数学分析 - Eufisky - The lost book

求和的一些估计式

计算$$\displaystyle\lim_{n\rightarrow\infty}\left({2\sqrt n}-\sum_{k=1}^n\frac{1}{\sqrt k}\right).$$


Use $\sqrt{n} = \sum_{k=1}^n \left( \sqrt{k} - \sqrt{k-1} \right)$, then
\begin{align*}&2 \sqrt{n} - \sum_{k=1}^n \frac{1}{\sqrt{k}} \\=& \sum_{k=1}^n \left( 2 \sqrt{k} - 2 \sqrt{k-1} - \frac{1}{\sqrt{k}} \right) = \sum_{k=1}^n \frac{1}{\sqrt{k}}  \left( \sqrt{k}-\sqrt{k-1} \right)^2\\=& \sum_{k=1}^n \frac{1}{\sqrt{k}} \left( \frac{(\sqrt{k}-\sqrt{k-1})(\sqrt{k}+\sqrt{k-1})}{(\sqrt{k}+\sqrt{k-1})} \right)^2 \\=& \sum_{k=1}^n \frac{1}{\sqrt{k} \left(\sqrt{k}+\sqrt{k-1}\right)^2}.\end{align*}
 
This shows the limit does exist and $\lim_{n \to \infty} \left( 2 \sqrt{n} - \sum_{k=1}^n \frac{1}{\sqrt{k}} \right) = \sum_{k=1}^\infty  \frac{1}{\sqrt{k} \left(\sqrt{k}+\sqrt{k-1}\right)^2}$.
 
The value of this sums equals $-\zeta\left(\frac{1}{2} \right) \approx 1.4603545$. This value is found by other means, though:
$$2 \sqrt{n} - \sum_{k=1}^n \frac{1}{\sqrt{k}} = 2 \sqrt{n} - \left( \zeta\left(\frac{1}{2}\right) - \zeta\left(\frac{1}{2}, n+1\right)\right) \sim -\zeta\left(\frac{1}{2}\right) - \frac{1}{2\sqrt{n}} + o\left( \frac{1}{n} \right) .$$ 
 

取整函数的求和问题

证明\[\sum\limits_{k = 1}^{n - 1} {\sin \left( {\frac{{\left( {2\left\lfloor {\sqrt {kn} } \right\rfloor  + 1} \right)\pi }}{{2n}}} \right)}  = \cot \left( {\frac{\pi }{{2n}}} \right)\cos \left( {\frac{\pi }{{2n}}} \right),\]

where $\lfloor\cdot\rfloor$ is greatest integer function.


Lemma Summation by Pasts (1)http://en.wikipedia.org/wiki/Summation_by_parts
 
$$\sum_{k=a}^b f_k\Delta g_k=f_kg_k\Bigg|_{k=a}^{b+1}-\sum_{k=a}^b g_{k+1}\Delta f_k=f_{b+1}g_{b+1}-f_ag_a-\sum_{k=a}^b g_{k+1}\Delta f_k$$
$\displaystyle \text{with difference operator }\Delta :\qquad \Delta f_k=f_{k+1}-f_k$.
 
My solution
 
$\text{Get }j=\left\lfloor \sqrt{kn}\right\rfloor\Rightarrow j\le \sqrt{kn}<j+1\Rightarrow \dfrac{j^2}{n}\le k<\dfrac{(j+1)^2}{n}$
 
Therefore if $\left\lfloor\dfrac{j^2}{n}\right\rfloor+1\le k\le \left\lfloor\dfrac{(j+1)^2}{n}\right\rfloor$ then $\sqrt{kn}=j$
 
Note:
 
$j=0\Rightarrow \left\lfloor\dfrac{j^2}{n}\right\rfloor+1=1$
 
$j=n-2\Rightarrow \left\lfloor\dfrac{(j+1)^2}{n}\right\rfloor=n-2$
 
$j=n-1\Rightarrow \left\lfloor\dfrac{(j+1)^2}{n}\right\rfloor=n>n-1$
 
So that sum became 
 
\begin{align*}S&=\sum_{k=1}^{n-1}\sin\left(\dfrac{\left(2\lfloor\sqrt{kn}\rfloor+1\right)\pi}{2n}\right)\\&=\sin\left(\dfrac{\left(2\lfloor\sqrt{(n-1)n}\rfloor+1\right)\pi}{2n}\right)+\sum_{k=1}^{n-2}\sin\left(\dfrac{\left(2\lfloor\sqrt{kn}\rfloor+1\right)\pi}{2n}\right)\\ &=\sin\left(\frac{(2n-1)\pi}{2n}\right)+\sum_{j=0}^{n-2}\left(\left\lfloor\frac{(j+1)^2}{n}\right\rfloor-\left\lfloor\frac{j^2}{n}\right\rfloor\right)\sin\left(\frac{(2j+1)\pi}{2n}\right)\\ &=\sin\left(\frac{\pi}{2n}\right)+\sum_{j=0}^{n-2}\Delta\left(\left\lfloor\frac{j^2}{n}\right\rfloor\right)\sin\left(\frac{(2j+1)\pi}{2n}\right)\end{align*}
Using the [1], we get
\begin{align*}S&=\sin\left(\frac{\pi}{2n}\right)+\left[\sin\left(\frac{(2j+1)\pi}{2n}\right)\left\lfloor\frac{j^2}{n}\right\rfloor\right]_{j=0}^{n-1}\\ &{}\quad -\sum_{j=0}^{n-2}\left\lfloor\frac{(j+1)^2}{n}\right\rfloor\left(\sin\left(\frac{(2j+3)\pi}{2n}\right)-\sin\left(\frac{(2j+1)\pi}{2n}\right)\right)\\ &=(n-1)\sin\left(\frac{\pi}{2n}\right)-2\sin\left(\frac{\pi}{2n}\right)\sum_{j=0}^{n-2}\left\lfloor\frac{(j+1)^2}{n}\right\rfloor\cos\left(\frac{(j+1)\pi}{n}\right)\\ &=(n-1)\sin\left(\frac{\pi}{2n}\right)-2\sin\left(\frac{\pi}{2n}\right)\underbrace{\sum_{j=1}^{n-1}\left\lfloor\frac{j^2}{n}\right\rfloor\cos\left(\frac{j\pi}{n}\right)}_{=A}\end{align*}
 
With sum A, using reverse summand property we get
\begin{align*}A&=\sum_{j=1}^{n-1}\left\lfloor\frac{j^2}{n}\right\rfloor\cos\left(\frac{j\pi}{n}\right)\\&=\sum_{j=1}^{n-1}\left\lfloor\frac{(n-j)^2}{n}\right\rfloor\cos\left(\frac{(n-j)\pi}{n}\right)\\ &=-\sum_{j=1}^{n-1}\left\lfloor n-2j+\frac{j^2}{n}\right\rfloor\cos\left(\frac{j\pi}{n}\right)\\ &=-A+\sum_{j=1}^{n-1}(2j-n)\cos\left(\frac{j\pi}{n}\right)\\ \Rightarrow A&=\frac{1}{2}\sum_{j=1}^{n-1}(2j-n)\cos\left(\frac{j\pi}{n}\right)\end{align*}
 
