqrt{\pi}=\frac{\pi}{2}\frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!},$$我们有

\begin{align*}I&=\frac{\pi ^4}{16}-\frac{3}{2}\left( \frac{\pi}{4} \right) ^4+\sum_{n=2}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-3 \right) !!}{2^nn!}\left( 4n-1 \right) \left[ \frac{\pi}{2}\frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!} \right] ^4}\\&=\frac{\pi ^4}{16}-\frac{3}{2}\left( \frac{\pi}{4} \right) ^4+\frac{\pi ^4}{16}\sum_{n=2}^{\infty}{\left( -1 \right) ^n\frac{4n-1}{2n-1}\left[ \frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!} \right] ^5}\\&=\frac{\pi ^4}{16}+\frac{\pi ^4}{16}\sum_{n=1}^{\infty}{\left( -1 \right) ^n\frac{4n-1}{2n-1}\left[ \frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!} \right] ^5}.\end{align*}
\begin{align*}\sum_{n=1}^{\infty}{\left( -1 \right) ^n\frac{4n-1}{2n-1}\left[ \frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!} \right] ^5}=& _5F_4\left( -\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1,1;-1 \right) \\&-\frac{1}{8}\,_5F_4\left( \frac{1}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};2,2,2,2;-1 \right) -1\end{align*}
可知
$$I=\frac{\pi ^4}{16}\left[ _5F_4\left( -\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1,1;-1 \right) -\frac{1}{8}\,_5F_4\left( \frac{1}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};2,2,2,2;-1 \right) \right] ,$$
其中$_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;z)$为Generalized Hypergeometric Function.

计算积分
$$\int_0^{\pi}{\sqrt{\tan \frac{\theta}{2}}\ln^2 \left( \sin \theta \right) \text{d}\theta}.$$
\large{\textbf{\textcolor{blue}{解}}}

\begin{align*}\int_0^{\pi}{\sqrt{\tan \frac{\theta}{2}}\ln^2 \left( \sin \theta \right) \text{d}\theta}&=\int_0^{\infty}{\frac{2\sqrt{t}}{1+t^2}\ln^2 \left( \frac{2t}{1+t^2} \right) \text{d}t}\hspace{0.5cm}t=\tan \frac{\theta}{2}\\&=\int_0^{\infty}{\frac{2\sqrt{1/t}}{1+t^2}\ln^2 \left( \frac{2}{t+1/t} \right) \text{d}t}\\&=\int_0^{\infty}{\frac{\sqrt{1/t}+\sqrt{1/t^3}}{t+1/t}\ln^2 \left( \frac{2}{t+1/t} \right) \text{d}t}\\&=\int_{-\infty}^{\infty}{\frac{2}{x^2+2}\ln^2 \left( \frac{2}{x^2+2} \right) \text{d}x}\\&=2\sqrt{2}\int_0^{\frac{\pi}{2}}{\ln^2 \left( \cos ^2u \right) \text{d}u}~~x=\sqrt2\tan u\\&=8\sqrt{2}\int_0^{\frac{\pi}{2}}{\ln ^2\sin u\text{d}u}=8\sqrt{2}\int_0^{\frac{\pi}{2}}{\left( -\ln 2-\sum_{k=1}^{\infty}{\frac{\cos \left( 2kx \right)}{k}} \right)^2 \text{d}u}\\&=8\sqrt{2}\left( \int_0^{\frac{\pi}{2}}{\ln ^22\text{d}u}+\sum_{n=1}^{\infty}{\frac{1}{k}\int_0^{\frac{\pi}{2}}{\frac{1+\cos 4kx}{2}\text{d}x}} \right)\\&=4\sqrt{2}\pi \ln ^22+2\sqrt{2}\pi \zeta \left( 2 \right) =\frac{\sqrt{2}}{3}\pi ^3+4\sqrt{2}\ln^2 2.\end{align*}


$f$是$[0,1]$上严格单增的凸实值连续函数,满足$f(0)=0,f(1)=1$,且$g(x)$满足$g(f(x))=x$对任意$x\in[0,1]$成立,证明

$$\int_0^1{f\left( x \right) g\left( x \right) \text{d}x}\le \frac{1}{3}.$$

\large{\textbf{\textcolor{blue}{证明}}} 首先我们有凸函数的等价定义:

