与双重对数函数有关的积分

Eufisky posted @ 2017年8月03日 16:02 in 数学分析 with tags 积分计算 , 1477 阅读

求$$\int_0^{\frac{\pi}{2}}{\frac{x^2}{1+\cos ^2x}dx}.$$

对于$|b|<a$,注意到

\begin{align*}\frac{a^2-b^2}{a^2-2ab\cos x+b^2}&=\frac{a}{a-e^{ix}b}+\frac{be^{-ix}}{a-e^{-ix}b}\\&=\sum_{n=0}^\infty \left(\frac{b}{a}\right)^ne^{inx}+\frac{be^{-ix}}{a}\sum_{n=0}^\infty\left(\frac{b}{a}\right)^ne^{-inx}\\&=1+\sum_{n=1}^\infty \left(\frac{b}{a}\right)^ne^{inx}+ \sum_{n=1}^{\infty}\left(\frac{b}{a}\right)^{n}e^{-inx}\\&=1+2\sum_{n=1}^\infty \left(\frac{b}{a}\right)^n\cos(n x),\end{align*}

因此

\begin{equation*}1+2\sum_{n=1}^\infty \left(\frac{b}{a}\right)^n\cos(n x)=\frac{a^2-b^2}{a^2+b^2-2ab\cos x},\qquad\qquad\mbox{对于}\, |b|<a.\end{equation*}

令$a=\frac{2+\sqrt{2}}{2}$和$b=\frac{-2+\sqrt{2}}{2}$,我们有

\begin{align*}1+2\sum_{n=1}^\infty \left(2\sqrt{2}-3\right)^n\cos(n x)=\frac{2\sqrt{2}}{3+\cos x}.\end{align*}

由$\displaystyle\int_0^{\pi}{x^2\cos \left( nx \right) dx}=2\pi \frac{\cos \left( \pi n \right)}{n^2}$可知

\begin{align*}I&=\int_0^{\frac{\pi}{2}}{\frac{x^2}{1+\cos ^2x}dx}=\int_0^{\frac{\pi}{2}}{\frac{x^2}{1+\frac{\cos 2x+1}{2}}dx}\\&=\int_0^{\frac{\pi}{2}}{\frac{2x^2}{3+\cos 2x}dx}=\frac{1}{4}\int_0^{\pi}{\frac{x^2}{3+\cos x}dx}\\&=\frac{1}{8\sqrt{2}}\int_0^{\pi}{x^2\left[ 1+2\sum_{n=1}^{\infty}{\left( 2\sqrt{2}-3 \right) ^n\cos \left( nx \right)} \right] dx}\\&=\frac{1}{8\sqrt{2}}\left[ \frac{\pi ^3}{3}+2\sum_{n=1}^{\infty}{\left( 2\sqrt{2}-3 \right) ^n\int_0^{\pi}{x^2\cos \left( nx \right) dx}} \right]\\&=\frac{\sqrt{2}}{48}\pi ^3+\frac{\sqrt{2}}{4}\pi \sum_{n=1}^{\infty}{\frac{\left( 2\sqrt{2}-3 \right) ^n\cos \left( \pi n \right)}{n^2}}\\&=\frac{\sqrt{2}}{48}\pi ^3+\frac{\sqrt{2}}{4}\pi \sum_{n=1}^{\infty}{\frac{\left( 3-2\sqrt{2} \right) ^n}{n^2}}\\&=\frac{\sqrt{2}}{48}\pi ^3+\frac{\sqrt{2}}{4}\pi \mathrm{Li}_2\left( 3-2\sqrt{2} \right).\end{align*}

求$$\int_0^{\pi}{\frac{x^2}{1+\sin ^2x}dx}.$$

令$t=x-\frac\pi2$,我们有

\begin{align*}J&=\int_0^{\pi}{\frac{x^2}{1+\sin ^2x}dx}=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\frac{\left( t+\frac{\pi}{2} \right) ^2}{1+\cos ^2t}dt}\\&=2\int_0^{\frac{\pi}{2}}{\frac{t^2}{1+\cos ^2t}dt}+\frac{\pi ^2}{2}\int_0^{\frac{\pi}{2}}{\frac{1}{1+\cos ^2t}dt}\\&=2\int_0^{\frac{\pi}{2}}{\frac{t^2}{1+\cos ^2t}dt}+\frac{\pi ^2}{2}\int_0^{\frac{\pi}{2}}{\frac{1}{\sin ^2t+2\cos ^2t}dt}\\&=2\int_0^{\frac{\pi}{2}}{\frac{t^2}{1+\cos ^2t}dt}+\frac{\pi ^2}{2}\int_0^{\frac{\pi}{2}}{\frac{1}{\tan ^2t+2}d\left( \tan t \right)}\\&=\frac{\sqrt{2}}{24}\pi ^3+\frac{\sqrt{2}}{2}\pi \mathrm{Li}_2\left( 3-2\sqrt{2} \right) +\frac{\sqrt{2}}{8}\pi ^3\\&=\frac{\sqrt{2}}{2}\pi \mathrm{Li}_2\left( 3-2\sqrt{2} \right) +\frac{\sqrt{2}}{6}\pi ^3.\end{align*}