We get
 
$\displaystyle\cos\left(\frac{j\pi}{n}\right)=\dfrac{1}{2\sin\left(\frac{\pi}{2n}\right)}\left[\sin\left(\frac{(2j+1)\pi}{2n}\right)-\sin\left(\frac{(2j-1)\pi}{2n}\right)\right]=\dfrac{1}{2\sin\left(\frac{\pi}{2n}\right)}\Delta\left[\sin\left(\frac{(2j-1)\pi}{2n}\right)\right]$
 
continue using [1] :)
 
$\displaystyle A =\left.\dfrac{(2j-n)}{4\sin\left(\frac{\pi}{2n}\right)}\sin\left(\frac{(2j-1)\pi}{2n}\right)\right|_{j=1}^{n}-\dfrac{1}{4\sin\left(\frac{\pi}{2n}\right)}\sum_{j=1}^{n-1}2\sin\left(\frac{(2j+1)\pi}{2n}\right)$
 
\begin{align*}A&=\frac{n-1}{2}+\dfrac{1}{4\sin^2\left(\frac{\pi}{2n}\right)}\sum_{j=1}^{n-1}\Delta\left[\cos\left(\frac{j\pi}{n}\right)\right]\\ &=\frac{n-1}{2}+\dfrac{1}{4\sin^2\left(\frac{\pi}{2n}\right)}\cdot\left.\cos\left(\frac{j\pi}{n}\right)\right|_{j=1}^n\\ &=\frac{n-1}{2}-\dfrac{1+\cos\left(\frac{\pi}{n}\right)}{4\sin^2\left(\frac{\pi}{2n}\right)}\\&=\frac{n-1}{2}-\dfrac{2\cos^2\left(\frac{\pi}{2n}\right)}{4\sin^2\left(\frac{\pi}{2n}\right)}\end{align*}
 
Therefore:
 
\begin{align*}S&=(n-1)\sin\left(\frac{\pi}{2n}\right)-2\sin\left(\frac{\pi}{2n}\right)A\\ &=(n-1)\sin\left(\frac{\pi}{2n}\right)-(n-1)\sin\left(\frac{\pi}{2n}\right)+\dfrac{\cos^2\left(\frac{\pi}{2n}\right)}{\sin\left(\frac{\pi}{2n}\right)}\\ &=\boxed{\displaystyle\cot\left(\frac{\pi}{2n}\right)\cos\left(\frac{\pi}{2n}\right)}\end{align*}
 
来源:http://math.stackexchange.com/questions/404573/sum-sum-sin-left-frac2-lfloor-sqrtkn-rfloor-1-pi2n-right?rq=1

又两个积分

计算$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\ln{(\ln{\tan{x}})}dx}.$$


Let $u = \ln \tan x$, so that $\frac{\pi}{4} < x< \frac{\pi}{2}$ is mapped to $u>0$, $x=\arctan(\exp(u))$ and $\mathrm{d}x = \frac{\mathrm{d}u}{2 \cosh(u)}$. Then
$$\int_{\pi/4}^{\pi/2} \ln( \ln(\tan x))\, \mathrm{d}x = \frac{1}{2} \int_0^\infty \frac{\ln (u)}{\cosh(u)} \mathrm{d}u = \frac{1}{2} \lim_{s \to 0^+}\frac{\mathrm{d}}{\mathrm{d} s} \int_0^\infty \frac{u^s}{\cosh(u)} \mathrm{d}u$$
The latter parametric integral is evaluated by expanding $\cosh(u)$ into exponential and using Euler's gamma-integral:
\begin{align*}\int_0^\infty \frac{u^s}{2\cosh(u)} \mathrm{d}u &= \int_0^\infty u^s \frac{\exp(-u)}{1+\exp(-2u)}\mathrm{d}u = \sum_{n=0}^\infty (-1)^n \int_0^\infty u^{s} \exp(-(2n+1)u) \,\mathrm{d}u \\&= \Gamma(s+1)  \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^{s+1}} = \Gamma(s+1) 2^{-2s-2} \left( \zeta\left(s+1,\frac{1}{4}\right) - \zeta\left(s+1,\frac{3}{4}\right) \right)\end{align*}
Near $s=0$:
$$\zeta(1+s,a) = \frac{1}{s} - \psi(a) - \gamma_1(a) s + \mathcal{o}(s)$$
where $\psi(a)$ is the digamma function (http://en.wikipedia.org/wiki/Digamma_function), and $\gamma_1(a)$ is the first generalized Stieltjes constant (http://en.wikipedia.org/wiki/Stieltjes_constants). Differentiating and taking the limit we have
\begin{align*}\int_{\pi/4}^{\pi/2} \ln( \ln(\tan x))\, \mathrm{d}x &= \frac{1}{4} \left( \left(\log(4) + \gamma\right)\left(\psi\left(\frac{1}{4}\right)-\psi\left(\frac{3}{4}\right)\right) - \left(\psi_1\left(\frac{1}{4}\right)-\psi_1\left(\frac{3}{4}\right)\right)\right) \\&= \frac{\pi}{4} \log \left( \frac{4 \pi^3}{\Gamma\left(\frac{1}{4}\right)^4} \right) \approx -0.260443\end{align*}
where the latter equality is given my Mathematica.

求\[\int_{0}^{\infty}\sin x \sin \sqrt{x}\,dx.\]


这个积分用Alpha给的是发散结果.下面的计算是论坛的结果:

I'm posting an asnwer (of the $2$ I have) using real analysis methods:

 

\begin{align*}\int_{0}^{\infty}\sin x \sin \sqrt{x}\,dx &\overset{\sqrt{x}=u}{=\! =\! =\!}2\int_{0}^{\infty}u\sin u \sin u^2 \,du \\&=-\int_{0}^{\infty}u\cos \left ( u^2+u \right )\,du+\int_{0}^{\infty}u\cos(u^2-u)\,du \\&\overset{u \mapsto u+1}{=\! =\! =\! =\!}-\int_{0}^{\infty}u\cos(u^2+u)\,du+\int_{-1}^{\infty}\left ( u+1 \right )\cos\left ( u^2+u \right )\,du \\&= \int_{0}^{\infty}\cos\left ( u^2+u \right )\,du+\int_{-1}^{0}\left ( u+1 \right )\cos\left ( u^2+u \right )\,du\\&\overset{u={\rm v}-\frac{1}{2}}{=\! =\! =\! =\!}\int_{1/2}^{\infty}\cos\left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v}+\int_{-1/2}^{1/2}\left ( {\rm v}+\frac{1}{2} \right )\cos \left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v} \\&= \int_{0}^{\infty}\cos \left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v}+\\& \left [ \int_{-1/2}^{1/2}{\rm v}\cos \left ( {\rm v}^2-\frac{1}{4} \right )\,d{\rm v}+\frac{1}{2}\int_{-1/2}^{0}\cos\left ( {\rm v}^2-\frac{1}{2} \right )\,d{\rm v}- \frac{1}{2}\int_{0}^{1/2}\cos\left ( {\rm v}^2-\frac{1}{2} \right )\,d{\rm v} \right ]\end{align*}

 

However, the equation in the bracket equals zero due to symmetry.

