设 $f\in C^1[a,b]$, 且 $f(a)=0$, 证明

$$\underset{x\in [a,b]}{sup}|f(x)|⩽\sqrt{b-a}\cdot \sqrt{{\int }_a^b(f^{\prime }(x){)}^{2}dx}.$$

特别的, 如果 $a=0,b=1$, 则有 $\underset{x\in [0,1]}{sup}|f(x)|⩽\sqrt{{\int }_0^1(f^{\prime }(x){)}^{2}dx}.$ (见问题1863)

References:

Paulo Ney de Souza, Jorge-Nuno Silva, Berkeley Problems in Mathematics,Third Edition.

假设 $g\in C^1[a,b]$, 且 $g(a)=g(b)=0$. 则有

$${\int }_a^b{g}^2(t)dt⩽(\frac{b-a}{\pi }{)}^2{\int }_a^b|g^{\prime }(t){|}^{2}dt.$$

Remark:

高维的 Wirtinger 不等式又称为球面上的 Poincaré 不等式.

References:

https://arxiv.org/abs/1407.6871
https://en.wikipedia.org/wiki/Wirtinger%27s_inequality_for_functions

Wirtinger's inequality is seen as the one-dimensional version of Friedrichs' inequality.

求 $${\int }_0^1\frac{\sqrt{1-x^4}}{1+x^2}dx.$$

[相关的积分]

证明:

$${\int }_0^1\frac{1}{\sqrt{1-x^4}}dx\cdot {\int }_0^1\frac{x^2}{\sqrt{1-x^4}}dx=\frac{\pi }{4}.$$

如果我们记 $f(x)=\frac{1}{\sqrt{1-x^4}}$, $g(x)=\frac{x^2}{\sqrt{1-x^4}}$, 则

$$\begin{array}{r}f(x)+g(x)=\frac{1+x^2}{\sqrt{1-x^4}}=\sqrt{\frac{1+x^2}{1-x^2}},\\ f(x)-g(x)=\frac{1-x^2}{\sqrt{1-x^4}}=\sqrt{\frac{1-x^2}{1+x^2}},\\ \end{array}$$

于是 $(f(x)+g(x))(f(x)-g(x))=1$, 因此原积分

$$\begin{array}{rl}{\int }_0^1\frac{\sqrt{1-x^4}}{1+x^2}dx& ={\int }_0^1\sqrt{\frac{1-x^2}{1+x^2}}\\ & ={\int }_0^1(f(x)-g(x))dx\\ & ={\int }_0^1\frac{1}{\sqrt{1-x^4}}dx-{\int }_0^1\frac{x^2}{\sqrt{1-x^4}}dx\end{array}$$

References:

吉米多维奇, 《数学分析习题集题解》(五) 3872.

[分析]

设 $x^2=\sin \theta$, $\theta \in [0,\frac{\pi }{2}]$, 则 $x=\sqrt{\sin \theta }$, $dx=\frac{1}{2\sqrt{\sin \theta }}\cos \theta d\theta$.

$$\begin{array}{rl}\text{原式}& ={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{1-{\sin }^2\theta }}{1+\sin \theta }\cdot \frac{\cos \theta }{2\sqrt{\sin \theta }}d\theta \\ & ={\int }_0^{\frac{\pi }{2}}\frac{{\cos }^2\theta }{2\sqrt{\sin \theta }(1+\sin \theta )}d\theta \\ & ={\int }_0^{\frac{\pi }{2}}\frac{1-{\sin }^2\theta }{2\sqrt{\sin \theta }(1+\sin \theta )}d\theta \\ & ={\int }_0^{\frac{\pi }{2}}\frac{1-\sin \theta }{2\sqrt{\sin \theta }}d\theta \\ & =\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{1}{\sqrt{\sin \theta }}d\theta -\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\sqrt{\sin \theta }d\theta ,\end{array}$$

其中积分

$${\int }_0^{\frac{\pi }{2}}\sqrt{\sin \theta }d\theta \stackrel{t=\sqrt{\sin \theta }}{=}2{\int }_0^1\frac{t^2}{\sqrt{1-t^4}}dt$$

是一个椭圆积分, 可以参考问题998 .

设 $I(x)={\int }_{\pi }^{2\pi }\frac{y\sin (xy)}{y-\sin y}dy$, 求 ${\int }_0^{1}I(x)dx$.

计算定积分

$${\int }_0^1\frac{x^2\arcsin x}{\sqrt{1-x^2}}dx$$

$${\int }_1^2\frac{1}{x(1+x^n)}dx$$

$${\int }_1^3\sqrt{(3-x)(x-1)}dx$$

$${\int }_0^1(1-x^2{)}^{8}dx$$

一般的, 证明积分

$${\int }_0^1(1-x^2{)}^{n}dx=\frac{(2n)!!}{(2n+1)!!}$$

除了这里利用递推关系求出, 还可以应用积分余项. 请参考梅加强著《数学分析》例5.7.1

验证

$${\int }_0^{+\mathrm{\infty }}\frac{1}{x(e^x-1)}dx=+\mathrm{\infty }$$

是否正确?

注意这个积分与 zeta 函数相关.

设 ${\sigma }_n(x)=\frac{1}{2}+\cos x+\cos 2x+\cdots +\cos nx$, 证明

$${\sigma }_n(x)=\frac{\sin \frac{2n+1}{2}x}{2\sin \frac{1}{2}x},\mathrm{\forall }x\ne 2k\pi .$$

而当 $x=2k\pi$ 时, ${\sigma }_n(x)=\frac{1}{2}+n$.