求$$\int_0^1{\int_0^1{\int_0^1{\int_0^1{\frac{\left( 1-x^2y^2z^2t^2 \right) dxdydzdt}{\sqrt{\left( 1-x^2 \right) \left( 1-y^2 \right) \left( 1-z^2 \right) \left( 1-t^2 \right) \left( 1+x^2y^2z^2t^2 \right)}}}}}}.$$

解.原积分等于$$I=\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\int_0^{\frac{\pi}{2}}{\frac{1-\cos ^2\alpha \cos ^2\beta \cos ^2\theta \cos ^2\gamma}{\sqrt{1+\cos ^2\alpha \cos ^2\beta \cos ^2\theta \cos ^2\gamma}}d\alpha d\beta d\theta d\gamma}}}}.$$

由于$$\frac{1}{\sqrt{1+x}}=\sum_{n=0}^{\infty}{\binom{-1/2}{n}x^n}=1+\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-1 \right) !!}{2^nn!}x^n},$$

因此

\begin{align*}\frac{1-x^2}{\sqrt{1+x^2}}&=\left( 1-x^2 \right) \left( 1+\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-1 \right) !!}{2^nn!}x^{2n}} \right) \\&=1+\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-1 \right) !!}{2^nn!}x^{2n}}-x^2-\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-1 \right) !!}{2^nn!}x^{2n+2}}\\&=1-x^2+\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-1 \right) !!}{2^nn!}x^{2n}}-\sum_{n=2}^{\infty}{\frac{\left( -1 \right) ^{n-1}\left( 2n-3 \right) !!}{2^{n-1}\left( n-1 \right) !}x^{2n}}\\&=1-x^2-\frac{x^2}{2}+\sum_{n=2}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-3 \right) !!}{2^nn!}\left( 2n-1+2n \right) x^{2n}}\\&=1-\frac{3x^2}{2}+\sum_{n=2}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-3 \right) !!}{2^nn!}\left( 4n-1 \right) x^{2n}}.\end{align*}

由于$$\int_0^{\frac{\pi}{2}}{\cos ^{2n}xdx}=\frac{\sqrt{\pi}\Gamma \left( n+\frac{1}{2} \right)}{2\Gamma \left( n+1 \right)}=\frac{\sqrt{\pi}}{2n!}\cdot \frac{\left( 2n-1 \right) !!}{2^n}\sqrt{\pi}=\frac{\pi}{2}\frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!},$$我们有

\begin{align*}I&=\frac{\pi ^4}{16}-\frac{3}{2}\left( \frac{\pi}{4} \right) ^4+\sum_{n=2}^{\infty}{\frac{\left( -1 \right) ^n\left( 2n-3 \right) !!}{2^nn!}\left( 4n-1 \right) \left[ \frac{\pi}{2}\frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!} \right] ^4}\\&=\frac{\pi ^4}{16}-\frac{3}{2}\left( \frac{\pi}{4} \right) ^4+\frac{\pi ^4}{16}\sum_{n=2}^{\infty}{\left( -1 \right) ^n\frac{4n-1}{2n-1}\left[ \frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!} \right] ^5}\\&=\frac{\pi ^4}{16}+\frac{\pi ^4}{16}\sum_{n=1}^{\infty}{\left( -1 \right) ^n\frac{4n-1}{2n-1}\left[ \frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!} \right] ^5}.\end{align*}

又由

\begin{align*}\sum_{n=1}^{\infty}{\left( -1 \right) ^n\frac{4n-1}{2n-1}\left[ \frac{\left( 2n-1 \right) !!}{\left( 2n \right) !!} \right] ^5}=& _5F_4\left( -\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1,1;-1 \right) \\&-\frac{1}{8}\,_5F_4\left( \frac{1}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};2,2,2,2;-1 \right) -1\end{align*}

可知

$$I=\frac{\pi ^4}{16}\left[ _5F_4\left( -\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};1,1,1,1;-1 \right) -\frac{1}{8}\,_5F_4\left( \frac{1}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};2,2,2,2;-1 \right) \right] ,$$

其中$_pF_q(a_1,\ldots,a_p;b_1,\ldots,b_q;z)$为Generalized Hypergeometric Function.

