zoukankan      html  css  js  c++  java
  • 一个含有超几何函数的积分

    [Largeint_0^{large frac{pi}{2}} ln^2left( anfrac{x}{2} ight){ _3F_2}left(frac12,1,1;frac32,frac32;sin^2 x ight)mathrm{d}x = frac{5 pi^5}{384} ]


    (Largemathbf{Proof:})
    The ({;}_3F_2) is just a deception.

    [{;}_3F_2left(frac{1}{2},1,1;frac{3}{2},frac{3}{2};sin^2(x) ight) =sum_{n=0}^inftyfrac{(2sin x)^{2n}}{(2n+1)^2dbinom{2n}{n}}= frac{1}{sin x}int_0^xfrac{ heta}{sin heta}mathrm{d} heta ]

    This is obtained by the simple observation

    [frac{left(dfrac{1}{2} ight)_n (1)_n^2}{left(dfrac{3}{2} ight)_n^2 n!}=frac{4^n}{dbinom{2n}{n} (2n+1)^2} ]

    Also, note that (displaystyle frac{mathrm{d}}{mathrm{d}x}lnleft( an frac{x}{2} ight)=frac{1}{sin x}). Therefore, by repeated integration by parts:

    [egin{align*} &int_0^{large frac{pi}{2}} ln^2left( anfrac{x}{2} ight){ _3F_2}left(frac12,1,1;frac32,frac32;sin^2 x ight)mathrm{d}x=frac{1}{12}int_0^{frac{pi}{2}}ln^4left( anfrac{x}{2} ight)mathrm{d}x\ &=frac{1}{6}int_0^1frac{ln^4 x}{1+x^2}mathrm{d}x=frac{1}{6}sum_{n=0}^{infty }left ( -1 ight )^{n}int_{0}^{1}x^{2n}ln^{4}xmathrm{d}x=frac{1}{6} cdot 24sum_{n=0}^{infty }frac{left ( -1 ight )^{n}}{left ( 2n+1 ight )^{5}}\&=frac{1}{6}cdot 24cdot eta left ( 5 ight )=frac{1}{6}cdot 24cdot frac{5pi ^{5}}{1536}=Largeoxed{color{blue}{dfrac{5pi ^{5}}{384}}} end{align*}]

  • 相关阅读:
    sql server 2016新特性 查询存储(Query Store)的性能影响
    Spring 事务管理详情介绍
    python爬虫之快速对js内容进行破解
    python爬虫的一个常见简单js反爬
    温习pycharm
    宋朝官员分析随堂理解笔记
    K-Means改进模式
    jupyter 饼图
    WebDriver常用的API使用详解
    从浏览器启动开始
  • 原文地址:https://www.cnblogs.com/Renascence-5/p/5453079.html
Copyright © 2011-2022 走看看