定理.函数$f$在区间$I$上是凸函数,当且仅当对任何$(x_1,x_2)\subset I$及任何$x\in(x_1,x_2)$,有
$$\frac{f\left( x \right) -f\left( x_1 \right)}{x-x_1}\le \frac{f\left( x_2 \right) -f\left( x_1 \right)}{x_2-x_1}\le \frac{f\left( x_2 \right) -f\left( x \right)}{x_2-x}.$$
注意到$g(x)=f^{-1}(x)$是$f(x)$的反函数,只需要证明$f(x)f^{-1}(x)\leqslant x^2$即可,即
$$\frac{f\left( x \right)}{x}\leqslant \frac{x}{f^{-1}\left( x \right)}.$$
由题意知$f(x)\leqslant x\leqslant f^{-1}(x)$对$x\in[0,1]$都成立,结合$f(0)=0,f(1)=1,f$是凸函数可知对任意$x\in[0,1]$,$\exists t\in[x,1]$,s.t.$f(t)=x$于是由上述凸函数等价定义可知
$$\frac{f\left( x \right)}{x}\le \frac{f\left( t \right)}{t}=\frac{x}{f^{-1}\left( x \right)}.$$
于是$\displaystyle{f\left( x \right) g\left( x \right) =f\left( x \right) f^{-1}\left( x \right) \le x^2}$,
$$\int_0^1{f\left( x \right) g\left( x \right) \text{d}x}\leqslant \int_0^1{x^2\text{d}x}=\frac{1}{3}.$$
等号成立当且仅当$f(x)=x$.

求最大的常数$b>0$,使得对任意$a>0$和一切$(1,+\infty)$上连续可导且单增的实值函数$f(x)$满足$f(x)\leqslant x^{2a}\ln^bx,x\in(1,+\infty)$就有积分$\displaystyle{\int_1^{\infty}\frac{x^{2a-2}}{f'(x)}\mathrm{d}x=+\infty}$.


\large{\textbf{\textcolor{blue}{解}}} 首先如果$b>1$,我们取$f(x)=x^{2a}\ln^{b}x$,则求导后很容易得到积分$$\displaystyle{\int_1^{\infty}\frac{x^{2a-2}}{f'(x)}\mathrm{d}x<+\infty}$$
因此$b\leqslant1$,下面验证$b=1$满足条件.
如果$f'(x)$有界,结论显然成立,不妨设$f'(x)$无界,这时$f(x)$单调趋于$+\infty$.对$\forall A>0$,由Cauchy不等式得
$$\left( \int_{\text{e}^{\frac{A}{2}}}^{\text{e}^A}{\frac{x^{2a-2}}{f'\left( x \right)}\text{d}x} \right) \left( \int_{\text{e}^{\frac{A}{2}}}^{\text{e}^A}{\frac{f'\left( x \right)}{x^{2a}\ln ^2x}\text{d}x} \right) \geqslant \left( \int_{\text{e}^{\frac{A}{2}}}^{\text{e}^A}{\frac{\text{d}x}{x\ln x}} \right) ^2=\ln ^22.$$
由$f(x)\leqslant x^{2a}\ln x$得$f(\mathrm{e}^x)\leqslant x\mathrm{e}^{2ax}$,因此
\begin{align*}\int_{\textrm{e}^{\frac{A}{2}}}^{\textrm{e}^A}{\frac{f'\left(x\right)}{x^{2a}\ln^2x}\textrm{d}x}&=\int_{\frac{A}{2}}^A{\frac{f'\left(\textrm{e}^t\right)\textrm{e}^{2at}}{t^2\textrm{e}^{2at}}\textrm{d}t}=\int_{\frac{A}{2}}^A{\frac{\textrm{d}\left[f\left(\textrm{e}^{2t}\right)\right]}{t^2\textrm{e}^{2at}}}\\&=\left.\frac{f\left(\textrm{e}^t\right)}{t^2\textrm{e}^{2at}}\right|_{\frac{A}{2}}^{A}+\int_{\frac{A}{2}}^A{\frac{2t^2\textrm{e}^{-2at}+2t\textrm{e}^{-2at}}{t^4}f\left(\textrm{e}^t\right)\textrm{d}t}\\&\leqslant\frac{f\left(\textrm{e}^A\right)}{A^2\textrm{e}^{2A}}+\int_{\frac{A}{2}}^A{\frac{2t^2\textrm{e}^{-2at}+2t\textrm{e}^{-2at}}{t^4}t\textrm{e}^{2at}\textrm{d}t}\\&\leqslant\frac{1}{A}+2\left(\ln 2+\frac{1}{A}\right)=2\ln 2+\frac{3}{A}.\end{align*}
取$A$充分大,则$\displaystyle{\int_{\textrm{e}^{\frac{A}{2}}}^{\textrm{e}^A}{\frac{f'\left(x\right)}{x^{2a}\ln^2x}\textrm{d}x}\leqslant2}$, 因此
$$\int_{\textrm{e}^{A/2}}^{\textrm{e}^A}{\frac{\textrm{d}x}{f'\left(x\right)}}\geqslant\frac{\ln^22}{2}$$
对任意充分大的$A$都成立,于是积分$\displaystyle{\int_1^{+\infty}\frac{1}{f'(x)}\mathrm{d}x}=+\infty$,因此最大的$b=1$.

