"로그 사인 적분 (log sine integrals)"의 두 판 사이의 차이

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* [[로그 사인 적분 (log sine integrals)]]<br>
 
* [[로그 사인 적분 (log sine integrals)]]<br>
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*  정의<br><math>\operatorname{Ls}_{a+b,a}(\theta):=-\int_{0}^{\theta}x^a\log^{b-1}}|2\sin \frac{x}{2}|\,dx</math><br><math>\operatorname{Ls}_{n}(\theta):=\operatorname{Ls}_{n,0}(\theta)=-\int_{0}^{\theta}\log^{n-1}}(2\sin \frac{x}{2})\,dx</math><br>
 
*  정의<br><math>\operatorname{Ls}_{a+b,a}(\theta):=-\int_{0}^{\theta}x^a\log^{b-1}}|2\sin \frac{x}{2}|\,dx</math><br><math>\operatorname{Ls}_{n}(\theta):=\operatorname{Ls}_{n,0}(\theta)=-\int_{0}^{\theta}\log^{n-1}}(2\sin \frac{x}{2})\,dx</math><br>
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<h5 style="MARGIN: 0px; LINE-HEIGHT: 2em;">로그사인 정적분</h5>
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<h5 style="margin: 0px; line-height: 2em;">로그사인 정적분</h5>
  
 
*  정적분 값의 계산 문제<br><math>\operatorname{Ls}_{n}(\pi)=-\int_{0}^{\pi}\log^{n-1}}(2\sin \frac{x}{2})\,dx</math><br>
 
*  정적분 값의 계산 문제<br><math>\operatorname{Ls}_{n}(\pi)=-\int_{0}^{\pi}\log^{n-1}}(2\sin \frac{x}{2})\,dx</math><br>
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<h5 style="MARGIN: 0px; LINE-HEIGHT: 2em;">special values</h5>
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<math>\int_{0}^{\frac{\pi}{4}}\ln (\sin t)dt =-\frac{\pi}{4}\ln 2-\frac{G}{2}</math>
 
<math>\int_{0}^{\frac{\pi}{4}}\ln (\sin t)dt =-\frac{\pi}{4}\ln 2-\frac{G}{2}</math>
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* [[중심이항계수(central binomial coefficient)]]<br><math>\sum_{n=1}^\infty \frac{1}{n^3\binom{2n}{n}}=4\int_{0}^{\frac{1}{2}}(\arcsin x)^2}\frac{dx}{x}=-2\int_{0}^{\pi/3}x\log(2\sin \frac{x}{2})\,dx</math><br>
 
* [[중심이항계수(central binomial coefficient)]]<br><math>\sum_{n=1}^\infty \frac{1}{n^3\binom{2n}{n}}=4\int_{0}^{\frac{1}{2}}(\arcsin x)^2}\frac{dx}{x}=-2\int_{0}^{\pi/3}x\log(2\sin \frac{x}{2})\,dx</math><br>
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* http://www.wolframalpha.com/input/?i=integrate+(log+(2sin(x/2)))^2+dx+from+0+to+pi/3<br>
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* [http://www.wolframalpha.com/input/?i=integrate+%28log+%282sin%28x/2%29%29%29%5E2+dx+from+0+to+pi/3 http://www.wolframalpha.com/input/?i=integrate+(log+(2sin(x/2)))^2+dx+from+0+to+pi/3]<br>
 
*   http://arxiv.org/abs/hep-ph/0411100v2<br>
 
*   http://arxiv.org/abs/hep-ph/0411100v2<br>
 
* http://mathworld.wolfram.com/RamanujanLog-TrigonometricIntegrals.html<br>
 
* http://mathworld.wolfram.com/RamanujanLog-TrigonometricIntegrals.html<br>
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* [[로그 탄젠트 적분(log tangent integral)]]<br>
 
* [[로그 탄젠트 적분(log tangent integral)]]<br>
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* 단어사전 http://www.google.com/dictionary?langpair=en|ko&q=
 
* 단어사전 http://www.google.com/dictionary?langpair=en|ko&q=
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* http://ko.wikipedia.org/wiki/
 
* http://ko.wikipedia.org/wiki/
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* http://www.wolframalpha.com/input/?i=
 
* http://www.wolframalpha.com/input/?i=
 
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
 
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
* [http://www.research.att.com/~njas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
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* [http://www.research.att.com/%7Enjas/sequences/index.html The On-Line Encyclopedia of Integer Sequences]<br>
 
** http://www.research.att.com/~njas/sequences/?q=
 
** http://www.research.att.com/~njas/sequences/?q=
  
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* [http://dx.doi.org/10.1016/S0377-0427(03)00438-2 On some log-cosine integrals related to ζ(3), ζ(4), and ζ(6)]<br>
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*  Borwein, David, Jonathan M Borwein, Armin Straub, and James Wan. 2011. Log-sine evaluations of Mahler measures, II. 1103.3035 (March 15). http://arxiv.org/abs/1103.3035. <br>  <br>
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* [http://dx.doi.org/10.1016/S0377-0427%2803%2900438-2 On some log-cosine integrals related to ζ(3), ζ(4), and ζ(6)]<br>
 
