数学分析 - Eufisky - The lost book

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

杂题

设函数$~f(x)$ 在$~[a,b]$ 上连续,但不为常数.求证:$~\exists \xi\in(a,b)$,使$~f(x)$ 在$~\xi$ 不取极值.

注意到 $f$ 不是常数函数, 并且 $f\in C[a,b],$ 所以 $f$ 的值域 $R$ 是一个有限闭区间.

先将 $f$ 开拓定义到整个 $\mathbf R$ 上: 对 $x\leq a,$ 让 $f(x)=f(a);$ 对 $x\geq b,$ 让 $f(x)=f(b).$ 让 $$C=\{f(x);\ \hbox{$x$ 是 $f$ 在 $\mathbf R$ 中的极小值点}\},$$ 下面来证明 $C$ 是一个至多可列集.

任给 $c\in C,$ 存在 $x\in\mathbf R$ 以及 $u_x,v_x\in\mathbf Q,$ 使得 $$u_x<x<v_x,\ f(y)\geq f(x)=c,\ \forall y\in(u_x,v_x).$$这样就得到了一个从 $C$ 到 $\mathbf Q\times\mathbf Q$ 的单射 $c\mapsto(u_x,v_x)$, 故 $C$ 是至多可列集.

类似可证 $f$ 极大值的全体也是至多可列集. 从而 $f$ 的极值的全体是 $R$ 的至多可列子集. 这就完成了证明.

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*}

而

\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*}

又

\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*}

和

\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

与调和数列有关的级数计算

求和$$\sum_{n=1}^{\infty}{\frac{H_n-H_{2n}}{n\left( 2n+1 \right)}}.$$

解.(向老师)首先不难得到

\begin{align*}\sum_{n=1}^{\infty}{\frac{H_n-H_{2n}}{n\left( 2n+1 \right)}}=&2\sum_{n=1}^{\infty}{\left( H_n-H_{2n} \right) \left( \frac{1}{2n}-\frac{1}{2n+1} \right)}\\=&2\sum_{n=1}^{\infty}{\left( \frac{1}{2n}-\frac{1}{2n+1} \right) \int_0^1{\frac{x^{2n}-x^n}{1-x}\text{d}x}}\\=&\int_0^1{\frac{\sqrt{x}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}-\ln \frac{1+x}{1-x}-\ln \left( 1+x \right)}{1-x}\text{d}x}\\&+\int_0^1{\left( \frac{1}{\sqrt{x}}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}-\frac{1}{x}\ln \frac{1+x}{1-x} \right) \text{d}x},\end{align*}

其中

$$\int_0^1{\frac{1}{\sqrt{x}}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}\text{d}x}=2\int_0^1{\ln \frac{1+t}{1-t}\text{d}t}=4\ln 2.$$

$$\int_0^1{\frac{1}{x}\ln \frac{1+x}{1-x}\text{d}x}=\mathrm{Li}_2\left( 1 \right) -\mathrm{Li}_2\left( -1 \right) =\frac{\pi ^2}{4}.$$

\begin{align*}\int_0^1{\frac{\left( \sqrt{x}-1 \right)}{1-x}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}\text{d}x}&=-2\int_0^1{\frac{t}{1+t}\ln \frac{1+t}{1-t}\text{d}t}\\&=2\int_0^1{\left[ \frac{\ln \left( 1+t \right)}{1+t}-\frac{\ln \left( 1-t \right)}{1+t} \right] \text{d}t}-2\int_0^1{\ln \frac{1+t}{1-t}\text{d}t}\\&=\ln ^22+2\mathrm{Li}_2\left( \frac{1}{2} \right) -4\ln 2=\frac{\pi^2}{6}-4\ln2.\end{align*}

又

\begin{align*}&\int_0^1{\frac{\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}-\ln \frac{1+x}{1-x}-\ln \left( 1+x \right)}{1-x}\text{d}x}=2\int_0^1{\frac{1}{1-x}\ln \frac{1+\sqrt{x}}{1-\sqrt{x}}\text{d}x}\\=&2\int_0^1{\frac{\ln \left( 1+\sqrt{x} \right) -\ln 2}{1-x}\text{d}x}-2\int_0^1{\frac{\ln \left( 1+x \right) -\ln 2}{1-x}\text{d}x},\end{align*}