 \textbf{证明:}不等式的左边可利用$Cauchy-Schwarz$不等式证之,下证不等式的右边成立.由题意可知,
\[\begin{split}& (1-\frac{f}{M})(1-\frac{m}{f}) \geq 0 \Longrightarrow  1+\frac{m}{M} \geq \frac{f}{M}+\frac{m}{f}\\\Longrightarrow & 1+\frac{m}{M} \geq \frac1{M}\int^{1}_{0}{f}dx +m\int^{1}_{0}{\frac1{f}}dx \geq 2\sqrt{\frac{m}{M}\int^{1}_{0} \frac1{f}dx\int^{1}_{0}{f}dx}\\\Longrightarrow & \int^{1}_{0} \frac1{f}dx\int^{1}_{0}{f}dx\leq \frac{(m+M)^2}{4mM}.\end{split}\]

二重积分计算

求$$\iint_D \sin x\sin y(\sin x+\sin y)e^{\sin x\sin y}d\sigma,$$其中$D:0<x<\pi/2,0<y<\pi/2$.

解.首先把待求积分写成

\begin{align*}&\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin x\sin y\left( \sin x+\sin y \right) \text{e}^{\sin x\sin y}\text{d}x\text{d}y}}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ^2x\sin y\text{e}^{\sin x\sin y}\text{d}x\text{d}y}}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ^2x\cos y\text{e}^{\sin x\cos y}\text{d}x\text{d}y}}\\=&2\iint_S{x\text{e}^x\text{d}x\text{d}y},\end{align*}
其曲面$S$是单位球面在第一象限的部分.
考虑到平面$x=0$处距离为$x$的宽度为$\mathrm{d}x$的球面窄条的面积,相当于是底边长为$y=\sqrt{1-x^2}$,宽为$\mathrm{d}s=\sqrt{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\mathrm{d}x=\frac{\mathrm{d}x}{\sqrt{1-x^2}}$,
因此
$$I=2\int_0^1{x\text{e}^x\frac{\pi}{2}\sqrt{1-x^2}\frac{\text{d}x}{\sqrt{1-x^2}}}=\pi \int_0^1{x\text{e}^x\text{d}x}=\pi.$$
解法二.利用分部积分及对称性可知
\begin{align*}&\iint_D{\sin x\sin y\left( \sin x+\sin y \right) e^{\sin x\sin y}d\sigma}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ^2x\sin ye^{\sin x\sin y}dxdy}}=2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\left( 1-\cos ^2x \right) \sin ye^{\sin x\sin y}dxdy}}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ye^{\sin x\sin y}dxdy}}-2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\cos x\frac{\partial e^{\sin x\sin y}}{\partial x}dxdy}}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ye^{\sin x\sin y}dxdy}}-2\int_0^{\frac{\pi}{2}}{\left[ \cos x\left. e^{\sin x\sin y} \right|_{0}^{\pi /2}-\left( \int_0^{\frac{\pi}{2}}{\left( -\sin x \right) e^{\sin x\sin y}dx} \right) \right] dy}\\=&2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin ye^{\sin x\sin y}dxdy}}-2\int_0^{\frac{\pi}{2}}{\left( -1 \right) dy}-2\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\sin xe^{\sin x\sin y}dxdy}}\\=&\pi .\end{align*}

 

Legendre 加倍公式的一个证明

Legendre 加倍公式的一个证明
\[\sqrt{\pi}\Gamma (2s)=2^{2s-1}\Gamma (s)\Gamma \left(s+\frac{1}{2}\right),s>0,\]
其中$\Gamma$是Gamma函数,
\[\Gamma (s)=\int_0^{+\infty}x^{s-1}e^{-x}\mathrm{d}x, s>0.\]
证.
\[I(s)=\int_0^1\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}}.\]
令$x=\tan^{2s} t$, 则$\mathrm{d}x=2s\tan^{2s-1}t\sec^2t\mathrm{d}t=\sin^{2s-1}t\cos^{-2s-1}t\mathrm{d}t$, $(1+x^{\frac{1}{s}})^{2s}=\sec^{4s}t$, 从而
\[\begin{array}{rl}I(s)&=\int_0^1\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}}\\&=2s\int_0^{\frac{\pi}{4}} (\sin t\cos t)^{2s-1}\mathrm{d}t\\&=s2^{1-2s} \int_0^{\frac{\pi}{2}}\sin^{2s-1}u\mathrm{d}u\\&=2^{-2s}sB(\frac{1}{2},s)\\&=2^{-2s}s\frac{\Gamma (\frac{1}{2})\Gamma (s)}{\Gamma(\frac{1}{2}+s)}\\&= 2^{-2s}\sqrt{\pi}s\frac{\Gamma (s)}{\Gamma(\frac{1}{2}+s)}.\end{array}\]
另一方面
\[I(s)=\int_0^1\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}}=\int_1^{+\infty}\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}},\]
从而
\[I(s)=\frac{1}{2}\int_0^{+\infty}\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}}=s\int_0^{\frac{\pi}{2}}(\sin t\cos t)^{2s-1}\mathrm{d}t=\frac{sB(s,s)}{2}=\frac{s\Gamma^2(s)}{2\Gamma (2s)}.\]  
因此
\[2^{-2s}\sqrt{\pi}s\frac{\Gamma (s)}{\Gamma(\frac{1}{2}+s)}=\frac{s\Gamma^2(s)}{2\Gamma (2s)}.\] 从而
\[\sqrt{\pi}\Gamma (2s)=2^{2s-1}\Gamma (s)\Gamma \left(s+\frac{1}{2}\right), s>0.\]
 