 

Hence:

\begin{align*}\int_{0}^{\infty}\sin x \sin x^2\,dx&=\int_{0}^{\infty}\cos\left ( {\rm v}^2-\frac{1}{2} \right )\,d{\rm v}\\&=\cos \frac{1}{4}\int_{0}^{\infty}\cos {\rm v}^2\,d{\rm v}+\sin \frac{1}{4}\int_{0}^{\infty}\sin {\rm v}^2\,d{\rm v}\\&\overset{(*)}{=}\int_{0}^{\infty}\sin {\rm v}^2\,d{\rm v}\left ( \cos \frac{1}{4}+\sin \frac{1}{4} \right )\\&=\frac{\sqrt{\pi}}{2}\sin \left ( \frac{3\pi-1}{4} \right )\;\; \;\;\;\; \square\end{align*}

 

$(*)$ We used the Frensel integrals stating that $\displaystyle \int_{0}^{\infty}\cos x^2 \,dx=\int_{0}^{\infty}\sin x^2 \,dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}$. 

 

一个初三竞赛几何题

博士论坛群里ZHX问的一道平面几何题,由余神提示,ZHX自己整理的解答:

2013武大数分压轴题

(13年武大数分)求$\displaystyle I = \iint\limits_\Sigma  {{{\left( {{x^2} + {y^2} + {z^2}} \right)}^{ - \frac{3}{2}}}{{\left( {\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} + \frac{{{z^2}}}{{{c^4}}}} \right)}^{ - \frac{1}{2}}}dS} $,其中$\sum$为椭球面: $\displaystyle \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1(a,b,c>0)$.


下面是自己的解答:

令$x = a\sin \varphi \cos \theta ,y = b\sin \varphi \sin \theta ,z = c\cos \varphi $,其中$0\leq \theta\leq 2\pi,0\leq\varphi \leq\pi$,经计算得到
\[\frac{{\partial \left( {y,z} \right)}}{{\partial \left( {\varphi ,\theta } \right)}} = bc{\sin ^2}\varphi \cos \theta ,\frac{{\partial \left( {z,x} \right)}}{{\partial \left( {\varphi ,\theta } \right)}} = ac{\sin ^2}\varphi \sin \theta ,\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {\varphi ,\theta } \right)}} = ab\sin \varphi \cos \varphi ,\]
所以
\begin{align*}EG - {F^2} &= {\left( {\frac{{\partial \left( {y,z} \right)}}{{\partial \left( {\varphi ,\theta } \right)}}} \right)^2} + {\left( {\frac{{\partial \left( {z,x} \right)}}{{\partial \left( {\varphi ,\theta } \right)}}} \right)^2} + {\left( {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {\varphi ,\theta } \right)}}} \right)^2}\\& = {\left( {abc} \right)^2}{\sin ^2}\varphi \left( {\frac{{{{\sin }^2}\varphi {{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\varphi {{\sin }^2}\theta }}{{{b^2}}} + \frac{{{{\cos }^2}\varphi }}{{{c^2}}}} \right).\end{align*}
而这时被积函数化为
\begin{align*}&{\left( {{x^2} + {y^2} + {z^2}} \right)^{ - \frac{3}{2}}}{\left( {\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}} + \frac{{{z^2}}}{{{c^4}}}} \right)^{ - \frac{1}{2}}}\\= &{\left( {{a^2}{{\sin }^2}\varphi {{\cos }^2}\theta  + {b^2}{{\sin }^2}\varphi {{\sin }^2}\theta  + {c^2}{{\cos }^2}\varphi } \right)^{ - \frac{3}{2}}}\\&{\left( {\frac{{{{\sin }^2}\varphi {{\cos }^2}\theta }}{{{a^2}}} + \frac{{{{\sin }^2}\varphi {{\sin }^2}\theta }}{{{b^2}}} + \frac{{{{\cos }^2}\varphi }}{{{c^2}}}} \right)^{ - \frac{1}{2}}}.\end{align*}
因此
\[I = abc\iint\limits_{\left[ {0,\pi } \right] \times \left[ {0,2\pi } \right]} {{{\left( {{a^2}{{\sin }^2}\varphi {{\cos }^2}\theta  + {b^2}{{\sin }^2}\varphi {{\sin }^2}\theta  + {c^2}{{\cos }^2}\varphi } \right)}^{ - \frac{3}{2}}}\sin \varphi d\varphi d\theta } \]
 
注意到这么一个事实,当$M+Nx^2$不取$0$且$M\neq 0$时,我们有
\[\int {{{\left( {M + N{x^2}} \right)}^{ - 3/2}}dx}  = \frac{1}{M} \cdot \frac{x}{{\sqrt {M + N{x^2}} }} + C.\]
 
\begin{align*}I &= abc\int_0^{2\pi } {d\theta } \int_0^\pi  {{{\left( {{a^2}{{\sin }^2}\varphi {{\cos }^2}\theta  + {b^2}{{\sin }^2}\varphi {{\sin }^2}\theta  + {c^2}{{\cos }^2}\varphi } \right)}^{ - \frac{3}{2}}}\sin \varphi d\varphi } \\&=  - abc\int_0^{2\pi } {d\theta } \int_0^\pi  {{{\left( {{a^2}{{\sin }^2}\varphi {{\cos }^2}\theta  + {b^2}{{\sin }^2}\varphi {{\sin }^2}\theta  + {c^2}{{\cos }^2}\varphi } \right)}^{ - \frac{3}{2}}}d\left( {\cos \varphi } \right)} \\&=  - abc\int_0^{2\pi } {d\theta } \int_0^\pi  {{{\left[ {\left( {{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta } \right) + \left( {{c^2} - {a^2}{{\cos }^2}\theta  - {b^2}{{\sin }^2}\theta } \right){{\cos }^2}\varphi } \right]}^{ - \frac{3}{2}}}d\left( {\cos \varphi } \right)} \\&= abc\int_0^{2\pi } {d\theta } \int_{ - 1}^1 {{{\left[ {\left( {{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta } \right) + \left( {{c^2} - {a^2}{{\cos }^2}\theta  - {b^2}{{\sin }^2}\theta } \right){x^2}} \right]}^{ - \frac{3}{2}}}dx} \\&= abc\int_0^{2\pi } {\frac{2}{{\left( {{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta } \right)c}}d\theta }  = 4ab\int_0^\pi  {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta } .\end{align*}
\begin{align*}&\int_0^\pi  {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta }  = \int_0^{\frac{\pi }{2}} {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta }  + \int_{\frac{\pi }{2}}^\pi  {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta } \\= &\int_0^{\frac{\pi }{2}} {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta }  + \int_0^{\frac{\pi }{2}} {\frac{1}{{{a^2}{{\sin }^2}\theta  + {b^2}{{\cos }^2}\theta }}d\theta } \\= &\int_0^{ + \infty } {\frac{1}{{{a^2} + {b^2}{x^2}}}d}  + \int_0^{ + \infty } {\frac{1}{{{a^2}{x^2} + {b^2}}}dx}  = \frac{1}{{ab}}\left. {\arctan \left( {\frac{b}{a}x} \right)} \right|_0^{ + \infty } + \frac{1}{{ab}}\left. {\arctan \left( {\frac{a}{b}x} \right)} \right|_0^{ + \infty }\\= &\frac{\pi }{{ab}}.\end{align*}
进而得到
\[I = 4ab\int_0^\pi  {\frac{1}{{{a^2}{{\cos }^2}\theta  + {b^2}{{\sin }^2}\theta }}d\theta }  = 4\pi .\]