其中

\begin{align*}\int_0^1{\frac{\ln \left( 1+\sqrt{x} \right) -\ln 2}{1-x}\text{d}x}&=2\int_0^1{\frac{t}{1-t^2}\ln \frac{1+t}{2}\text{d}t}\\&=\int_0^1{\frac{1}{1-t}\ln \frac{1+t}{2}\text{d}t}-\int_0^1{\frac{1}{1+t}\ln \frac{1+t}{2}\text{d}t}\\&=-\mathrm{Li}_2\left( \frac{1}{2} \right) +\frac{1}{2}\ln ^22=\ln ^22-\frac{\pi ^2}{12}.\end{align*}

$$\int_0^1{\frac{\ln \left( 1+x \right) -\ln 2}{1-x}\text{d}x}=-\mathrm{Li}_2\left( \frac{1}{2} \right) =\frac{\ln ^22}{2}-\frac{\pi ^2}{12}.$$

最后得到

\begin{align*}\sum_{n=1}^{\infty}{\frac{H_n-H_{2n}}{n\left( 2n+1 \right)}}=4\ln 2-\frac{\pi ^2}{4}+\frac{\pi ^2}{6}-4\ln 2+2\left( \ln ^22-\frac{\pi ^2}{12} \right) -2\left( \frac{\ln ^22}{2}-\frac{\pi ^2}{12} \right) =\ln ^22-\frac{\pi ^2}{6}.\end{align*}

几个重要定理

1.Mittag-Leffler's theorem.

设$\Omega$是平面内的开集, $A\subset \Omega$, $A$在$\Omega$内没有极限点,且对每个$a\in A$,对应有一个正整数$m(\alpha)$和一个有理函数$$P_\alpha (z)=\sum_{j=1}^{m(\alpha)}c_{j,\alpha}(z-\alpha)^{-j},$$则在$\Omega$内存在一个亚纯函数$f$,它在每个$\alpha\in A$处的主要部分是$P_\alpha$且在$\Omega$内没有其它极点.详见Rudin实分析与复分析P216.

这里有亚纯函数极展开的一些例子.

\begin{align*}\frac{1}{\sin \left( z \right)}&=\sum_{n\in \mathbb{Z}}{\frac{\left( -1 \right) ^n}{z-n\pi}}=\frac{1}{z}+2z\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{z^2-\left( n\,\pi \right) ^2}},\\\cot \left( z \right) &\equiv \frac{\cos \left( z \right)}{\sin \left( z \right)}=\sum_{n\in \mathbb{Z}}{\frac{1}{z-n\pi}}=\frac{1}{z}+2z\sum_{k=1}^{\infty}{\frac{1}{z^2-\left( k\,\pi \right) ^2}},\\\frac{1}{\sin ^2\left( z \right)}&=\sum_{n\in \mathbb{Z}}{\frac{1}{\left( z-n\,\pi \right) ^2}},\\\frac{1}{z\sin \left( z \right)}&=\frac{1}{z^2}+\sum_{n\ne 0}{\frac{\left( -1 \right) ^n}{\pi n\left( z-\pi n \right)}}=\frac{1}{z^2}+\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n\,\pi}}\frac{2z}{z^2-\left( n\,\pi \right) ^2}.\end{align*}

2.Ramanujan's Master Theorem

假设$x=0$的一些邻域中有$$F(x)=\sum_{k=0}^\infty\frac{\phi (k)(-x)^k}{k!}$$对某些函数（称为解析或可积的）$\phi (k)$成立.那么$$\int_0^\infty x^{n-1}F(x)dx=\Gamma (n)\phi (-n).$$

参考：这里以及

Ramanujan's Proof

This proof was given by none other than Ramanujan.