来自:http://www.math.org.cn/forum.php?mod=viewthread&tid=32538

两道积分习题

1.设$f:[0,1]\to\mathbb{R}$连续,求极限$$\lim\limits_{n\rightarrow \infty}\int_0^1\int_0^1\cdots\int_0^1 f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)dx_1dx_2\cdots dx_n.$$

解法一.设$|f|$最大值为$M$.对任何$\varepsilon>0$,存在$\delta>0$,使得当$|x-1/2|<\delta$时,有$$\left|f(x)-f(\frac{1}{2})\right|<\varepsilon.$$

\begin{align*}&\int_{[0,1]^n}\left| f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)-f(\frac{1}{2})\right|dx_1dx_2\cdots dx_n\\\leq &\int_{\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|\geq\delta}\left| f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)-f(\frac{1}{2})\right|dx_1dx_2\cdots dx_n\\+&\int_{\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|<\delta}\left| f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)-f(\frac{1}{2})\right|dx_1dx_2\cdots dx_n\\\leq&2M\int_{\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|\geq\delta}dx_1dx_2\cdots dx_n+\varepsilon\\\leq&\frac{2M}{\delta^2}\int_{\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|\geq\delta}\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|^2dx_1dx_2\cdots dx_n+\varepsilon\\\leq&\frac{2M}{\delta^2}\int_{[0,1]^n}\left|\frac{x_1+x_2+\cdots+x_n}{n}-\frac{1}{2}\right|^2dx_1dx_2\cdots dx_n+\varepsilon\\=&\frac{M}{6n\delta^2}+\varepsilon.\end{align*}

因此$$\limsup_{n\rightarrow\infty}\int_{[0,1]^n}\left| f\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)-f(\frac{1}{2})\right|dx_1dx_2\cdots dx_n\leq \varepsilon.$$

令$\varepsilon\rightarrow0$即可.

 

解法二.由科尔莫格罗夫强大数定律得$$\frac{{{X_1} + {X_2} + \cdots + {X_n}}}{n}\mathop \to \limits^{a.s.} E\left( {{X_i}} \right) = \frac{1}{2}\left( {n \to + \infty } \right).$$

又因为$f(x)$连续有界,由控制收敛定理可知

$$\mathop {\lim }\limits_{n \to \infty } E\left( {f\left( {\frac{{{X_1} + {X_2} + \cdots + {X_n}}}{n}} \right)} \right) = E\left( {\mathop {\lim }\limits_{n \to \infty } f\left( {\frac{{{X_1} + {X_2} + \cdots + {X_n}}}{n}} \right)} \right) = E\left( {f\left( {\mathop {\lim }\limits_{n \to \infty } \frac{{{X_1} + {X_2} + \cdots + {X_n}}}{n}} \right)} \right) = f\left( {\frac{1}{2}} \right).$$


2.求证$$\int_0^1\prod_{n=1}^\infty(1-x^n)dx=\frac{4\pi\sqrt3}{\sqrt{23}}\frac{\sinh\frac{\pi\sqrt{23}}3}{\cosh\frac{\pi\sqrt{23}}2}.$$

解.注意到Pentagonal number theorem,我们知$$\int_{0}^{1}\prod_{n\geq1}\left(1-x^{n}\right)dx=\sum_{k\in\mathbb{Z}}\left(-1\right)^{k}\int_{0}^{1}x^{k\left(3k-1\right)/2}dx=\sum_{k\in\mathbb{Z}}\left(-1\right)^{k}\frac{2}{3k^{2}-k+2}.$$再利用求和公式可知$$\sum_{n\in\mathbb{Z}}\left(-1\right)^{n}f\left(n\right)=-\sum\left\{ \pi\csc\left(\pi z\right)f(z) \textrm{ 在 } f\left(z\right)\textrm{ 的极点上的留数}\right\}.$$而极点为$z=\frac{1}{6}\left(1\pm i\sqrt{23}\right)$,由此求得.