** Mark W. Coffey, 2003
 
** Mark W. Coffey, 2003
* [http://www.cs.cmu.edu/~adamchik/articles/Srivastava/ch_sr.pdf Multiple Gamma and Related Functions]<br>
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* [http://www.cs.cmu.edu/%7Eadamchik/articles/Srivastava/ch_sr.pdf Multiple Gamma and Related Functions]<br>
 
** J. Choi, H. M. Srivastava, V.S. Adamchik , Applied Mathematics and Computation, 134 (2003), 515-533
 
** J. Choi, H. M. Srivastava, V.S. Adamchik , Applied Mathematics and Computation, 134 (2003), 515-533
 
* '''[Borwein1995]'''[http://www.jstor.org/stable/2160718 On an Intriguing Integral and Some Series Related to ζ(4)]<br>
 
* '''[Borwein1995]'''[http://www.jstor.org/stable/2160718 On an Intriguing Integral and Some Series Related to ζ(4)]<br>
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*  도서내검색<br>
 
*  도서내검색<br>
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*  네이버 뉴스 검색 (키워드 수정)<br>
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2011년 3월 17일 (목) 13:23 판

이 항목의 스프링노트 원문주소

 

 

개요
  • 정의
    \(\operatorname{Ls}_{a+b,a}(\theta):=-\int_{0}^{\theta}x^a\log^{b-1}}|2\sin \frac{x}{2}|\,dx\)
    \(\operatorname{Ls}_{n}(\theta):=\operatorname{Ls}_{n,0}(\theta)=-\int_{0}^{\theta}\log^{n-1}}(2\sin \frac{x}{2})\,dx\)
  • 클라우센 함수의 일반화로 볼 수 있다
    \(\operatorname{Cl}_2(\theta)=-\int_0^{\theta} \ln |2\sin \frac{t}{2}| \,dt=\sum_{n=1}^{\infty}\frac{\sin (n\theta)}{n^2}\)

 

 

\(\int_{0}^{1-e^{i\theta}}\log^{n-1}z\frac{dz}{1-z}=-i\int_{0}^{\theta}(\frac{i}{2}(x-\pi)+\log|2\sin \frac{x}{2}|)^{n-1}\,dx \)\(=-\int_{0}^{\theta}x^a\log^{b-1}}|2\sin \frac{x}{2}|\,dx\)

 

 

로그사인 정적분
  • 정적분 값의 계산 문제
    \(\operatorname{Ls}_{n}(\pi)=-\int_{0}^{\pi}\log^{n-1}}(2\sin \frac{x}{2})\,dx\)
  • 지수생성함수
    \(I(x)=\int_{0}^{\pi}e^{x\log(2\sin \frac{1}{2}\theta)}d\theta =\sum_{n=0}^{\infty}\int_{0}^{\pi}\frac{x^n}{n!}\log^n(2\sin\frac{1}{2}\theta)d\theta=-\sum_{n=0}^{\infty}\frac{x^n}{n!}\operatorname{Ls}_{n+1}(\pi)\)

 

  • 정적분의 점화식
    \(\operatorname{Ls}_{m+2}(\pi)=(-1)^{m}m![\pi(1-2^{-m})\zeta(m+1)-\sum_{k=2}^{m-1}(-1)^{k}\frac{1-2^{k-m}}{k!}\zeta(m-k+1)\operatorname{Ls}_{k+1}(\pi)\)
  • 이 정적분은 \(\ln 2\)와 \(\zeta(n), n\geq 2\) 의 다항식으로 표현할 수 있다[Bowman1947]
  • 다음 정리로부터 이러한 결과들을 이해할 수 있다

 

(정리) [Lewin1958]

\(I(x)=\frac{\pi\Gamma(1+x)}{(\Gamma(1+\frac{1}{2}x))^2}\)

\(\log I(x)=\log {\pi}+\sum_{k=2}^{\infty}(-1)^k (1-2^{1-k})\frac{\zeta(k)}{k}x^k\)

 

(증명)

오일러 베타적분 의 결과를 이용하자. 