另外有更好的方法:(Hansschwarzkopf)

注意到$\Sigma$ 在点$(x,y,z)$处的单位外法向量是
$$n=\frac{\left(\frac{x}{a^2},\frac{y}{b^2},\frac{z}{c^2}\right)}{\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}},$$
且$1=x\cdot \frac{x}{a^2}+y\cdot\frac{y}{b^2}+z\cdot\frac{z}{c^2}$.
从而原积分可写成第二型曲面积分
$$\iint\limits_\Sigma \frac{xdydz+yd zd x+zdxdy}{\sqrt{(x^2+y^2+z^2)^3}}.$$
作小球面$S_\varepsilon: x^2+y^2+z^2=\varepsilon^2$. 运用Gauss公式可知
$$\iint\limits_\Sigma \frac{xd yd z+yd zd x+zd xd y}{\sqrt{(x^2+y^2+z^2)^3}} =\iint\limits_{S_\varepsilon} \frac{xdyd z+ydzd x+zd xd y}{\sqrt{(x^2+y^2+z^2)^3}}=4\pi.$$ 即
$$\iint\limits_\Sigma\frac{d S}{\sqrt{(x^2+y^2+z^2)^3}\sqrt{\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4}}}=4\pi.$$

 

一道杂志征解题的解答

这道题来自MAA的杂志The American Mathematical Monthly, Vol. 122, No. 5 (May 2015), pp. 500-507,可以参考链接http://www.jstor.org/stable/10.4169/amer.math.monthly.122.5.500?seq=1#page_scan_tab_contents


求\[\int_0^\infty  {\frac{1}{x}dx} \int_0^x {\frac{{\cos \left( {x - y} \right) - \cos x}}{y}dy} .\]


解.(翻译而来)令$f\left( {x,y} \right) = \frac{{\cos \left( {x - y} \right) - \cos x}}{y}$.对$x>0$,我们有\[\int_0^x {f\left( {x,y} \right)dy}  = \int_0^1 {\frac{{\cos \left( {1 - t} \right)x - \cos x}}{y}dt}  = x\int_0^1 {\frac{1}{t}} \int_{1 - t}^1 {\sin ux\, dudt} .\]

因而对$R>0$,

\[\int_0^R {\frac{1}{x}} \int_0^x {f\left( {x,y} \right)\,dydx}  = \int_0^R {\int_0^1 {\frac{1}{t}} \int_{1 - t}^1 {\sin ux\,dudtdx} } .\]

而$|\sin ux|\leq1$,该三重积分是绝对收敛的.由Fubini定理可知积分能交换次序

\begin{align*}&\int_0^R {\frac{1}{x}} \int_0^x {f\left( {x,y} \right)\,dydx}  = \int_0^1 {\int_{1 - t}^1 {\frac{1}{t}} \int_0^R {\sin ux\,dxdudt} } \\=& \int_0^1 {\frac{{1 - \cos Ru}}{u}\int_{1 - u}^1 {\frac{1}{t}} \,dtdu} \\= & - \int_0^1 {\frac{{\ln \left( {1 - u} \right)}}{u}\,du}  + \int_0^1 {\frac{{\ln \left( {1 - u} \right)}}{u}\cos Ru\,du} .\end{align*}
我们知$|\ln (1-u)/u|\in L^1([0,1])$,由Riemann-Lebesgue引理可知
\[\mathop {\lim }\limits_{R \to \infty } \int_0^1 {\frac{{\ln \left( {1 - u} \right)}}{u}\cos Ru\,du}  = 0.\]
由于${\sum\limits_{n = 1}^\infty  {\frac{{{t^{n - 1}}}}{n}} }$一致收敛,故可逐项积分.因此我们有
\begin{align*}&\mathop {\lim }\limits_{R \to \infty } \int_0^R {\frac{1}{x}} \int_0^x {f\left( {x,y} \right)dydx}  =  - \int_0^1 {\frac{{\ln \left( {1 - u} \right)}}{u}du} \\= &\int_0^1 {\sum\limits_{n = 1}^\infty  {\frac{{{t^{n - 1}}}}{n}} dt}  = \sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}}  = \frac{{{\pi ^2}}}{6}.\end{align*}

 

哆嗒数学网里代数龙发的一系列级数题

练习题1.证明:$$\sum\limits_{n=1}^{\infty}\frac{1}{(n+1)\sqrt[p]{n}}\leq p,\,\,(p\ge1).$$

证:由Lagrange中值定理,我们有

\[\sqrt[p]{{n + 1}} - \sqrt[p]{n} = \frac{1}{p}{\xi ^{1/p - 1}} \ge \frac{1}{p}{\left( {n + 1} \right)^{1/p - 1}},\quad \xi  \in \left( {n,n + 1} \right).\]

因此\[\frac{1}{{\left( {n + 1} \right)\sqrt[p]{n}}} = \frac{{\sqrt[p]{{n + 1}} - \sqrt[p]{n}}}{{\sqrt[p]{n} \cdot \sqrt[p]{{n + 1}}}} \cdot \frac{{{{\left( {n + 1} \right)}^{1/p - 1}}}}{{\sqrt[p]{{n + 1}} - \sqrt[p]{n}}} \le p\frac{{\sqrt[p]{{n + 1}} - \sqrt[p]{n}}}{{\sqrt[p]{n} \cdot \sqrt[p]{{n + 1}}}} = p\left( {\frac{1}{{\sqrt[p]{n}}} - \frac{1}{{\sqrt[p]{{n + 1}}}}} \right).\]

立即有

\[\sum\limits_{n = 1}^\infty  {\frac{1}{{\left( {n + 1} \right)\sqrt[p]{n}}}}  \le p\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{\sqrt[p]{n}}} - \frac{1}{{\sqrt[p]{{n + 1}}}}} \right)}  = p.\]


练习题2.设$\displaystyle S_n=\sum\limits_{k=1}^{n}a_k, p>1,c>1$,证明:$$\sum\limits_{n=1}^{\infty}\frac{S_n^p}{n^c}\le K\sum\limits_{n=1}^{\infty}\frac{(na_n)^p}{n^c},$$并求出$K$的最优值.


练习题3.设$a_n$是有界的正数列,$p>0$,证明:

$$\frac{1}{a_1^p}+\sum\limits_{n=1}^{\infty}\frac{a_1a_2 \cdots a_n}{a_{n+1}^p} \ge \sum\limits_{n=0}^{\infty}(\frac{p}{p+1})^{n-p}.$$

练习题4.设$(0,+\infty)$上的函数列$f_n$由下式定义:$$f_1(x)=x,f_{n+1}(x)=(f_n(x)+\frac{1}{n})f_n(x).$$证明:存在唯一的正数$a$,使得对于所有$n$,$$0<f_n(x)<f_{n+1}(a)<1.$$


练习题5.$\displaystyle\sum\limits_{n=1}^{\infty}a_n$为正项收敛级数,$\displaystyle r_n=\sum\limits_{k=n}^{\infty}a_k,0<p<1$,证明:$$\sum\limits_{n=1}^{\infty}\frac{a_n}{r_n^p}<\frac{1}{1-p}\left(\sum\limits_{n=1}^{\infty}a_n \right)^{1-p}.$$


练习题6.设$a>0,a_n$是一个数列,并且$a_n>0,a_{n+1}\ge a_n$,证明:$$\sum\limits_{n=1}^{\infty}\frac{a_n-a_{n-1}}{a_na_{n-1}^a}$$收敛.