Ahmed’s integrals 和 Coxeter’s integrals

最后得到Ahmed’s integrals的一个简单表达式.
 
1. 主要结果
 
一般地,我们将含参数$p,q,r$的Ahmed’s integral定义成
$$A(p, q, r) := \int_{0}^{1} \frac{\arctan q \sqrt{p^{2}x^{2} + 1} }{q \sqrt{p^{2}x^{2} + 1}} \frac{p q r}{(r^{2}+1)p^{2}x^{2} + 1} \, dx.$$
则下面的定理成立:
 
定理. 对任意$p, q, r > 0$,记$\rho, \alpha, \beta, \gamma$为
\begin{align*} \rho &= \frac{1-pqr}{1+pqr}, & \alpha &= 2\arctan\left( \frac{qr}{\sqrt{r^{2}+1}} \right), \\ \beta &= 2\arctan\left( \frac{rp}{\sqrt{p^{2}+1}} \right), & \gamma &= 2\arctan\left( r\sqrt{q^{2}+1} \right). \end{align*}
我们有
\begin{align*} A(p, q, r) &= \frac{1}{8} \left( \alpha(2\pi-\alpha) + \beta(2\pi-\beta) – \gamma(2\pi-\gamma) \right) \\ &\qquad – \frac{1}{2}\Re \left( \operatorname{Li}_{2}(\rho) – \operatorname{Li}_{2}(\rho e^{i\alpha}) – \operatorname{Li}_{2}(\rho e^{i\beta}) + \operatorname{Li}_{2}(\rho e^{i\gamma}) \right). \end{align*}
基于此定理,只要不产生混乱,我们都可写成
$$A(p, q, r) = A(\rho \mid \alpha, \beta, \gamma)$$
在此,当然, $\rho, \alpha, \beta, \gamma$是出现在上述定理中的参数.
 
2. 推论
 
虽然一般的结果包含在dilogarithmic的一支,但他们或者$\rho = 0$或者当$\rho= -1$时,碰巧可以简化.
 
推论 1.将$\rho, \alpha, \beta, \gamma$记为主定理那样的数.如果$pqr = 1$, 则
$$A(p, q, r) = \frac{1}{8} \left( \alpha(2\pi-\alpha) + \beta(2\pi-\beta) – \gamma(2\pi-\gamma) \right).$$
在定理中令$\rho = 0$便可得到.
 
比如, 传统上的Ahmed’s integral可以这样计算:
\begin{align*} \int_{0}^{1} \frac{\arctan\sqrt{x^{2} + 2}}{\sqrt{x^{2} + 2}} \, \frac{dx}{x^{2}+1} &= A\left( \frac{1}{\sqrt{2}}, \sqrt{2}, 1 \right) \\ &= A\left( 0 \, \middle| \, \frac{\pi}{2}, \frac{\pi}{3}, \frac{2\pi}{3} \right) \\ &= \frac{5\pi^{2}}{96}. \end{align*}
 
推论 2. 将$\rho, \alpha, \beta, \gamma$记为主定理那样的数. 如果$pqr \to \infty$, 则
$$A(p, q, r) = \frac{\pi}{4} \left( \alpha + \beta – \gamma \right).$$
证.当$pqr \to \infty$, 我们有$\rho \to -1$.由等式
$$\Re\operatorname{Li}_{2}(-e^{i\theta}) = \operatorname{Li}_{2}(-1) + \frac{\theta^{2}}{4} \quad |\theta| \leq \pi,$$
推出
\begin{align*} A(-1 \mid \alpha, \beta, \gamma) &= \frac{\pi}{4} \left( \alpha + \beta – \gamma \right). \end{align*}
 
3. 在Coxeter’s integrals上的应用
 
命题. 假设$ a \geq |b|$并且对$\theta \in (0, \beta)$,有$a \cos\theta + b > 0$. 则我们有
\begin{align*} &\int_{0}^{\beta} \arctan \sqrt{\frac{\cos\theta + 1}{a\cos\theta + b}} \, d\theta \\ &= 2 A \left( \sqrt{\frac{a-b}{2} \cdot \frac{1-\cos\beta}{a\cos\beta+ b}}, \sqrt{\frac{2}{a+b}}, \sqrt{\frac{a+b}{a-b}} \right) \\ &= 2 A \left( \frac{1-k}{1+k} \, \middle| \, \alpha, \beta, \gamma\right) \end{align*}
其中 $k, \alpha, \gamma$是通过
$$k = \left( \frac{1-\cos\beta}{a\cos\beta+ b} \right)^{1/2}, \quad \alpha = \arccos\left(\frac{a-1}{a+1}\right), \quad \gamma = \arccos\left(-\frac{1+b}{1+a}\right).$$定义的.
比如,一个经典的Coxeter’s integrals可以这样计算:
\begin{align*} \int_{0}^{\frac{\pi}{2}} \arccos \left( \frac{\cos \theta}{1+2\cos \theta} \right) \; d\theta &= 2 \int_{0}^{\frac{\pi}{2}} \arctan \sqrt{\frac{\cos \theta + 1}{3\cos \theta + 1}} \, d\theta \\ &= 4 A \left( 0 \, \middle| \, \frac{\pi}{3}, \frac{\pi}{2}, \frac{2\pi}{3} \right) \\ &= \frac{5\pi^{2}}{24}. \end{align*}
 