\(\int_0^{\frac{\pi}{2}}\sin^{p}\theta{d\theta}= \frac{1}{2}B(\frac{p+1}{2},\frac{1}{2})=\frac{\sqrt{\pi}\Gamma(\frac{p}{2}+\frac{1}{2})}{2\Gamma(\frac{p}{2}+1)}\)

 \(I(x)=\int_{0}^{\pi}e^{x\log(2\sin \frac{1}{2}\theta)}d\theta =\int_{0}^{\pi}(2\sin \frac{1}{2}\theta)^{x}\,d\theta=2^{x+1}\int_{0}^{\pi/2}\sin^{x}t\,dt=\sqrt{\pi}\frac{2^x\Gamma(\frac{x}{2}+\frac{1}{2})}{\Gamma(\frac{x}{2}+1)}\)

여기서 감마함수의 곱셈공식 \(2^{2z}\Gamma(z) \; \Gamma\left(z + \frac{1}{2}\right) = 2\sqrt{\pi}\;\Gamma(2z)\) 을 이용하면, 우변을 정리하여 원하는 식을 얻는다. 

한편, 

\(\log I(x)=\log {\pi}+\sum_{k=2}^{\infty}(-1)^k (1-2^{1-k})\frac{\zeta(k)}{k}x^k\) 를 구하려면, 로그감마 함수의 테일러전개를 이용하면 된다.  \(\log\Gamma(1+x) =-\gamma x+\sum_{k=2}^{\infty}(-1)^k \frac{\zeta(k)}{k}x^k\) ■

 

  • 노트*

\(\frac{1}{\pi}\int_{0}^{\pi}2^{x}\sin^{x}\frac{1}{2}\theta\,d\theta=\frac{1}{\pi}\int_{0}^{\pi}2^{x}\cos^{x}\frac{1}{2}\theta\,d\theta=\frac{\Gamma(1+x)}{\Gamma(1+\frac{1}{2}x)\Gamma(1+\frac{1}{2}x)}\)

좀 더 일반적으로

\(\frac{1}{\pi}\int_{0}^{\pi}2^{x}\cos^{x}\frac{1}{2}\theta\cos y\theta \,d\theta=\frac{\Gamma(1+x)}{\Gamma(1+\frac{1}{2}x+y)\Gamma(1+\frac{1}{2}x-y)}\) 가 성립한다. [Borwein1995]

 

 

special values

\(\int_{0}^{\frac{\pi}{4}}\ln (\sin t)dt =-\frac{\pi}{4}\ln 2-\frac{G}{2}\)

\(\int_{0}^{\frac{\pi}{4}}\ln (\cos t)dt =-\frac{\pi}{4}\ln 2+\frac{G}{2}\)

\(\int_{0}^{\frac{\pi}{4}}t\ln (\sin t)dt =\frac{35}{128}\zeta(3)-\frac{\pi G}{8}-\frac{\pi^2}{32}\log 2\)

(여기서 G는 카탈란 상수)

\(\int_{0}^{\pi/3}\log^2(2\sin \frac{x}{2})\,dx=\frac{7\pi^3}{108}\)

\(\int_{0}^{\pi/3}x\log^2(2\sin \frac{x}{2})\,dx=\frac{17\pi^4}{6480}\)

\(\int_{0}^{\pi}\log(2\sin \frac{x}{2})\,dx=0\)

\(\int_{0}^{\pi/2}\log(\sin x)\,dx=-\frac{\pi\log 2}{2}\)

\(\int_{0}^{\pi/2}x\log(\sin x)\,dx=\frac{7}{16}\zeta(3)-\frac{\pi^2}{8}\log 2\)

\(\int_{0}^{\frac{\pi}{2}}x^2 \ln (\sin x)dx=-\frac{\pi^3}{24}\ln 2+\frac{3}{16} \zeta(3)\)

\(\int_{0}^{\pi/2}\log^2(\sin x)\,dx=\frac{\pi}{2}(\log 2)^2+\frac{\pi^3}{24}\)

\(\int_{0}^{\pi}\log^2(2\sin \frac{x}{2})\,dx=\frac{\pi^3}{12}\)

\(\int_{0}^{\pi}x^2\log^2(2\cos \frac{x}{2})\,dx=\frac{11\pi^5}{180}\)

\(\int_{0}^{\pi}\log^3(2\sin \frac{x}{2})\,dx=-\frac{3\pi}{2}\zeta(3)\)

\(\int_{0}^{\pi}\log^4(2\sin \frac{x}{2})\,dx=\frac{19\pi^5}{240}\)

\(\int_{0}^{\pi}\log^5(2\sin \frac{x}{2})\,dx=-\frac{45\pi}{2}\zeta(5)-\frac{5\pi^3}{4}\zeta(3)\)

\(\int_{0}^{\pi}\log^6(2\sin \frac{x}{2})\,dx=\frac{45\pi}{2}\zeta^2(3)+\frac{275\pi^7}{1344}\)

 

 

 

메모

 

 

재미있는 사실

 

 

 

역사

 

 

 

메모

 

 

 

관련된 항목들

 

 

수학용어번역

 

 

사전 형태의 자료

 

 

관련논문

 

관련도서

 

 

관련기사

 

 

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