Recall Euler's integral representation of the Gamma Function -

$$\int_0^\infty e^{-mx}x^{n-1}dx = m^{-n}\Gamma(n).$$

where $m,n>0$. Let $m=r^k$ with $r>0$, multiply both sides by $\frac{f^{(k)}h^k}{k!}$ and sum on $k, 0 \leq k <\infty$, to obtain

$$\sum_{k=0}^\infty \frac{f^{(k)}(a)h^k}{k!}\int_0^\infty e^{-r^k x}x^{n-1}dx=\Gamma(n) \sum_{k=0}^\infty \frac{f^{(k)}(a)(hr^{-n})^k}{k!}.$$

Next, expand $e^{-r^k x}, 0\leq k<\infty$, in its Maclaurin Series, invert the order of summation and integration, and apply Taylor's Theorem to deduce that

$$\int_0^\infty x^{n-1}\sum_{j=0}^\infty \frac{f(h r^j+a)}{j!}(-x)^j dx = \Gamma(n)f(hr^{-n}+a).$$

Now define $f(hr^m+a)=\varphi(m)$, where $m$ is real and $a,h$ and $r$ are regarded as constants. Then

$$\int_0^\infty x^{n-1}\sum_{j=0}^\infty \frac{\varphi(j) (-x)^j}{j!}dx= \Gamma(n) \varphi(-n).$$

This completes Ramanujan's proof. Ramanujan was very fond of this clever, original technique and he used it many contexts.

例：证明$$\int_{0}^{\infty}{\left(\sum_{k=0}^{\infty}{\frac{(-1)^{k}P_{k+1}}{k!}x^{k}} \right)dx}=2,$$

其中$P_{k+1}$表示第$k+1$个素数,记$P_1=3$.

由于$$\int_0^\infty x^{s-1} \left( \sum_{k=0}^\infty \frac{P_{k+1}}{k!}(-x)^k\right)dx =\Gamma(s) P_{1-s},$$

令$s=1$我们有$$\int_0^\infty \left( \sum_{k=0}^\infty \frac{P_{k+1}}{k!}(-x)^k\right)dx = P_{0}=2.$$

3.Glasser's Master Theorem

对任意可积函数$F(x)$和形如

$$\phi(x)=|a|x-\sum_{n=1}^N\frac{|\alpha_n|}{x-\beta_n}$$

的$\phi(x)$,恒等式$$PV \int_{-\infty}^\infty F(\phi (x))dx=PV \int_{-\infty}^\infty F(x)dx$$成立,其中$a,\{\alpha_n\}_{n=1}^N$和$\{\beta_n\}_{n=1}^N$为任意常数.这里, $PV$表示Cauchy主值.这是从Cauchy的著名结果$$PV\int_{-\infty}^\infty F(u)dx=\int_{-\infty}^\infty F(x)dx$$

归纳出来的,其中$u=x-1/x$.

例.求$$\int_{0}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x.$$

$$I=\int_{0}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x,$$

$$I=\frac{1}{2}\int_{-\infty}^{\infty} \left[\left(\frac{2015}{2015+x}+\cdots +\frac{2}{2+x}+\frac{1}{1+x}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x,$$

$$I=\frac{1}{2}\int_{-\infty}^{\infty} \left[\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)^{2016}+1 \right] ^{-1}\mathrm{d}x.$$

Now, letting $f(x)=\frac{1}{x^{2016}+1}$, and noting that $f(x)=f(-x)$,

$$I=\frac{1}{2}\int_{-\infty}^{\infty} f\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)\mathrm{d}x=\frac{1}{2}\int_{-\infty}^{\infty} f\left(-\left(\sum^{2015}_{i=1}\frac{i}{x+i}-x\right)\right)\mathrm{d}x$$

$$I=\frac{1}{2}\int_{-\infty}^{\infty} f\left(x-\sum^{2015}_{i=1}\frac{i}{x-(-i)}\right)\mathrm{d}x \tag {1}$$