许多其它的Coxeter’s integrals可以直接由此命题获得, 结合推论1或推论2. 一些更复杂的情形可能需要polylogarithmic作为阶梯. 例如,
\begin{align*} \int_{0}^{\frac{\pi}{3}} \arccos \left( \frac{\cos \theta}{1+2\cos \theta} \right) \; d\theta &= 2 \int_{0}^{\frac{\pi}{3}} \arctan \sqrt{\frac{\cos \theta + 1}{3\cos \theta + 1}} \, d\theta \\ &= 4 A \left( \frac{\sqrt{5}-1}{\sqrt{5}+1} \, \middle| \, \frac{\pi}{3}, \frac{\pi}{3}, \frac{2\pi}{3} \right). \end{align*}
为了简化此表达式需要以下的polylogarithmic阶梯工具,
$$\operatorname{Li}_{2}(\rho) – \operatorname{Li}_{2}(\rho^{2}) – \operatorname{Li}_{2}(\rho^{3}) + \frac{1}{3}\operatorname{Li}_{2}(\rho^{6}), \quad \rho = \frac{\sqrt{5}-1}{\sqrt{5}+1}.$$
由此可知表达式得到的值为 $\pi^{2}/45$, 因此我们最后得到
$$\int_{0}^{\frac{\pi}{3}} \arccos \left( \frac{\cos \theta}{1+2\cos \theta} \right) \; d\theta = \frac{2\pi^{2}}{15}.$$
 
译自:http://www.sos440.net/?p=169
主页:Sangchul Lee http://www.math.ucla.edu/~sos440/m/index.html

下面给出自己的一些结果:

\begin{align*}\int_0^{\frac{\pi }{3}} {\arccos \frac{{1 - \cos x}}{{2\cos x}}dx}  &= 2\int_0^{\frac{\pi }{3}} {\arctan \sqrt {\frac{{ - 1 + 3\cos x}}{{1 + \cos x}}} dx}  = 2\int_0^{\frac{\pi }{3}} {\left( {\frac{\pi }{2} - \arctan \sqrt {\frac{{\cos x + 1}}{{3\cos x - 1}}} } \right)dx} \\&= \frac{{{\pi ^2}}}{3} - 2\int_0^{\frac{\pi }{3}} {\arctan \sqrt {\frac{{\cos x + 1}}{{3\cos x - 1}}} dx}  = \frac{{{\pi ^2}}}{3} - 4A\left( {1\left| {\frac{\pi }{3},\frac{\pi }{3},\frac{\pi }{2}} \right.} \right).\end{align*}
其中用到了\[\cos x = \frac{{{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}}}{{{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2}}} = \frac{{1 - {{\tan }^2}\frac{x}{2}}}{{1 + {{\tan }^2}\frac{x}{2}}},\arctan x + \arctan \frac{1}{x} = \frac{\pi }{2}.\]
\begin{align*}A\left( {0\left| {\frac{\pi }{3},\frac{\pi }{3},\frac{\pi }{2}} \right.} \right) = &\frac{1}{8}\left[ {\frac{\pi }{3}\left( {2\pi  - \frac{\pi }{3}} \right) + \frac{\pi }{3}\left( {2\pi  - \frac{\pi }{3}} \right) - \frac{\pi }{2}\left( {2\pi  - \frac{\pi }{2}} \right)} \right]\\=& \frac{{13{\pi ^2}}}{{288}} .\end{align*}
因此\[\int_0^{\frac{\pi }{3}} {\arccos \frac{{1 - \cos x}}{{2\cos x}}dx}  = \frac{{{11\pi ^2}}}{72}.\]

\[\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx}  = \int_0^{\frac{\pi }{2}} {\arctan \sqrt {\frac{{\cos x + 1}}{{\cos x}}} dx}  = 2A\left( { - 1\left| {\frac{\pi }{2},\frac{\pi }{2},\frac{{2\pi }}{3}} \right.} \right)\]