证:首先可以确定给定的级数是正项级数.

(1)当$0<a<1$时,我们利用Lagrange中值定理,有\[\frac{{a_n^a - a_{n - 1}^a}}{{{a_n} - {a_{n - 1}}}} = a{\xi ^{a - 1}} \ge aa_n^{a - 1},\quad \xi  \in \left( {{a_{n - 1}},{a_n}} \right).\]

因此\[\frac{{{a_n} - {a_{n - 1}}}}{{{a_n}a_{n - 1}^a}} = \frac{{a_n^a - a_{n - 1}^a}}{{a_n^aa_{n - 1}^a}} \cdot \left( {\frac{{{a_n} - {a_{n - 1}}}}{{a_n^a - a_{n - 1}^a}} \cdot a_n^{a - 1}} \right) \le \frac{1}{a}\frac{{a_n^a - a_{n - 1}^a}}{{a_n^aa_{n - 1}^a}} = \frac{1}{a}\left( {\frac{1}{{a_{n - 1}^a}} - \frac{1}{{a_n^a}}} \right).\]

故\[\sum\limits_{n = 1}^\infty  {\frac{{{a_n} - {a_{n - 1}}}}{{{a_n}a_{n - 1}^a}}}  \le \frac{1}{a}\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{a_{n - 1}^a}} - \frac{1}{{a_n^a}}} \right)}  = \frac{1}{a}\left( {\frac{1}{{a_0^a}} - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{a_n^a}}} \right).\]

由于$\{a_n\}$是单增的正数列,则${\mathop {\lim }\limits_{n \to \infty } \frac{1}{{a_n^a}}}$必定存在,由此可知原正项级数收敛;

(2)当$a\geq1$时,由\[\sum\limits_{n = 1}^\infty  {\frac{{{a_n} - {a_{n - 1}}}}{{{a_n}a_{n - 1}^a}}}  = \sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{a_{n - 1}^a}} - \frac{{a_{n - 1}^{1 - a}}}{{{a_n}}}} \right)}  \le \sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{a_{n - 1}^a}} - \frac{1}{{a_n^a}}} \right)}  = \frac{1}{a}\left( {\frac{1}{{a_0^a}} - \mathop {\lim }\limits_{n \to \infty } \frac{1}{{a_n^a}}} \right)\]同样可知原正项级数收敛.

综上,级数$\sum\limits_{n=1}^{\infty}\frac{a_n-a_{n-1}}{a_na_{n-1}^a}$收敛.


练习题7.设$\displaystyle S(x)=\sum\limits_{n=1}^{\infty}\frac{2n}{(n^2 +x^2)^2}$,证明:$$\frac{1}{x^2 +\frac{1}{2\zeta(3)}}<S(x)<\frac{1}{x^2 +\frac{1}{6}},$$其中$\displaystyle \zeta(3)=\sum\limits_{n=1}^{\infty}\frac{1}{n^3}.$


练习题8.给定序列$\{a_n\}$,且$a_n$满足$a_1=2,a_2=8,a_n=4a_{n-1}-a_{n-2}(n=3,4,\ldots)$,证明:$$\sum\limits_{n=1}^{\infty}\text{arccot}\,\,a_n^2=\frac{\pi}{12}.$$


证.由${a_n} + {a_{n - 2}} = 4{a_{n - 1}}$可知\[{a_n}\left( {{a_n} + {a_{n - 2}}} \right) = 4{a_{n - 1}}{a_n} = {a_{n - 1}}\left( {{a_{n + 1}} + {a_{n - 1}}} \right),\]递推得\[a_n^2 - {a_{n + 1}}{a_{n - 1}} = a_{n - 1}^2 - {a_n}{a_{n - 2}} = \cdots = a_2^2 - {a_3}{a_1} = 4.\]

 

注意到$\mathrm{arccot\,} x$的一个公式

\[\mathrm{arccot\,} x-\mathrm{arccot\,} y=\mathrm{arccot\,}\left( \frac{1+xy}{y-x}\right).\]

因此有

\begin{align*}\mathrm{arccot\,} a_n^2 &= \mathrm{arccot\,} \frac{{{a_n} \cdot 4{a_n}}}{4} = \mathrm{arccot\,} \frac{{{a_n}\left( {{a_{n + 1}} + {a_{n - 1}}} \right)}}{4} =\mathrm{arccot\,} \frac{{{a_n}\left( {{a_{n + 1}} + {a_{n - 1}}} \right)}}{{a_n^2 - {a_{n + 1}}{a_{n - 1}}}}\\& = \mathrm{arccot\,} \frac{{1 + \frac{{{a_{n + 1}}}}{{{a_{n - 1}}}}}}{{\frac{{{a_n}}}{{{a_{n - 1}}}} - \frac{{{a_{n + 1}}}}{{{a_n}}}}} = \mathrm{arccot\,} \frac{{{a_{n + 1}}}}{{{a_n}}} -\mathrm{arccot\,} \frac{{{a_n}}}{{{a_{n - 1}}}}.\end{align*}

易得\[\mathop {\lim }\limits_{n \to \infty } \frac{{{a_{n + 1}}}}{{{a_n}}} = 2 + \sqrt 3 .\]

\[\sum\limits_{n = 1}^\infty {\mathrm{arccot\,} a_n^2} = \mathrm{arccot\,} a_1^2 + \sum\limits_{n = 2}^\infty {\mathrm{arccot\,} a_n^2} = \mathop {\lim }\limits_{n \to \infty } \mathrm{arccot\,} \frac{{{a_{n + 1}}}}{{{a_n}}} - \mathrm{arccot\,} \frac{{{a_2}}}{{{a_1}}} + \mathrm{arccot\,} a_1^2 = \frac{\pi }{{12}}.\]


练习题9.设$\displaystyle a_n=\arctan \frac{1}{n^2 +n +1}$,证明: $$\sum\limits_{k=1}^{\infty}\frac{a_k^{1/2}}{k^2} \le \sqrt{\frac{\pi}{3}}.$$


证.注意到

\begin{align*}\sum\limits_{k = 1}^\infty  {{a_k}}  &= \sum\limits_{k = 1}^\infty  {\arctan \frac{1}{{{k^2} + k + 1}}}  = \sum\limits_{k = 1}^\infty  {\left( {\arctan \frac{1}{k} - \arctan \frac{1}{{k + 1}}} \right)}  = \frac{\pi }{4}\\\sum\limits_{k = 1}^\infty  {\frac{1}{{{k^4}}}}  &= \zeta \left( 4 \right) = \frac{{{\pi ^4}}}{{90}}.\end{align*}

由Cauchy-Schwarz不等式可知

\[\sum\limits_{k = 1}^N {\frac{1}{{{k^4}}}}  \cdot \sum\limits_{k = 1}^N {{a_k}}  \ge {\left( {\sum\limits_{k = 1}^N {\frac{{a_k^{1/2}}}{{{k^2}}}} } \right)^2}.\]

