# 一个涉及Beta函数积分不等式的多种推广Some Extensions of an Integral Inequality with Beta Function

DOI: 10.12677/PM.2021.111005, PDF, HTML, XML, 下载: 72  浏览: 160

Abstract: We analyse the integral inequality of a postgraduate examination at Nankai University, a method of proof is given, and on the basis several extensions have been made to it, it is helpful to guiding students to carry out mathematics research.

${\int }_{0}^{1}{|{f}^{\prime }\left(x\right)|}^{\text{3}}\text{d}x\ge {\left(\frac{128}{3\pi }\right)}^{2}$. (1)

${\int }_{0}^{x}f\left(t\right)\text{d}t=xf\left(x\right)-{\int }_{0}^{x}t{f}^{\prime }\left(t\right)\text{d}t$,

${\int }_{0}^{1}{\int }_{0}^{x}f\left(t\right)\text{d}t\text{d}x={\int }_{0}^{1}xf\left(x\right)\text{d}x-{\int }_{0}^{1}{\int }_{0}^{x}t{f}^{\prime }\left(t\right)\text{d}t\text{d}x$,

${\int }_{0}^{1}\left(1-t\right)f\left(t\right)\text{d}t=1-{\int }_{0}^{1}t\left(1-t\right){f}^{\prime }\left(t\right)\text{d}t$,

${\int }_{0}^{1}\left(1-t\right)f\left(t\right)\text{d}t={\int }_{0}^{1}f\left(t\right)\text{d}t-{\int }_{0}^{1}tf\left(t\right)\text{d}t=1-1=0$,

$1={\int }_{0}^{1}t\left(1-t\right){f}^{\prime }\left(t\right)\text{d}t\le {\left({\int }_{0}^{1}{\left[t\left(1-t\right)\right]}^{\frac{3}{2}}\text{d}t\right)}^{\frac{2}{3}}{\left({\int }_{0}^{1}{|{f}^{\prime }\left(t\right)|}^{\text{3}}\text{d}t\right)}^{\frac{1}{3}}$,

$\begin{array}{c}{\int }_{0}^{1}{\left[t\left(1-t\right)\right]}^{\frac{3}{2}}\text{d}t={\int }_{0}^{1}{t}^{\frac{5}{2}-1}{\left(1-t\right)}^{\frac{5}{2}-1}\text{d}t=B\left(\frac{5}{2},\frac{5}{2}\right)=\frac{\Gamma \left(\frac{5}{2}\right)\Gamma \left(\frac{5}{2}\right)}{\Gamma \left(5\right)}\\ =\frac{1}{4!}{\Gamma }^{2}\left(\frac{3}{2}+1\right)=\frac{1}{4!}{\left(\frac{3}{2}\right)}^{2}{\Gamma }^{2}\left(\frac{3}{2}\right)=\frac{1}{4!}\frac{9}{4}{\Gamma }^{2}\left(\frac{1}{2}+1\right)\\ =\frac{1}{4!}\frac{9}{4}{\left(\frac{1}{2}\right)}^{2}{\Gamma }^{2}\left(\frac{1}{2}\right)=\frac{3}{128}\Gamma \left(\frac{1}{2}\right)\Gamma \left(1-\frac{1}{2}\right)\\ =\frac{3}{128}\frac{\pi }{\mathrm{sin}\pi /2}=\frac{3\pi }{128}\end{array}$,

$1\le {\left(\frac{3\pi }{128}\right)}^{2}{\int }_{0}^{1}{|{f}^{\prime }\left(x\right)|}^{\text{3}}\text{d}x$,

${\int }_{0}^{1}{|{f}^{\prime }\left(x\right)|}^{n}\text{d}x\ge {\left[\frac{n}{4\left(n+1\right)\left(3n-1\right)}\right]}^{1-n}{B}^{1-n}\left(\frac{1}{n-1},\frac{1}{n-1}\right)$.