计算积分

$$\int_0^{\pi}{\sqrt{\tan \frac{\theta}{2}}\ln^2 \left( \sin \theta \right) \text{d}\theta}.$$

\large{\textbf{\textcolor{blue}{解}}}

\begin{align*}\int_0^{\pi}{\sqrt{\tan \frac{\theta}{2}}\ln^2 \left( \sin \theta \right) \text{d}\theta}&=\int_0^{\infty}{\frac{2\sqrt{t}}{1+t^2}\ln^2 \left( \frac{2t}{1+t^2} \right) \text{d}t}\hspace{0.5cm}t=\tan \frac{\theta}{2}\\&=\int_0^{\infty}{\frac{2\sqrt{1/t}}{1+t^2}\ln^2 \left( \frac{2}{t+1/t} \right) \text{d}t}\\&=\int_0^{\infty}{\frac{\sqrt{1/t}+\sqrt{1/t^3}}{t+1/t}\ln^2 \left( \frac{2}{t+1/t} \right) \text{d}t}\\&=\int_{-\infty}^{\infty}{\frac{2}{x^2+2}\ln^2 \left( \frac{2}{x^2+2} \right) \text{d}x}\\&=2\sqrt{2}\int_0^{\frac{\pi}{2}}{\ln^2 \left( \cos ^2u \right) \text{d}u}~~x=\sqrt2\tan u\\&=8\sqrt{2}\int_0^{\frac{\pi}{2}}{\ln ^2\sin u\text{d}u}=8\sqrt{2}\int_0^{\frac{\pi}{2}}{\left( -\ln 2-\sum_{k=1}^{\infty}{\frac{\cos \left( 2kx \right)}{k}} \right)^2 \text{d}u}\\&=8\sqrt{2}\left( \int_0^{\frac{\pi}{2}}{\ln ^22\text{d}u}+\sum_{n=1}^{\infty}{\frac{1}{k}\int_0^{\frac{\pi}{2}}{\frac{1+\cos 4kx}{2}\text{d}x}} \right)\\&=4\sqrt{2}\pi \ln ^22+2\sqrt{2}\pi \zeta \left( 2 \right) =\frac{\sqrt{2}}{3}\pi ^3+4\sqrt{2}\ln^2 2.\end{align*}

$f$是$[0,1]$上严格单增的凸实值连续函数,满足$f(0)=0,f(1)=1$,且$g(x)$满足$g(f(x))=x$对任意$x\in[0,1]$成立,证明

$$\int_0^1{f\left( x \right) g\left( x \right) \text{d}x}\le \frac{1}{3}.$$

\large{\textbf{\textcolor{blue}{证明}}} 首先我们有凸函数的等价定义:

定理.函数$f$在区间$I$上是凸函数,当且仅当对任何$(x_1,x_2)\subset I$及任何$x\in(x_1,x_2)$,有

$$\frac{f\left( x \right) -f\left( x_1 \right)}{x-x_1}\le \frac{f\left( x_2 \right) -f\left( x_1 \right)}{x_2-x_1}\le \frac{f\left( x_2 \right) -f\left( x \right)}{x_2-x}.$$

注意到$g(x)=f^{-1}(x)$是$f(x)$的反函数,只需要证明$f(x)f^{-1}(x)\leqslant x^2$即可,即

$$\frac{f\left( x \right)}{x}\leqslant \frac{x}{f^{-1}\left( x \right)}.$$

由题意知$f(x)\leqslant x\leqslant f^{-1}(x)$对$x\in[0,1]$都成立,结合$f(0)=0,f(1)=1,f$是凸函数可知对任意$x\in[0,1]$,$\exists t\in[x,1]$,s.t.$f(t)=x$于是由上述凸函数等价定义可知

$$\frac{f\left( x \right)}{x}\le \frac{f\left( t \right)}{t}=\frac{x}{f^{-1}\left( x \right)}.$$

于是$\displaystyle{f\left( x \right) g\left( x \right) =f\left( x \right) f^{-1}\left( x \right) \le x^2}$,

$$\int_0^1{f\left( x \right) g\left( x \right) \text{d}x}\leqslant \int_0^1{x^2\text{d}x}=\frac{1}{3}.$$

等号成立当且仅当$f(x)=x$.

