谢惠民下册上的一道题:求

I am here,because U are here.

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.

（13年武大数分）求$\displaystyle I = \iint\limits_\Sigma  {{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{ - \frac{3}{2}}}{{\left( {\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} + \frac{{{z^2}}}{{{c^4}}}} \right)}^{ - \frac{1}{2}}}dS} $,其中$\sum$为椭球面: $\displaystyle \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1(a,b,c>0)$.

下面是自己的解答：

前几天徐半仙问了我谢惠民下册P325页上一道难度稍大的全微分题目，利用今晚美好的独处时间（笑哭），做了下，解答如下：

对于以下一阶微分形式$\omega $,求函数$M(x,y)\neq0$, 使得在适当的区域内$M\omega $为全微分,并求其原函数:

（1） $\displaystyle \omega = \left[ { - y\sqrt {{x^2} + {y^2} + 1} - x\left( {{x^2} + {y^2}} \right)} \right]dx + \left[ {x\sqrt {{x^2} + {y^2} + 1} - y\left( {{x^2} + {y^2}} \right)} \right]dy$;

（2） $\displaystyle \omega = x\left[ {{{\left( {ay + bx} \right)}^3} + a{y^3}} \right]dx + y\left[ {{{\left( {ay + bx} \right)}^3} + b{x^3}} \right]dy$.

解：（1）取\[M = \frac{1}{{\left( {{x^2} + {y^2}} \right)\sqrt {{x^2} + {y^2} + 1} }},\]我们有

\[M\omega = \left( { - \frac{y}{{{x^2} + {y^2}}} - \frac{x}{{\sqrt {{x^2} + {y^2} + 1} }}} \right)dx + \left( {\frac{x}{{{x^2} + {y^2}}} - \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }}} \right)dy.\]

则有$P = - \frac{y}{{{x^2} + {y^2}}} - \frac{x}{{\sqrt {{x^2} + {y^2} + 1} }},Q = \frac{x}{{{x^2} + {y^2}}} - \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }}$,且\[\frac{{\partial P}}{{\partial y}} = \frac{{{y^2} - {x^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} + \frac{{xy}}{{{{\left( {{x^2} + {y^2} + 1} \right)}^{3/2}}}} = \frac{{\partial Q}}{{\partial x}}.\]

此时原函数为

\begin{align*}\varphi \left( {x,y} \right) = &\int_{{x_0}}^x {P\left( {x,{y_0}} \right)dx} + \int_{{y_0}}^y {Q\left( {x,y} \right)dy} + C'\\= &\int_{{x_0}}^x {\left( { - \frac{{{y_0}}}{{{x^2} + y_0^2}} - \frac{x}{{\sqrt {{x^2} + y_0^2 + 1} }}} \right)dx} + \int_{{y_0}}^y {\left( {\frac{x}{{{x^2} + {y^2}}} - \frac{y}{{\sqrt {{x^2} + {y^2} + 1} }}} \right)dy} + C'\\= &\left( { - \arctan \frac{x}{{{y_0}}} - \sqrt {{x^2} + y_0^2 + 1} + \arctan \frac{{{x_0}}}{{{y_0}}} + \sqrt {x_0^2 + y_0^2 + 1} } \right)\\&+ \left( {\arctan \frac{y}{x} - \sqrt {{x^2} + {y^2} + 1} - \arctan \frac{{{y_0}}}{x} + \sqrt {{x^2} + y_0^2 + 1} } \right) + C'\\=& \arctan \frac{y}{x} - \sqrt {{x^2} + {y^2} + 1} + C.\end{align*}

值得一提的是：本题的积分因子是通过Wolfram Alpha求解出ODE，然后分别对$x,y$求偏导得来的.

（2）丁同仁书上一定理：

齐次方程$P(x,y)dx+Q(x,y)dy=0$有积分因子$M=\frac{1}{xP+yQ}$.