令$N\to\infty$,我们有\[\sum\limits_{k = 1}^\infty  {\frac{{a_k^{1/2}}}{{{k^2}}}}  \le \sqrt {\frac{{{\pi ^4}}}{{90}} \cdot \frac{\pi }{4}}  = \sqrt {\frac{{{\pi ^5}}}{{360}}}  < \sqrt {\frac{\pi }{3}} .\]

也可通过放缩实现\[\sum\limits_{k = 1}^\infty  {\frac{1}{{{k^4}}}}  = 1 + \sum\limits_{k = 2}^\infty  {\frac{1}{{{k^4}}}}  < 1 + \sum\limits_{k = 2}^\infty  {\frac{1}{{\left( {k + 1} \right)\left( {k + 2} \right)}}}  = \frac{4}{3}.\]


练习题10.设$\displaystyle a_n > 0, S_n=\sum\limits_{k=1}^na_k$,证明:

(1)$\displaystyle\sum\limits_{n=1}^{\infty}\frac{n}{S_n} \le 2 \sum\limits_{n=1}^{\infty}\frac{1}{a_n}$;
(2)$\displaystyle\sum\limits_{n=1}^{\infty}\frac{2n+1}{S_n} \le 4 \sum\limits_{n=1}^{\infty}\frac{1}{a_n}$.

证.(1)由柯西不等式我们得

\[\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} \sum\limits_{m = 1}^n {{a_m}}  \ge {\left( {1 + 2 +  \cdots  + n} \right)^2} = \frac{1}{4}{n^2}{\left( {n + 1} \right)^2},\]

即\[\frac{n}{{{a_1} + {a_2} +  \cdots  + {a_n}}} \le \frac{4}{{n{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} .\]

因此

\begin{align*}\sum\limits_{n = 1}^\infty  {\frac{n}{{{a_1} + {a_2} +  \cdots  + {a_n}}}}  &\le 4\sum\limits_{n = 1}^\infty  {\frac{1}{{n{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} }  = 4\sum\limits_{m = 1}^\infty  {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty  {\frac{1}{{n{{\left( {n + 1} \right)}^2}}}} } \\&\le 4\sum\limits_{m = 1}^\infty  {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty  {\frac{1}{2}\left[ {\frac{1}{{{n^2}}} - \frac{1}{{{{\left( {n + 1} \right)}^2}}}} \right]} }  = 2\sum\limits_{m = 1}^\infty  {\frac{1}{{{a_m}}}} .\end{align*}

这里用到了\[\frac{1}{{n{{\left( {n + 1} \right)}^2}}} \le \frac{1}{2}\frac{{2n + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}} = \frac{1}{2}\left[ {\frac{1}{{{n^2}}} - \frac{1}{{{{\left( {n + 1} \right)}^2}}}} \right].\]

 

注意到$a_n=n^\alpha,\alpha>1$时有

\[\mathop {\lim }\limits_{\alpha \to 1} \frac{{\sum\limits_{n = 1}^\infty {\frac{n}{{{a_1} + {a_2} + \cdots + {a_n}}}} }}{{\sum\limits_{j = 1}^\infty {\frac{1}{{{a_j}}}} }} = \mathop {\lim }\limits_{\alpha \to 1} \mathop {\lim }\limits_{N \to \infty } \frac{{\sum\limits_{n = 1}^N {\frac{n}{{{1^\alpha } + {2^\alpha } + \cdots + {n^\alpha }}}} }}{{\sum\limits_{n = 1}^N {\frac{1}{{{n^\alpha }}}} }} = \mathop {\lim }\limits_{\alpha \to 1} \mathop {\lim }\limits_{N \to \infty } \frac{{\frac{N}{{\frac{1}{{\alpha + 1}}{N^{\alpha + 1}} + O\left( {{N^\alpha }} \right)}}}}{{\frac{1}{{{N^\alpha }}}}} = 2.\]

(2)如法炮制.由柯西不等式我们得

\[\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} \sum\limits_{m = 1}^n {{a_m}} \ge {\left( {1 + 2 + \cdots + n} \right)^2} = \frac{1}{4}{n^2}{\left( {n + 1} \right)^2},\]

\[\frac{{2n + 1}}{{{a_1} + {a_2} + \cdots + {a_n}}} \le \frac{{4\left( {2n + 1} \right)}}{{{n^2}{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} .\]

因此

\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{2n + 1}}{{{a_1} + {a_2} + \cdots + {a_n}}}} &\le 4\sum\limits_{n = 1}^\infty {\frac{{2n + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}}\sum\limits_{m = 1}^n {\frac{{{m^2}}}{{{a_m}}}} } = 4\sum\limits_{m = 1}^\infty {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty {\frac{{2n + 1}}{{{n^2}{{\left( {n + 1} \right)}^2}}}} } \\&= 4\sum\limits_{m = 1}^\infty {\frac{{{m^2}}}{{{a_m}}}\sum\limits_{n = m}^\infty {\left[ {\frac{1}{{{n^2}}} - \frac{1}{{{{\left( {n + 1} \right)}^2}}}} \right]} } = 4\sum\limits_{m = 1}^\infty {\frac{1}{{{a_m}}}}.\end{align*}


练习题11.设$\displaystyle a_n \ge 0, n=1,2,\ldots,\sum\limits_{n=1}^{\infty}a_n < \infty$,证明:

$$\sum\limits_{n=1}^{\infty}(a_1a_2\cdots a_n)^{\frac{1}{n}} \le e \sum\limits_{n=1}^{\infty}a_n$$,且证明$e$是最优值.

此题再拓展下求证:$$\sum\limits_{n=1}^{\infty}(a_1a_2\cdots a_n)^{\frac{1}{n}} \le e \sum\limits_{n=1}^{\infty}[1-\frac{1}{2(n+1)}]a_n.$$

 


练习题12.如果正项级数$\displaystyle \sum\limits_{n=1}^{\infty}\frac{1}{p_n}$收敛,证明:级数$\displaystyle \sum\limits_{n=1}^{\infty}\frac{n^2}{(p_1+p_2+\cdots+p_n)^2}p_n$也收敛.


练习题13.设$\displaystyle \sum\limits_{n=1}^{\infty}a_n$为正项级数,且$\displaystyle \sum\limits_{k=1}^{n}(a_k-a_n)$对$n$有界,$a_n$单调递减趋于$0$,证明:级数$\displaystyle \sum\limits_{n=1}^{\infty}a_n$收敛.


练习题14.设级数$\displaystyle \sum\limits_{n=1}^{\infty}a_n$收敛, $\displaystyle \sum\limits_{n=1}^{\infty}(b_{n+1}-b_n)$绝对收敛,证明:级数$\displaystyle \sum\limits_{n=1}^{\infty}a_nb_n$收敛.


练习题15.设$a_n>0,\left\{ a_n-a_{n+1}\right\}$为一个严格递减的数列.如果$\sum_{n=1}^{\infty}a_n$收敛。试证:$$\lim\limits_{n \to \infty}\left( \cfrac{1}{a_{n+1}}-\cfrac{1}{a_n}\right)=+\infty.$$


练习题16.能否构造一个收敛数列$\sum\limits_{n=1}^{\infty}a_n$,使得级数$\sum\limits_{n=1}^{\infty}a_n^3$发散.


