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与双重对数函数有关的积分 - Eufisky - The lost book
二重积分计算
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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

这里

π20x21+cos2xdx.


对于|b|<a,注意到

a2b2a22abcosx+b2=aaeixb+beixaeixb=n=0(ba)neinx+beixan=0(ba)neinx=1+n=1(ba)neinx+n=1(ba)neinx=1+2n=1(ba)ncos(nx),
因此
1+2n=1(ba)ncos(nx)=a2b2a2+b22abcosx,对于|b|<a.

a=2+22b=2+22,我们有

1+2n=1(223)ncos(nx)=223+cosx.
π0x2cos(nx)dx=2πcos(πn)n2可知
I=π20x21+cos2xdx=π20x21+cos2x+12dx=π202x23+cos2xdx=14π0x23+cosxdx=182π0x2[1+2n=1(223)ncos(nx)]dx=182[π33+2n=1(223)nπ0x2cos(nx)dx]=248π3+24πn=1(223)ncos(πn)n2=248π3+24πn=1(322)nn2=248π3+24πLi2(322).

π0x21+sin2xdx.


t=xπ2,我们有

J=π0x21+sin2xdx=π2π2(t+π2)21+cos2tdt=2π20t21+cos2tdt+π22π2011+cos2tdt=2π20t21+cos2tdt+π22π201sin2t+2cos2tdt=2π20t21+cos2tdt+π22π201tan2t+2d(tant)=224π3+22πLi2(322)+28π3=22πLi2(322)+26π3.

10101010(1x2y2z2t2)dxdydzdt(1x2)(1y2)(1z2)(1t2)(1+x2y2z2t2).

解.原积分等于I=π20π20π20π201cos2αcos2βcos2θcos2γ1+cos2αcos2βcos2θcos2γdαdβdθdγ.

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*}
\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)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[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)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)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.
f(x)\leqslant x^{2a}\ln xf(\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*}
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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