${\int }_{0}^{1}x\left(1-x\right){f}^{\prime }\left(x\right)\text{d}x=1$,

$1={\int }_{0}^{1}x\left(1-x\right){f}^{\prime }\left(x\right)\text{d}x\le {\left({\int }_{0}^{1}{|{f}^{\prime }\left(x\right)|}^{n}\text{d}x\right)}^{\frac{1}{n}}{\left({\int }_{0}^{1}{\left[x\left(1-x\right)\right]}^{\frac{n}{n-1}}\text{d}x\right)}^{\frac{n-1}{n}}$,

$1\le {\int }_{0}^{1}{|{f}^{\prime }\left(x\right)|}^{n}\text{d}x{\left({\int }_{0}^{1}{\left[x\left(1-x\right)\right]}^{\frac{n}{n-1}}\text{d}x\right)}^{n-1}$,

$\begin{array}{c}{\int }_{0}^{1}{\left[x\left(1-x\right)\right]}^{\frac{n}{n-1}}\text{d}x=B\left(\frac{n}{n-1}+1,\frac{n}{n-1}+1\right)=\frac{{\Gamma }^{2}\left(\frac{n}{n-1}+1\right)}{\Gamma \left(\frac{2n}{n-1}+2\right)}\\ =\frac{{\Gamma }^{2}\left(\frac{1}{n-1}+2\right)}{\Gamma \left(\frac{2}{n-1}+4\right)}=\frac{n{\Gamma }^{2}\left(\frac{1}{n-1}\right)}{4\left(n+1\right)\left(3n-1\right)\Gamma \left(\frac{2}{n-1}\right)}\\ =\frac{n}{4\left(n+1\right)\left(3n-1\right)}B\left(\frac{1}{n-1},\frac{1}{n-1}\right),\end{array}$

${\int }_{0}^{1}{|{f}^{\prime }\left(x\right)|}^{n}\text{d}x\ge {\left({\int }_{0}^{1}{\left[x\left(1-x\right)\right]}^{\frac{n}{n-1}}\text{d}x\right)}^{1-n}={\left[\frac{n}{4\left(n+1\right)\left(3n-1\right)}\right]}^{1-n}{B}^{1-n}\left(\frac{1}{n-1},\frac{1}{n-1}\right)$.

${\int }_{0}^{1}{x}^{k}{f}^{\left(k-1\right)}\left(x\right)\text{d}x=k!$, $k=1,2,\cdots ,2\left(n+1\right)$.

${\int }_{0}^{1}{|{f}^{\left(2n+1\right)}\left(x\right)|}^{m}\text{d}x\ge {\left[\left(2n+1\right)!\right]}^{m}{B}^{1-m}\left(\frac{m}{m-1}\left(2n+1\right)+1,\frac{m}{m-1}+1\right)$.

$\begin{array}{c}{\int }_{0}^{x}f\left(t\right)\text{d}t=xf\left(x\right)-{\int }_{0}^{x}t{f}^{\prime }\left(t\right)\text{d}t=xf\left(x\right)-\frac{1}{2}{x}^{2}{f}^{\prime }\left(x\right)+\frac{1}{2}{\int }_{0}^{x}{t}^{2}{f}^{″}\left(t\right)\text{d}t\\ =xf\left(x\right)-\frac{1}{2!}{x}^{2}{f}^{\prime }\left(x\right)+\frac{1}{3!}{x}^{3}{f}^{″}\left(x\right)-\frac{1}{3!}{\int }_{0}^{x}{t}^{3}{f}^{‴}\left(t\right)\text{d}t=\cdots \\ =\underset{k=1}{\overset{2n+1}{\sum }}{\left(-1\right)}^{k+1}\frac{{x}^{k}}{k!}{f}^{\left(k-1\right)}\left(x\right)-\frac{1}{\left(2n+1\right)!}{\int }_{0}^{x}{t}^{2n+1}{f}^{\left(2n+1\right)}\left(t\right)\text{d}t,\end{array}$

$\begin{array}{l}{\int }_{0}^{1}{\int }_{0}^{x}f\left(x\right)\text{d}t\text{d}x\\ =\underset{k=1}{\overset{2n+1}{\sum }}{\left(-1\right)}^{k+1}\frac{1}{k!}{\int }_{0}^{1}{x}^{k}{f}^{\left(k-1\right)}\left(x\right)\text{d}x-\frac{1}{\left(2n+1\right)!}{\int }_{0}^{1}{\int }_{0}^{x}{t}^{2n+1}{f}^{\left(2n+1\right)}\left(t\right)\text{d}t\text{d}x\\ =1-\frac{1}{\left(2n+1\right)!}{\int }_{0}^{1}\left(1-t\right){t}^{2n+1}{f}^{\left(2n+1\right)}\left(t\right)\text{d}t,\end{array}$