练习题17.设$\lim \limits_{n\rightarrow +\infty}x_n=+\infty$,正项级数$\sum\limits_{n=1}^{\infty}y_n$收敛,设$n_0$是某一自然数,

若当$n>n_0$时有$x_n <x_{n+1},x_n< \frac{1}{2}(x_{n-1}+x_{n+1}),y_{n+1}< y_n$,
求证:$$\lim \limits_{n\rightarrow +\infty}\frac{x_ny_n}{x_{n+1}-x_n}=0.$$

练习题18.设$\sum\limits_{n=1}^{\infty}a_n$是一正项收敛级数,且有$a_{n+1}< \frac{1}{2}(a_n+a_{n+2}),\,\frac{1}{a_{n+2}}-\frac{1}{a_{n+1}}\le \frac{1}{3}(\frac{1}{a_{n+3}}-\frac{1}{a_{n}})$,

求极限$$\lim\limits_{n \rightarrow +\infty}\frac{\displaystyle a_na_{n+2}(a_n-a_{n+1})}{\displaystyle a_na_{n+1}-2a_na_{n+2}+a_{n+1}a_{n+2}}.$$

裴礼文上的一道积分不等式

证明:对$n\geq 3$有$$\int_{0}^{\frac{\pi}{2}}\left|\frac{\sin{(2n+1)t}}{\sin{t}}\right|dt<\pi\left(1+\frac{\ln{n}}{2}\right).$$


Poof.For all $x$,

$$\left|\frac{\sin((2n+1)x)}{\sin(x)}\right|=\left|\sum_{k=-n}^{n}e^{i2kx}\right|\le2n+1$$
Note that $\displaystyle\sum_{k=-n}^{n}e^{i2kx}=1+2\sum_{k=1}^n\cos(2kx)$.
 
For $0\le x\le\pi/2$, we have $\sin(x)\ge2x/\pi$. Therefore,
$$\left|\frac{\sin((2n+1)x)}{\sin(x)}\right|\le\frac\pi{2x}$$
Thus,
\begin{align*}\int_0^{\pi/2}\left|\frac{\sin((2n+1)x)}{\sin(x)}\right|\,\mathrm{d}x&\le\int_0^{\pi/(4n+2)}(2n+1)\,\mathrm{d}x+\int_{\pi/(4n+2)}^{\pi/2}\frac\pi{2x}\,\mathrm{d}x\\&=\frac\pi2+\frac\pi2\log\left(2n+1\right)\end{align*}
For $n\ge3$, $2n+1\le\frac73n$. Therefore,
\begin{align*}\frac\pi2+\frac\pi2\log\left(2n+1\right)&\le\frac\pi2\left(1+\log\left(\frac73\right)+\log(n)\right)\\[6pt]&\le\pi\left(1+\frac{\log(n)}{2}\right).\end{align*}

Show that for $p>1$ and $x \ge 0$,$$\dfrac{2}{\pi}\int_{x}^{px}\left(\dfrac{\sin{t}}{t}\right)^2\,\mathrm dt\le 1-\dfrac{1}{p}$$

Proof.This is quite a difficult problem, and I found it very enjoyable.  Here is the solution I found:  

First, we give some simple bounds when $x$ is large, or $px$ is small.  If $x\geq\frac{2}{\pi},$ then by using the bound $|\sin(t)|\leq1$,
we have that
$$\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt\leq\frac{2}{\pi}\int_{x}^{px}\frac{1}{t^{2}}dt=\frac{2}{\pi x}\left(1-\frac{1}{p}\right)\leq1-\frac{1}{p}.$$
Similarly, if $px\leq\frac{\pi}{2}$, then since $\frac{\text{sin}(t)}{t}\leq1$,
it follows that 
$$\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt\leq\frac{2}{\pi}\left(px-x\right)=\frac{2xp}{\pi}\left(1-\frac{1}{p}\right)\leq\left(1-\frac{1}{p}\right).$$
Now, assume that $0\leq x\leq\frac{2}{\pi}$, and that $px\geq\frac{\pi}{2}$.
Then notice that 
$$\frac{2}{\pi}\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt=1-\frac{2}{\pi}\int_{px}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt-\frac{2}{\pi}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt$$
since $\int_{0}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt=\frac{\pi}{2}$.
We will now find a bound on the other two terms. Working over an interval
of length $\pi$, by pulling out a lower bound for $\frac{1}{t^{2}}$,
we have that for any $y$ 
 
$$\int_{y}^{y+\pi}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{1}{\left(y+\pi\right)^{2}}\int_{0}^{\pi}\sin^{2}(t)dt\geq\frac{\pi}{2}\int_{y+\pi}^{y+2\pi}\frac{1}{t^{2}}dt,$$
and so 
$$\frac{2}{\pi}\int_{px}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\int_{px+\pi}^{\infty}\frac{1}{t^{2}}dt=\frac{1}{px+\pi}.$$
Since the function $\frac{\sin(t)}{t}$ is monotonically decreasing
on the interval $\left[0,\frac{2}{\pi}\right],$ it follows that for
$x\leq\frac{2}{\pi}$ we have 
 
 
$$\frac{1}{x}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{\pi}{2}\int_{0}^{\frac{2}{\pi}}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{\pi}{2}\cdot\frac{5}{3\pi},$$
 
and hence 
 
$$\frac{2}{\pi}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{5x}{3\pi}.$$ 
 
Now, notice that since $px\geq\frac{\pi}{2},$ and $p>1$, by plugging them in directly, we have that 
 
$$\frac{5\left(xp\right)^{2}}{3\pi}+\frac{2}{3}px+p-\pi>\frac{5\pi}{12}+\frac{\pi}{3}+1-\pi=1-\frac{\pi}{4}>0.$$ 
 
Rearranging the above by dividing through by both $(px+\pi)$  and $p$, we obtain the inequality
 
$$\frac{5}{3\pi}x+\frac{1}{px+\pi}>\frac{1}{p},$$ 
for $px\geq\frac{\pi}{2}$, and $p>1$. It then follows that
$$\frac{2}{\pi}\int_{px}^{\infty}\left(\frac{\sin(t)}{t}\right)^{2}dt+\frac{2}{\pi}\int_{0}^{x}\left(\frac{\sin(t)}{t}\right)^{2}dt\geq\frac{1}{p},$$
 for $x\leq\frac{2}{\pi},$ and $px\geq\frac{\pi}{2}$, and hence
we have shown that for all $x\geq0$, and all $p>1$, 
$$\int_{x}^{px}\left(\frac{\sin(t)}{t}\right)^{2}dt\leq1-\frac{1}{p},$$
as desired. 

Show that if $f\in \mathcal C^{n+1}([a,b])$ and $f(a)=f^{'}(a)=\cdots=f^\left(n\right)(a)=0,$ then the following statements are ture:
 
$\mathbf a)$
 
$ \forall r\in[1,\infty),$the inequality $$\left(\int_{a}^{b}|f(x)|^rdx\right)^{\frac{1}{r}} \leq \frac{(b-a)^{n+\frac{1}{r}}}{n!(nr+1)^{\frac{1}{r}}}\int_{a}^{b}|f^{(n+1)}(x)|dx$$holds.
 
$\mathbf b)$
 
$ \forall r\in[1,\infty),$the inequality $$\left(\int_{a}^{b}|f(x)|^rdx\right)^{\frac{1}{r}} \leq \frac{2^{\frac{1}{r}}(b-a)^{n+\frac{1}{r}+\frac{1}{2}}}{n!\sqrt{2n+1}(2nr+r+1)^{\frac{1}{r}}}\left(\int_{a}^{b}|f^{(n+1)}(x)|^{2}dx\right)^{\frac{1}{2}}$$holds.