求最大的常数$b>0$,使得对任意$a>0$和一切$(1,+\infty)$上连续可导且单增的实值函数$f(x)$满足$f(x)\leqslant x^{2a}\ln^bx,x\in(1,+\infty)$就有积分$\displaystyle{\int_1^{\infty}\frac{x^{2a-2}}{f'(x)}\mathrm{d}x=+\infty}$.

\large{\textbf{\textcolor{blue}{解}}} 首先如果$b>1$,我们取$f(x)=x^{2a}\ln^{b}x$,则求导后很容易得到积分$$\displaystyle{\int_1^{\infty}\frac{x^{2a-2}}{f'(x)}\mathrm{d}x<+\infty}$$

因此$b\leqslant1$,下面验证$b=1$满足条件.

如果$f'(x)$有界,结论显然成立,不妨设$f'(x)$无界,这时$f(x)$单调趋于$+\infty$.对$\forall A>0$,由Cauchy不等式得

$$\left( \int_{\text{e}^{\frac{A}{2}}}^{\text{e}^A}{\frac{x^{2a-2}}{f'\left( x \right)}\text{d}x} \right) \left( \int_{\text{e}^{\frac{A}{2}}}^{\text{e}^A}{\frac{f'\left( x \right)}{x^{2a}\ln ^2x}\text{d}x} \right) \geqslant \left( \int_{\text{e}^{\frac{A}{2}}}^{\text{e}^A}{\frac{\text{d}x}{x\ln x}} \right) ^2=\ln ^22.$$

由$f(x)\leqslant x^{2a}\ln x$得$f(\mathrm{e}^x)\leqslant x\mathrm{e}^{2ax}$,因此

\begin{align*}\int_{\textrm{e}^{\frac{A}{2}}}^{\textrm{e}^A}{\frac{f'\left(x\right)}{x^{2a}\ln^2x}\textrm{d}x}&=\int_{\frac{A}{2}}^A{\frac{f'\left(\textrm{e}^t\right)\textrm{e}^{2at}}{t^2\textrm{e}^{2at}}\textrm{d}t}=\int_{\frac{A}{2}}^A{\frac{\textrm{d}\left[f\left(\textrm{e}^{2t}\right)\right]}{t^2\textrm{e}^{2at}}}\\&=\left.\frac{f\left(\textrm{e}^t\right)}{t^2\textrm{e}^{2at}}\right|_{\frac{A}{2}}^{A}+\int_{\frac{A}{2}}^A{\frac{2t^2\textrm{e}^{-2at}+2t\textrm{e}^{-2at}}{t^4}f\left(\textrm{e}^t\right)\textrm{d}t}\\&\leqslant\frac{f\left(\textrm{e}^A\right)}{A^2\textrm{e}^{2A}}+\int_{\frac{A}{2}}^A{\frac{2t^2\textrm{e}^{-2at}+2t\textrm{e}^{-2at}}{t^4}t\textrm{e}^{2at}\textrm{d}t}\\&\leqslant\frac{1}{A}+2\left(\ln 2+\frac{1}{A}\right)=2\ln 2+\frac{3}{A}.\end{align*}

取$A$充分大,则$\displaystyle{\int_{\textrm{e}^{\frac{A}{2}}}^{\textrm{e}^A}{\frac{f'\left(x\right)}{x^{2a}\ln^2x}\textrm{d}x}\leqslant2}$, 因此

$$\int_{\textrm{e}^{A/2}}^{\textrm{e}^A}{\frac{\textrm{d}x}{f'\left(x\right)}}\geqslant\frac{\ln^22}{2}$$

对任意充分大的$A$都成立,于是积分$\displaystyle{\int_1^{+\infty}\frac{1}{f'(x)}\mathrm{d}x}=+\infty$,因此最大的$b=1$.

\textbf{证明：}不等式的左边可利用$Cauchy-Schwarz$不等式证之，下证不等式的右边成立.由题意可知，
\[\begin{split}& (1-\frac{f}{M})(1-\frac{m}{f}) \geq 0 \Longrightarrow 1+\frac{m}{M} \geq \frac{f}{M}+\frac{m}{f}\\\Longrightarrow & 1+\frac{m}{M} \geq \frac1{M}\int^{1}_{0}{f}dx +m\int^{1}_{0}{\frac1{f}}dx \geq 2\sqrt{\frac{m}{M}\int^{1}_{0} \frac1{f}dx\int^{1}_{0}{f}dx}\\\Longrightarrow & \int^{1}_{0} \frac1{f}dx\int^{1}_{0}{f}dx\leq \frac{(m+M)^2}{4mM}.\end{split}\]

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