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.