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 , 1511 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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