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$\iint_\Sigma \frac z{\rho (x,y,z)}dS.$

$I=\iint_{\Sigma}{\frac{z}{\rho \left( x,y,z \right)}dS}=\frac{1}{2}\iint_{\Sigma}{z\sqrt{x^2+y^2+z^2}dS}.$

\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*}

\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*}

\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

\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*}

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

(1)研究

## 含奇点的第二型曲面积分计算

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

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

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

\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*}

$$\frac{\partial J}{\partial t}=\frac{1}{4}\left( \frac{\partial J}{\partial x} \right) ^2-x^2-\frac{1}{2}x^4,\qquad \text{其中}J\left[ x\left( 1 \right) ,1 \right] =0$$

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

\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*}

\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*}

\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*}

\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*}

\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*}

\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*}

$$I=\frac{\pi ^4}{16}\left[ _5F_4\left( -\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1,1;-1 \right) -\frac{1}{8}\,_5F_4\left( \frac{1}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};2,2,2,2;-1 \right) \right] ,$$

$$\int_0^{\pi}{\sqrt{\tan \frac{\theta}{2}}\ln^2 \left( \sin \theta \right) \text{d}\theta}.$$
\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:

$$\frac{f\left( x \right) -f\left( x_1 \right)}{x-x_1}\le \frac{f\left( x_2 \right) -f\left( x_1 \right)}{x_2-x_1}\le \frac{f\left( x_2 \right) -f\left( x \right)}{x_2-x}.$$

$$\frac{f\left( x \right)}{x}\leqslant \frac{x}{f^{-1}\left( x \right)}.$$

$$\frac{f\left( x \right)}{x}\le \frac{f\left( t \right)}{t}=\frac{x}{f^{-1}\left( x \right)}.$$

$$\int_0^1{f\left( x \right) g\left( x \right) \text{d}x}\leqslant \int_0^1{x^2\text{d}x}=\frac{1}{3}.$$

\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$$\displaystyle{\int_1^{\infty}\frac{x^{2a-2}}{f'(x)}\mathrm{d}x<+\infty}$$

$$\left( \int_{\text{e}^{\frac{A}{2}}}^{\text{e}^A}{\frac{x^{2a-2}}{f'\left( x \right)}\text{d}x} \right) \left( \int_{\text{e}^{\frac{A}{2}}}^{\text{e}^A}{\frac{f'\left( x \right)}{x^{2a}\ln ^2x}\text{d}x} \right) \geqslant \left( \int_{\text{e}^{\frac{A}{2}}}^{\text{e}^A}{\frac{\text{d}x}{x\ln x}} \right) ^2=\ln ^22.$$

\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*}

$$\int_{\textrm{e}^{A/2}}^{\textrm{e}^A}{\frac{\textrm{d}x}{f'\left(x\right)}}\geqslant\frac{\ln^22}{2}$$

\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

\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*}

$$I=2\int_0^1{x\text{e}^x\frac{\pi}{2}\sqrt{1-x^2}\frac{\text{d}x}{\sqrt{1-x^2}}}=\pi \int_0^1{x\text{e}^x\text{d}x}=\pi.$$

\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
$\sqrt{\pi}\Gamma (2s)=2^{2s-1}\Gamma (s)\Gamma \left(s+\frac{1}{2}\right),s>0，$

$\Gamma (s)=\int_0^{+\infty}x^{s-1}e^{-x}\mathrm{d}x, s>0.$

$I(s)=\int_0^1\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}}.$

$\begin{array}{rl}I(s)&=\int_0^1\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}}\\&=2s\int_0^{\frac{\pi}{4}} (\sin t\cos t)^{2s-1}\mathrm{d}t\\&=s2^{1-2s} \int_0^{\frac{\pi}{2}}\sin^{2s-1}u\mathrm{d}u\\&=2^{-2s}sB(\frac{1}{2},s)\\&=2^{-2s}s\frac{\Gamma (\frac{1}{2})\Gamma (s)}{\Gamma(\frac{1}{2}+s)}\\&= 2^{-2s}\sqrt{\pi}s\frac{\Gamma (s)}{\Gamma(\frac{1}{2}+s)}.\end{array}$

$I(s)=\int_0^1\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}}=\int_1^{+\infty}\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}},$

$I(s)=\frac{1}{2}\int_0^{+\infty}\frac{\mathrm{d}x}{(1+x^{\frac{1}{s}})^{2s}}=s\int_0^{\frac{\pi}{2}}(\sin t\cos t)^{2s-1}\mathrm{d}t=\frac{sB(s,s)}{2}=\frac{s\Gamma^2(s)}{2\Gamma (2s)}.$

$2^{-2s}\sqrt{\pi}s\frac{\Gamma (s)}{\Gamma(\frac{1}{2}+s)}=\frac{s\Gamma^2(s)}{2\Gamma (2s)}.$ 从而
$\sqrt{\pi}\Gamma (2s)=2^{2s-1}\Gamma (s)\Gamma \left(s+\frac{1}{2}\right), s>0.$

## 两道积分习题

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

\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*}

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

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

## Ahmed’s integrals 和 Coxeter’s integrals

1. 主要结果

$$A(p, q, r) := \int_{0}^{1} \frac{\arctan q \sqrt{p^{2}x^{2} + 1} }{q \sqrt{p^{2}x^{2} + 1}} \frac{p q r}{(r^{2}+1)p^{2}x^{2} + 1} \, dx.$$

\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*}

\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*}

$$A(p, q, r) = A(\rho \mid \alpha, \beta, \gamma)$$

2. 推论

$$A(p, q, r) = \frac{1}{8} \left( \alpha(2\pi-\alpha) + \beta(2\pi-\beta) – \gamma(2\pi-\gamma) \right).$$

\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*}

$$A(p, q, r) = \frac{\pi}{4} \left( \alpha + \beta – \gamma \right).$$

$$\Re\operatorname{Li}_{2}(-e^{i\theta}) = \operatorname{Li}_{2}(-1) + \frac{\theta^{2}}{4} \quad |\theta| \leq \pi,$$

\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

\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*}

$$k = \left( \frac{1-\cos\beta}{a\cos\beta+ b} \right)^{1/2}, \quad \alpha = \arccos\left(\frac{a-1}{a+1}\right), \quad \gamma = \arccos\left(-\frac{1+b}{1+a}\right).$$定义的.

\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*}

\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*}

$$\operatorname{Li}_{2}(\rho) – \operatorname{Li}_{2}(\rho^{2}) – \operatorname{Li}_{2}(\rho^{3}) + \frac{1}{3}\operatorname{Li}_{2}(\rho^{6}), \quad \rho = \frac{\sqrt{5}-1}{\sqrt{5}+1}.$$

$$\int_{0}^{\frac{\pi}{3}} \arccos \left( \frac{\cos \theta}{1+2\cos \theta} \right) \; d\theta = \frac{2\pi^{2}}{15}.$$

\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*}

\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*}

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

\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*}

\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*}

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

## 又一道二元反常积分题

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

## 又两个积分

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.

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 $\displaystyle \int_{0}^{\infty}\cos x^2 \,dx=\int_{0}^{\infty}\sin x^2 \,dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$.