Using Glasser's Master Theorem,

$$I=\frac{1}{2}\int^{\infty}_{-\infty} f(x)\ \mathrm{d}x=\frac{1}{2}\int^{\infty}_{-\infty} \frac{1}{x^{2016}+1}\ \mathrm{d}x=\int^{\infty}_{0} \frac{1}{x^{2016}+1}\ \mathrm{d}x \tag {2}$$

Now we know that $$B(a,b)=\int^{\infty}_0\frac{t^{a-1}}{(1+t)^{a+b}}dt$$

From $(2)$,after substituting $x^{2016} =t$,

$$I=\frac{1}{2016}\int^{\infty}_{0}\frac{t^{\frac{1}{2016}-1}}{(1+t)^{\frac{1}{2016}+\frac{2015}{2016}}}dt=\frac{1}{2016}B(\frac{1}{2016},\frac{2015}{2016})$$

Therefore $$\color{red}{I=\frac{1}{2016}\frac{\Gamma(\frac{1}{2016})\Gamma(\frac{2015}{2016})}{\Gamma(1)}=\frac{\pi}{2016\sin(\frac{\pi}{2016})}\approx1.0000004047320180811575}$$

这里

Rogers L-Function

Watson's Triple Integrals

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*}

又由

\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*}

误差函数Erf专题

误差函数(entire function)

1.int e^(-x^2-2y^2), y=0..x, x=0..\infty用WolframAlpha计算二重积分.

\[\int_0^\infty {\int_0^x {{e^{ - {x^2} - 2{y^2}}}} } dydx = \frac{{\arctan \left( {\sqrt 2 } \right)}}{{2\sqrt 2 }}.\]

求\[I\left( b \right) = \int_0^\infty {\int_0^x {{e^{ - {a^2}{x^2} - {b^2}{y^2}}}} } dydx,\qquad a,b\geq 0.\]

解.事实上,我们有

\begin{align*}I\left( b \right) &= \int_0^\infty {\int_0^x {{e^{ - {a^2}{x^2} - {b^2}{y^2}}}} } dydx = \int_0^\infty {{e^{ - {a^2}{x^2}}}dx\int_0^x {{e^{ - {b^2}{y^2}}}} } dy\\&= \frac{1}{b}\int_0^\infty {{e^{ - {a^2}{x^2}}}dx\int_0^{bx} {{e^{ - {y^2}}}} } dy = \frac{{\sqrt \pi }}{{2b}}\int_0^\infty {{e^{ - {a^2}{x^2}}} \mathrm{erf}\left( {bx} \right)dx}. \end{align*}

对$b$求导我们有

\begin{align*}I'\left( b \right) &= - \frac{1}{{{b^2}}}\int_0^\infty {{e^{ - {a^2}{x^2}}}dx\int_0^{bx} {{e^{ - {y^2}}}} } dy + \frac{1}{b}\int_0^\infty {x{e^{ - \left( {{a^2} + {b^2}} \right){x^2}}}dx} \\&= - \frac{1}{b}I\left( b \right) + \frac{1}{{2b\left( {{a^2} + {b^2}} \right)}}.\end{align*}

又$I(0)=\frac1{2a^2}$,解得\[I\left( b \right) = \frac{1}{{2ab}}\arctan \frac{b}{a},\]即\[\int_0^\infty {{e^{ - {a^2}{x^2}}}\mathrm{erf}\left( {bx} \right)dx} = \frac{1}{{a\sqrt \pi }}\arctan \frac{b}{a}.\]

法医秦明，完美案件

解常微分方程

求解ODE$$\frac{{{d^2}r}}{{d{t^2}}} + 2h\frac{{dr}}{{dt}} + w_0^2r = - \frac{e}{m}E.$$

参考：http://wenku.baidu.com/link?url=1z4AQfIr20D9-JVBbLLwyrYOwG2nVBubGuEVfCW8Urqz1rR2ABfF0lewEsWfWBwCp7W-0Z9fh4LRFh6wbEHhFEQwTiTtB0Nd51SBE1lTiFC

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.记

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

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

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