\begin{align*}A\left( { - 1\left| {\frac{\pi }{2},\frac{\pi }{2},\frac{{2\pi }}{3}} \right.} \right) = &\frac{1}{8}\left[ {\frac{\pi }{2}\left( {2\pi  - \frac{\pi }{2}} \right) + \frac{\pi }{2}\left( {2\pi  - \frac{\pi }{2}} \right) - \frac{{2\pi }}{3}\left( {2\pi  - \frac{{2\pi }}{3}} \right)} \right]\\&- \frac{1}{2}{\mathop{\rm Re}\nolimits} \left[ { \operatorname{Li}_2\left( { - 1} \right) - \operatorname{Li}_2\left( { - {e^{i\frac{\pi }{2}}}} \right) - \operatorname{Li}_2\left( { - {e^{i\frac{\pi }{2}}}} \right) + \operatorname{Li}_2\left( { - {e^{i\frac{{2\pi }}{3}}}} \right)} \right]\\= &\frac{{11{\pi ^2}}}{{144}} - \frac{1}{2}\left[ { - \frac{{{\pi ^2}}}{{12}} + \frac{{{\pi ^2}}}{{48}} + \frac{{{\pi ^2}}}{{48}} + \frac{{{\pi ^2}}}{{36}}} \right] = \frac{{{\pi ^2}}}{{12}}.\end{align*}
这里用到了$\operatorname{Li}_2\left( i \right) =  - \frac{{{\pi ^2}}}{{48}} + iC$, 其中$C$是Catalan constant.利用\[\operatorname{Li}_s\left( z \right) = \sum\limits_{k = 1}^\infty  {\frac{{{z^k}}}{{{k^s}}}}  = z + \frac{{{z^2}}}{{{2^s}}} + \frac{{{z^3}}}{{{3^s}}} +  \cdots \]和\[\operatorname{Li}_2\left( x \right) + \operatorname{Li}_2\left( {1 - x} \right) = \frac{1}{6}{\pi ^2} - \ln x\ln \left( {1 - x} \right),\]我们有\[{\mathop{\rm Re}\nolimits} \left\{ { \operatorname{Li}_2\left( z \right)} \right\} = \frac{1}{2}\left\{ { \operatorname{Li}_2\left( z \right) + \operatorname{Li}_2\left( {\bar z} \right)} \right\}\]以及
\begin{align*}&{\mathop{\rm Re}\nolimits} \left\{ {\operatorname{Li}_2\left( {{e^{i\frac{\pi }{3}}}} \right)} \right\} = {\mathop{\rm Re}\nolimits} \left\{ {\operatorname{Li}_2\left( {{e^{ - i\frac{{2\pi }}{3}}}} \right)} \right\} = \frac{1}{2}\left\{ { \operatorname{Li}_2\left( {{e^{i\frac{\pi }{3}}}} \right) + \operatorname{Li}_2\left( {1 - {e^{i\frac{\pi }{3}}}} \right)} \right\}\\= &\frac{1}{2}\left\{ {\frac{1}{6}{\pi ^2} - \ln {e^{i\frac{\pi }{3}}}\ln {e^{ - i\frac{\pi }{3}}}} \right\} = \frac{1}{2}\left\{ {\frac{1}{6}{\pi ^2} - \left( {i\frac{\pi }{3}} \right)\left( { - i\frac{\pi }{3}} \right)} \right\} = \frac{{{\pi ^2}}}{{36}}.\end{align*}
因此

\[\int_0^{\frac{\pi }{2}} {\arccos \sqrt {\frac{{\cos x}}{{1 + 2\cos x}}} dx}  = \frac{{{\pi ^2}}}{6}.\]

又一道二元反常积分题

网友WB问我一道积分题\[\int_0^\infty  {\int_0^\infty  {{e^{ - a\sqrt {{x^2} + {y^2}} }}\cos \alpha x\cos \beta ydxdy} } .\]


经极坐标代换后得到

\[\int_0^\infty  {dr} \int_0^{\frac{\pi }{2}} {r{e^{ - ar}}\cos \left( {\alpha r\cos \theta } \right)\cos \left( {\beta r\sin \theta } \right)d\theta } .\]

然后利用积化和差公式感觉可以继续往下算.

又两个积分

计算$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\ln{(\ln{\tan{x}})}dx}.$$