${\int }_{0}^{1}{\int }_{0}^{x}f\left(x\right)\text{d}t\text{d}x={\int }_{0}^{1}\left(1-t\right)f\left(t\right)\text{d}t={\int }_{0}^{1}f\left(t\right)\text{d}t-{\int }_{0}^{1}tf\left(t\right)\text{d}t=1-1=0$,

$\begin{array}{c}\left(2n+1\right)!={\int }_{0}^{1}\left(1-t\right){t}^{2n+1}{f}^{\left(2n+1\right)}\left(t\right)\text{d}t\\ \le {\left({{\int }_{0}^{1}|{f}^{\left(2n+1\right)}\left(x\right)|}^{m}\text{d}x\right)}^{\frac{1}{m}}{\left({\int }_{0}^{1}{\left[{x}^{2n+1}\left(1-x\right)\right]}^{\frac{m}{m-1}}\text{d}x\right)}^{\frac{m-1}{m}}\\ \le {\left({{\int }_{0}^{1}|{f}^{\left(2n+1\right)}\left(x\right)|}^{m}\text{d}x\right)}^{\frac{1}{m}}{B}^{\frac{m-1}{m}}\left(\frac{m}{m-1}\left(2n+1\right)+1,\frac{m}{m-1}+1\right),\end{array}$

${\int }_{0}^{1}{|{f}^{\left(2n+1\right)}\left(x\right)|}^{m}\text{d}x\ge {\left[\left(2n+1\right)!\right]}^{m}{B}^{1-m}\left(\frac{m}{m-1}\left(2n+1\right)+1,\frac{m}{m-1}+1\right)$.

${\int }_{0}^{1}f\left(x\right)\text{d}x=1$, ${\int }_{0}^{1}{x}^{k}{f}^{\left(k-1\right)}\left(x\right)\text{d}x=k!$, $k=1,2,\cdots ,2\left(n+1\right)$

${\int }_{0}^{1}{|{f}^{\left(2n+1\right)}\left(x\right)|}^{2n+3}\text{d}x\ge {\left[2\left(n+2\right)\right]}^{2\left(n+1\right)}{\left[\left(2n+1\right)!\right]}^{2n+3}$.

$\left(2n+1\right)!={\int }_{0}^{1}\left(1-x\right){x}^{2n+1}{f}^{\left(2n+1\right)}\left(x\right)\text{d}x$,

$\begin{array}{c}{\left[\left(2n+1\right)!\right]}^{2n+3}={\left({\int }_{0}^{1}\left(1-x\right)xx\cdots x{f}^{\left(2n+1\right)}\left(x\right)\text{d}x\right)}^{2n+3}\\ \le \left(\underset{k=1}{\overset{2n+1}{\sum }}{\int }_{0}^{1}{x}^{2n+3}\text{d}x\right){\int }_{0}^{1}{\left(1-x\right)}^{2n+3}\text{d}x{\int }_{0}^{1}{|{f}^{\left(2n+1\right)}\left(x\right)|}^{2n+3}\text{d}x\\ ={\left[\frac{1}{2\left(n+2\right)}\right]}^{2\left(n+1\right)}{\int }_{0}^{1}{|{f}^{\left(2n+1\right)}\left(x\right)|}^{2n+3}\text{d}x,\end{array}$

${\int }_{0}^{1}{|{f}^{\left(2n+1\right)}\left(x\right)|}^{\text{2}n+\text{3}}\text{d}x\ge {\left[2\left(n+2\right)\right]}^{2\left(n+1\right)}{\left[\left(2n+1\right)!\right]}^{2n+3}$.

 [1] 数学手册编写组. 数学手册[M]. 北京: 人民教育出版社, 1997: 20-22. [2] 陈纪修, 於崇华, 金路. 数学分析(下册) [M]. 北京: 高等教育出版社, 2000: 372-382.