定理的证明:作变换$y=ux$,则由$P\left( {x,y} \right)dx + Q\left( {x,y} \right)dy = 0$是齐次方程,我们有$$P\left( {x,ux} \right)dx + Q\left( {x,ux} \right)\left( {udx + xdu} \right) = \left[ {{x^m}P\left( {1,u} \right) + u{x^m}Q\left( {1,u} \right)} \right]dx + {x^{m + 1}}Q\left( {1,u} \right)du = 0.$$

方程两边同乘\[\frac{1}{{xP + yQ}} = \frac{1}{{{x^{m + 1}}\left[ {P\left( {1,u} \right) + uQ\left( {1,u} \right)} \right]}},\]则有

\[\frac{1}{x}dx + \frac{{Q\left( {1,u} \right)}}{{P\left( {1,u} \right) + uQ\left( {1,u} \right)}}du = 0.\]显然此方程为全微分方程.证毕.

取\[M = \frac{1}{{xP + yQ}} = \frac{1}{{\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)}}.\]

则有

\[P' = \frac{{x{{\left( {ay + bx} \right)}^3} + ax{y^3}}}{{\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)}},Q' = \frac{{y{{\left( {ay + bx} \right)}^3} + b{x^3}y}}{{\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)}}.\]

我猜此时一定成立\[\frac{{\partial P'}}{{\partial y}} = \frac{{\partial Q'}}{{\partial x}}.\]

事实上

\begin{align*}\frac{{\partial P'}}{{\partial y}} = & - 2xy{\left( {ay + bx} \right)^6} - 2{x^3}y{\left( {ay + bx} \right)^4} + \left( {5a{x^3}{y^2} + ax{y^4}} \right){\left( {ay + bx} \right)^3}\\& - 3{a^2}x{y^3}\left( {{x^2} + {y^2}} \right){\left( {ay + bx} \right)^2} + a{x^3}{y^4}\left( {ay + bx} \right) - {a^2}{x^3}{y^5}\\\frac{{\partial Q'}}{{\partial x}} = &- 2xy{\left( {ay + bx} \right)^6} - 2x{y^3}{\left( {ay + bx} \right)^4} + \left( {5b{x^2}{y^3} + b{x^4}y} \right){\left( {ay + bx} \right)^3}\\& - 3{b^2}{x^3}y\left( {{x^2} + {y^2}} \right){\left( {ay + bx} \right)^2} + b{x^4}{y^3}\left( {ay + bx} \right) - {b^2}{x^5}{y^3}.\end{align*}

于是

\begin{align*}\varphi \left( {x,y} \right) &= \int_{{x_0}}^x {P'\left( {x,{y_0}} \right)dx} + \int_{{y_0}}^y {Q'\left( {x,y} \right)dy} + C'\\&= \frac{1}{2}\ln \left[ {\left( {{x^2} + {y^2}} \right){{\left( {ay + bx} \right)}^3} + {x^2}{y^2}\left( {ay + bx} \right)} \right] - \frac{3}{2}\ln \left( {ay + bx} \right) + C.\end{align*}

事实上,我们还可取\[M = \frac{1}{{{{\left( {ay + bx} \right)}^3}}},\]由此得到

\[\varphi \left( {x,y} \right) = \frac{{{x^2} + {y^2}}}{2} + \frac{{{x^2}{y^2}}}{{2{{\left( {ay + bx} \right)}^2}}} + C.\]

“数学是你们的选择，你们随时都可以放弃。但当数学仍是你们的选择时，就必须为此负责。”

——S.Lang对他学生上课前说的话

（1）设$f$在$D:x^2+y^2\leq1$上二阶连续可微,且\[\Delta f=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}=x^2+y^2,\]求\[\iint\limits_D\left(\frac{x}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial x}+ \frac{y}{\sqrt{x^2+y^2}}\frac{\partial f}{\partial y}\right)\mathrm{d}x\mathrm{d}y.\]

解：(Hansschwarzkopf)根据Gauss公式