Let $u = \ln \tan x$, so that $\frac{\pi}{4} < x< \frac{\pi}{2}$ is mapped to $u>0$, $x=\arctan(\exp(u))$ and $\mathrm{d}x = \frac{\mathrm{d}u}{2 \cosh(u)}$. Then
$$\int_{\pi/4}^{\pi/2} \ln( \ln(\tan x))\, \mathrm{d}x = \frac{1}{2} \int_0^\infty \frac{\ln (u)}{\cosh(u)} \mathrm{d}u = \frac{1}{2} \lim_{s \to 0^+}\frac{\mathrm{d}}{\mathrm{d} s} \int_0^\infty \frac{u^s}{\cosh(u)} \mathrm{d}u$$
The latter parametric integral is evaluated by expanding $\cosh(u)$ into exponential and using Euler's gamma-integral:
\begin{align*}\int_0^\infty \frac{u^s}{2\cosh(u)} \mathrm{d}u &= \int_0^\infty u^s \frac{\exp(-u)}{1+\exp(-2u)}\mathrm{d}u = \sum_{n=0}^\infty (-1)^n \int_0^\infty u^{s} \exp(-(2n+1)u) \,\mathrm{d}u \\&= \Gamma(s+1)  \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^{s+1}} = \Gamma(s+1) 2^{-2s-2} \left( \zeta\left(s+1,\frac{1}{4}\right) - \zeta\left(s+1,\frac{3}{4}\right) \right)\end{align*}
Near $s=0$:
$$\zeta(1+s,a) = \frac{1}{s} - \psi(a) - \gamma_1(a) s + \mathcal{o}(s)$$
where $\psi(a)$ is the digamma function (http://en.wikipedia.org/wiki/Digamma_function), and $\gamma_1(a)$ is the first generalized Stieltjes constant (http://en.wikipedia.org/wiki/Stieltjes_constants). Differentiating and taking the limit we have
\begin{align*}\int_{\pi/4}^{\pi/2} \ln( \ln(\tan x))\, \mathrm{d}x &= \frac{1}{4} \left( \left(\log(4) + \gamma\right)\left(\psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4}\right)\right) - \left(\psi_1\left(\frac{1}{4}\right)-\psi_1\left(\frac{3}{4}\right)\right)\right) \\&= \frac{\pi}{4} \log \left( \frac{4 \pi^3}{\Gamma\left(\frac{1}{4}\right)^4} \right) \approx -0.260443\end{align*}
where the latter equality is given my Mathematica.

求\[\int_{0}^{\infty}\sin x \sin \sqrt{x}\,dx.\]


这个积分用Alpha给的是发散结果.下面的计算是论坛的结果:

I'm posting an asnwer (of the $2$ I have) using real analysis methods:

 

\begin{align*}\int_{0}^{\infty}\sin x \sin \sqrt{x}\,dx &\overset{\sqrt{x}=u}{=\! =\! =\!}2\int_{0}^{\infty}u\sin u \sin u^2 \,du \\&=-\int_{0}^{\infty}u\cos \left ( u^2+u \right )\,du+\int_{0}^{\infty}u\cos(u^2-u)\,du \\&\overset{u \mapsto u+1}{=\! =\! =\! =\!}-\int_{0}^{\infty}u\cos(u^2+u)\,du+\int_{-1}^{\infty}\left ( u+1 \right )\cos\left ( u^2+u \right )\,du \\&= \int_{0}^{\infty}\cos\left ( u^2+u \right )\,du+\int_{-1}^{0}\left ( u+1 \right )\cos\left ( u^2+u \right )\,du\\&\overset{u={\rm v}-\frac{1}{2}}{=\! =\! =\! =\!}\int_{1/2}^{\infty}\cos\left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v}+\int_{-1/2}^{1/2}\left ( {\rm v}+\frac{1}{2} \right )\cos \left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v} \\&= \int_{0}^{\infty}\cos \left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v}+\\& \left [ \int_{-1/2}^{1/2}{\rm v}\cos \left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v}+\frac{1}{2}\int_{-1/2}^{0}\cos\left ( {\rm v}^2-\frac{1}{2} \right )\,d{\rm v}- \frac{1}{2}\int_{0}^{1/2}\cos\left ( {\rm v}^2-\frac{1}{2} \right )\,d{\rm v} \right ]\end{align*}

 

However, the equation in the bracket equals zero due to symmetry.

 

Hence:

\begin{align*}\int_{0}^{\infty}\sin x \sin x^2\,dx&=\int_{0}^{\infty}\cos\left ( {\rm v}^2-\frac{1}{2} \right )\,d{\rm v}\\&=\cos \frac{1}{4}\int_{0}^{\infty}\cos {\rm v}^2\,d{\rm v}+\sin \frac{1}{4}\int_{0}^{\infty}\sin {\rm v}^2\,d{\rm v}\\&\overset{(*)}{=}\int_{0}^{\infty}\sin {\rm v}^2\,d{\rm v}\left ( \cos \frac{1}{4}+\sin \frac{1}{4} \right )\\&=\frac{\sqrt{\pi}}{2}\sin \left ( \frac{3\pi-1}{4} \right )\;\; \;\;\;\; \square\end{align*}

 

$(*)$ We used the Frensel integrals stating that $\displaystyle \int_{0}^{\infty}\cos x^2 \,dx=\int_{0}^{\infty}\sin x^2 \,dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$.