"로그 탄젠트 적분(log tangent integral)"의 두 판 사이의 차이

수학노트
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* https://docs.google.com/file/d/0B8XXo8Tve1cxNmRFU0Vyak14NGM/edit
 
* https://docs.google.com/file/d/0B8XXo8Tve1cxNmRFU0Vyak14NGM/edit
* http://www.wolframalpha.com/input/?i=
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* [http://www.wolframalpha.com/input/?i=integrate_0%5E%28pi%29+x+cos+x+%2F%281%2Bsin%5E2+x%29 http://www.wolframalpha.com/input/?i=integrate_0^(pi)+x+cos+x+%2F(1%2Bsin^2+x)]<br>
* http://functions.wolfram.com/
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* [http://www.wolframalpha.com/input/?i=log%5E2+%281%2Bsqrt%282%29%29+-pi%5E2%2F4 http://www.wolframalpha.com/input/?i=log^2+(1%2Bsqrt(2))+-pi^2%2F4]<br>
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
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* [http://www.wolframalpha.com/input/?i=integrate+%28tan+x%2B1%29/sqrt%28tan%5E2+x+%2B1%29dx http://www.wolframalpha.com/input/?i=integrate+(tan+x%2B1)/sqrt(tan^2+x+%2B1)dx]<br>
* [http://people.math.sfu.ca/%7Ecbm/aands/toc.htm Abramowitz and Stegun Handbook of mathematical functions]
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* [http://www.wolframalpha.com/input/?i=integrate+%28tan+x-1%29/sqrt%28tan%5E2+x+%2B1%29dx http://www.wolframalpha.com/input/?i=integrate+(tan+x-1)/sqrt(tan^2+x+%2B1)dx]<br>
* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]
 
* [http://numbers.computation.free.fr/Constants/constants.html Numbers, constants and computation]
 
* [https://docs.google.com/open?id=0B8XXo8Tve1cxMWI0NzNjYWUtNmIwZi00YzhkLTkzNzQtMDMwYmVmYmIxNmIw 매스매티카 파일 목록]
 
  
 
 
 
 

2013년 1월 27일 (일) 01:41 판

이 항목의 수학노트 원문주소

 

 

개요

\[\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\frac{\pi}{2}\ln \left(\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}\right)\]\[\int_{0}^{\pi} (\ln \tan \frac{x}{4})^2\,dx=\frac{\pi^3}{4}\]\[\int_0^{\infty}\frac{(\ln x)^2}{1+x^2} dx =\int_{0}^{\pi/2}(\ln \tan x)^2\,dx = \frac{ \pi^3}{8}\]

 

 

증명

(보조정리)

\(\Gamma(s)\beta(s)=\int_{\pi/4}^{\pi/2} \ln^{s-1}\tan x\, dx\)

여기서 \(\Gamma(s)\)는 감마함수,\(\beta(s)\)는 디리클레 베타함수.

 

(증명)

\(F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}\) 라 하자.

\(\Gamma(s)F(s)=\int_0^{\infty}(\sum_{n=1}^{\infty}f(n)e^{-nt})t^{s-1}\,dt\)

\(z=e^{-t}\) 로 치환하면,

\(\Gamma(s)F(s)=\int_0^{1}(\sum_{n=1}^{\infty}f(n)z^n)(\log\frac{1}{z})^{s-1}\,\frac{dz}{z}\)

 

만약 \(f(n+q)=f(n)\) 을 만족하면 (가령 디리클레 캐릭터의 경우)

\(p(z)=\sum_{n=1}^{q-1}f(n)z^n\)라면,  \(\sum_{n=1}^{\infty}f(n)z^n=\frac{p(z)}{1-z^q}\) 로 쓸 수 있다.

 

이를 이용하면, 

\(\Gamma(s)F(s)=\int_0^{1}\frac{p(z)(\log\frac{1}{z})^{s-1}}{1-z^q}\,\frac{dz}{z}\) 를 얻는다.

\(f\)가 \(f(3)=-1\)인 주기가 4인 디리클레 캐릭터라면, \(q=4\), \(p(z)=z-z^3\)

따라서

\(\Gamma(s)\beta(s)=\int_0^{1}\frac{(\log\frac{1}{z})^{s-1}}{1+z^2} \,dz=\int_1^{\infty}\frac{(\log u)^{s-1}}{1+u^2} \,du=\int_{\pi/4}^{\pi/2} \ln^{s-1}\tan x\, dx\) ■

 

 

(따름정리1)

\(\int_{\pi/4}^{\pi/2} \ln \tan x\, dx=G\), G는 카탈란 상수.

(증명)

위에서 얻은 보조정리에 \(s=2\)를 적용하면, 

\(\int_{\pi/4}^{\pi/2} \ln^{2-1}\tan x\, dx=\Gamma(2)\beta(2)=G\) ■

 

 

(따름정리2)

\(\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\frac{\pi}{2}\ln \left(\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}\right)\)

 

(증명)

\(\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\frac{d}{ds}(\Gamma(s)\beta(s))|_{s=1}\)임을 보이자.

\(\frac{d}{ds}(\Gamma(s)\beta(s))=\frac{d}{ds}\int_1^{\infty}\frac{(\log u)^{s-1}}{1+u^2} \,du=\int_1^{\infty}\frac{(\log u)^{s-1}}{1+u^2}\log \log u \,du\)

\(s=1\) 일때,

\(\Gamma'(1)\beta(1)+\Gamma(1)\beta'(1)=\int_1^{\infty}\log \log u \,\frac{du}{1+u^2}=\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx\)

이제 다이감마 함수(digamma function)와 디리클레 베타함수에서 얻은 결과를 사용하자. 

\(\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}\), \(\psi(1) = -\gamma\,\!\). 따라서 \(\Gamma(1)=-\gamma\).

\(\beta'(1)=\frac{\pi}{4}\gamma+\frac{\pi}{2}\ln(\frac{\Gamma(3/4)}{\Gamma(1/4)}\sqrt{2\pi})\).

 

그러므로

\(\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\Gamma'(1)\beta(1)+\Gamma(1)\beta'(1)= -\frac{\pi}{4}\gamma+\beta'(1)=\frac{\pi}{2}\ln \left(\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}\right)\)

임이 증명된다. ■

 

 

역사

 

 

메모

\(\frac{24}{7\sqrt{7}}\int_{\pi/3}^{\pi/2}\ln|\frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}|\,dt=L_{-7}(2)=1.15192547054449\cdots\)


 

 

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