Proof.We have by Taylor's Theorem with Integral form of the Remainder

\begin{align*}f(x) = \int_a^x\dfrac{f^{(n+1)}(t)}{n!}(x-t)^ndt\end{align*}
 
Then we have
\begin{align*}\int_a^b |f(x)|^rdx &= \int_a^b \left|\int_a^x\dfrac{f^{(n+1)}(t)}{n!}(x-t)^ndt\right|^rdx \\&\leq \int_a^b \left(\int_a^x \left|\dfrac{f^{(n+1)}(t)}{n!}\right| \left|(x-t)^n \right|dt\right)^rdx \\&\leq \int_a^b \left(\int_a^x \left|\dfrac{f^{(n+1)}(t)}{n!}\right|dt (x-a)^n\right)^rdx \\&\leq \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|dt\right)^r\int_a^b \left( (x-a)^n\right)^rdx \\& = \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|dt\right)^r\frac{(b-a)^{nr+1}}{nr+1} .\end{align*}
 
So we get
\begin{align*}\left(\int_a^b |f(x)|^rdx\right)^{1/r} \leq \left(\frac{(b-a)^{nr+1}}{nr+1}\right)^{1/r} \int_a^b \left|\dfrac{f^{(n+1)}(x)}{n!}\right|dx\end{align*}
which is $\mathbf a)$
 
To get $\mathbf b)$ we can proceed similarly using Holder's inequality
 
\begin{align*}\int_a^b |f(x)|^rdx &= \int_a^b \left|\int_a^x\dfrac{f^{(n+1)}(t)}{n!}(x-t)^ndt\right|^rdx \\&\leq \int_a^b \left(\int_a^x \left|\dfrac{f^{(n+1)}(t)}{n!}\right|^2dt \int_a^x \left|(x-t)^{2n} \right|dt\right)^{r/2} dx \\&\leq \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|^2dt\right)^{r/2} \int_a^b  \left(\int_a^x \left|(x-t)^{2n} \right|dt\right)^{r/2} dx \\&= \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|^2dt\right)^{r/2} \int_a^b  \left(\frac{(x-a)^{2n+1}}{2n+1}\right)^{r/2} dx\\& = \left(\int_a^b \left|\dfrac{f^{(n+1)}(t)}{n!}\right|^2dt\right)^{r/2} \left(\frac{1}{2n+1}\right)^{r/2} \left(\frac{(b-a)^{nr+\frac{r}{2} +1}}{nr+\frac{r}{2} +1}\right)\end{align*}

Let $f$ be a twice continuously differentiable function from $[0,1]$ into $R$,Give that
$$f(0)+2f(\frac{1}{2})+f(1)=0$$
show that
$$\int_{0}^{1}(f''(x))^2dx\ge 1920\left(\int_{0}^{1}f(x)dx\right)^2$$

Proof.1) Let $g_1(x)=x(x-1/2)$, $g_2(x)=(x-1)(x-1/2)$. By two integration by parts, we have

 
$$\int_0^{1/2}f^{\prime\prime}(x)g_1(x)dx=-\frac{f(1/2)+f(0)}{2}+2\int_0^{1/2}f(x)dx$$
and
$$\int_{1/2}^{1}f^{\prime\prime}(x)g_2(x)dx=-\frac{f(1/2)+f(1)}{2}+2\int_0^{1/2}f(x)dx$$
Hence
$$\int_0^{1/2}f^{\prime\prime}(x)g_1(x)dx+\int_{1/2}^{1}f^{\prime\prime}(x)g_2(x)dx=2\int_0^1f(x)dx$$
 
2) By Cauchy-Schwarz:
 
$$(\int_{0}^{1/2}f^{\prime\prime}(x)g_1(x)dx)^2\leq (\int_{0}^{1/2}f^{\prime\prime}(x)^2dx)\frac{1}{15.2^6}$$
 
$$(\int_{1/2}^{1}f^{\prime\prime}(x)g_2(x)dx)^2\leq (\int_{1/2}^{1}f^{\prime\prime}(x)^2dx)\frac{1}{15.2^6}$$
 
3) Use now $\sqrt{U}+\sqrt{V}\leq \sqrt{2}\sqrt{U+V}$ with $\displaystyle U=\int_{0}^{1/2}f^{\prime\prime}(x)^2dx$ and $\displaystyle V=\int_{1/2}^{1}f^{\prime\prime}(x)^2dx$ to finish the proof. 

Let $f$ be a positive-valued,concave function on $[0,1]$,Prove that
$$6\left(\int_{0}^{1}f(x)dx\right)^2\le 1+ 8\int_{0}^{1}f^3(x)dx.$$

Proof.Let $c=\int_0^1 f(x)\,dx$ and $g=f/c$, so $\int_0^1 g(x)\,dx=1$. Then by Holder's inequality,

$$1\le \left(\int_0^1 g(x)^3\,dx\right)^{1/3}\left(\int_0^1 1^{3/2}\,dx\right)^{2/3} .$$
Therefore $\int_0^1 f(x)^3\,dx=c^3\int_0^1g(x)^3\,dx\ge c^3$, and 
$$8\int_0^1 f(x)^3\,dx + 1-6\left(\int_0^1 f(x)\,dx\right)^2\ge 8c^3+1-6c^2 =: h(c).$$
For $c>0$ the right-hand side is minimized when $0=h'(c)=24c^2-12c$, meaning $c=1/2$ (noting $h'(c)<0$ for $c<1/2$ and $h'(c)>0$ for $c>1/2$). Thus $$h(1/2)=8(1/2)^3+1-6(1/2)^2=1/2\le h(c)$$ for all $c>0$.
Actually, then it follows  
$$6\left(\int_0^1 f(x)\,dx\right)^2 \le \frac12 + 8\int_0^1f(x)^3\,dx.$$
Concavity of $f$ is not needed.

链接:http://math.stackexchange.com/questions/763253/how-prove-this-integral-inequality-6-left-int-01fxdx-right2-le-1-8-i?rq=1

三角多项式不等式

逻辑丁的提问:证明\[\sum\limits_{k = 1}^{+\infty} {\frac{{\sin kx}}{{{k^a}}}}  > 0,x \in \left( {0,\pi } \right),a \in \left( {0,\frac{1}{2}} \right]\]证明在$(0,\pi)$上勒贝格可积.

一个很好的函数

一个白衣书生出游,于湖光山色之中寻得一寺,寺中有一老僧。二人相谈甚欢,便携手入院。老僧见书生谈吐不凡,遂生考较之意。见院内瓜果藤蔓,老僧出上联曰:”一阶石桌两个闲人三尺小院,四顾春色静看五月石榴。”书生羽扇轻摇,蛋定一笑:”这有何难:五维空间四次齐次三角函数,二重积分必然一致连续。

哆塔微博上告知了一个很好的实函数\[y = x\left( {\sqrt {\cos \left( {2\pi x} \right) - 1}  + 1} \right) + 0 \cdot \ln x.\]

图象是$(1,1),(2,2),\cdots,(n,n),\cdots$这些离散的点.