# 两道积分习题

Eufisky posted @ 2016年8月05日 03:21 in 数学分析 with tags 积分计算 多元 , 1491 阅读

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

2023年7月10日 04:27

Bravo TV is one of the most famous pay television networks in the United States. This channel is one of the NBC Universal families and acts as an alternative for Comcast. Bravo TV broadcasts worldwide and mostly wants to focus on movies and events. bravotv.com/link Activation Code is the password for activating the entertainment service. You will be capable of activating your Smart TV using this method.

CBSE 3rd Class Syll 说:
2023年8月20日 23:46

CBSE Curriculum is based on the National Curriculum Framework and Provides Opportunities for Students to Achieve Excellence in Learning, CBSE Provides the Syllabus for 3rd Class, This new Syllabus CBSE 3rd Class Syllabus 2024 are Designed Strategically by a Team of Subject Experts and are Prescribed by the Ministry of Human Resource Development, formerly Ministry of Education, is Responsible for the Development of Human Resources in India, Primary Level Syllabus for the Children of has been developed with the Supervision of the Central Board of Secondary Education.

(输入验证码)
or Ctrl+Enter