"로그 탄젠트 적분(log tangent integral)"의 두 판 사이의 차이
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<math>\int_{0}^{\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}\,dx=\pi\ln2</math> | <math>\int_{0}^{\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}\,dx=\pi\ln2</math> | ||
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* [http://www.springerlink.com/content/p2k0106727416271/?p=03915f5244d74523b6d36406299c80d5&pi=6 A class of logarithmic integrals]<br> | * [http://www.springerlink.com/content/p2k0106727416271/?p=03915f5244d74523b6d36406299c80d5&pi=6 A class of logarithmic integrals]<br> | ||
** Luis A. Medina1 and Victor H. Moll, The Ramanujan Journal, Volume 20, Number 1 / 2009년 10월 | ** Luis A. Medina1 and Victor H. Moll, The Ramanujan Journal, Volume 20, Number 1 / 2009년 10월 | ||
* [http://link.aip.org/link/?JMAPAQ/49/093508/1 Evaluation of a ln tan integral arising in quantum field theory]<br> | * [http://link.aip.org/link/?JMAPAQ/49/093508/1 Evaluation of a ln tan integral arising in quantum field theory]<br> | ||
** Mark W. Coffey, J. Math. Phys. 49, 093508 (2008) | ** Mark W. Coffey, J. Math. Phys. 49, 093508 (2008) | ||
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* '''[Borwein and Boradhurst 1998]'''[http://arxiv.org/abs/hep-th/9811173 Determinations of rational Dedekind-zeta invariants of hyperbolic manifolds and Feynman knots and links]<br> | * '''[Borwein and Boradhurst 1998]'''[http://arxiv.org/abs/hep-th/9811173 Determinations of rational Dedekind-zeta invariants of hyperbolic manifolds and Feynman knots and links]<br> | ||
** J.M. Borwein, D.J. Broadhurst, 1998 | ** J.M. Borwein, D.J. Broadhurst, 1998 | ||
* [http://doi.acm.org/10.1145/258726.258736 A class of logarithmic integrals]<br> | * [http://doi.acm.org/10.1145/258726.258736 A class of logarithmic integrals]<br> | ||
** Victor Adamchik, 1997 | ** Victor Adamchik, 1997 | ||
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* '''[Vardi1988]'''[http://www.jstor.org/stable/2323562 Integrals, an Introduction to Analytic Number Theory]<br> | * '''[Vardi1988]'''[http://www.jstor.org/stable/2323562 Integrals, an Introduction to Analytic Number Theory]<br> | ||
** Ilan Vardi, The American Mathematical Monthly, Vol. 95, No. 4 (Apr., 1988), pp. 308-315 | ** Ilan Vardi, The American Mathematical Monthly, Vol. 95, No. 4 (Apr., 1988), pp. 308-315 | ||
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2010년 6월 18일 (금) 09:35 판
이 항목의 스프링노트 원문주소
개요
- 로그 사인 적분 (log sine integrals)과 밀접하게 관련되어 있음
- 다음과 같은 정적분값의 계산
\(\int_{\pi/4}^{\pi/2} \ln \tan x\, dx=G\), \(G\)는 카탈란 상수
[[카탈란 상수|]]\(\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\frac{\pi}{2}\ln{\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}\)
증명
'[Vardi1988] '참조
(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)
\(\int_{\pi/4}^{\pi/2} \ln \tan x\, dx=G\)
(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)
\(\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\frac{\pi}{2}\ln{\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}\)
(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 Digamma 함수와 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
\(\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\Gamma'(1)\beta(1)+\Gamma(1)\beta'(1)= -\frac{\pi}{4}\gamma+\beta'(1)=\frac{\pi}{2}\ln{\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}\)
임이 증명된다. ■
메모
\(\int_{0}^{\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}\,dx=\pi\ln2\)
- 로그 사인 적분 (log sine integrals)
- http://cjackal.tistory.com/109
- http://www.artofproblemsolving.com/Forum/viewtopic.php?t=340081
\(\int_{0}^{\infty}\ln(1+e^{-x})}\,dx=\frac{\pi^2}{12}\)
[[란덴변환(Landen's transformation)|]]
재미있는 사실
역사
메모
- http://www.wolframalpha.com/input/?i=integrate_0^(pi)+x+cos+x+%2F(1%2Bsin^2+x)
- http://www.wolframalpha.com/input/?i=log^2+(1%2Bsqrt(2))+-pi^2%2F4
- http://www.wolframalpha.com/input/?i=integrate+(tan+x%2B1)/sqrt(tan^2+x+%2B1)dx
- http://www.wolframalpha.com/input/?i=integrate+(tan+x-1)/sqrt(tan^2+x+%2B1)dx
관련된 다른 주제들
- 등차수열의 소수분포에 관한 디리클레 정리
- 디리클레 급수
- Hurwitz 제타함수
- 감마함수
- Dilogarithm
- 란덴변환(Landen's transformation)
- 르장드르 카이 함수
- 모듈라 군, j-invariant and the singular moduli
- 카탈란 상수
수학용어번역
사전 형태의 자료
- http://ko.wikipedia.org/wiki/
- http://en.wikipedia.org/wiki/
- http://www.wolframalpha.com/input/?i=
- NIST Digital Library of Mathematical Functions
관련논문
-
- A class of logarithmic integrals
- Luis A. Medina1 and Victor H. Moll, The Ramanujan Journal, Volume 20, Number 1 / 2009년 10월
- Evaluation of a ln tan integral arising in quantum field theory
- Mark W. Coffey, J. Math. Phys. 49, 093508 (2008)
- [Borwein and Boradhurst 1998]Determinations of rational Dedekind-zeta invariants of hyperbolic manifolds and Feynman knots and links
- J.M. Borwein, D.J. Broadhurst, 1998
- A class of logarithmic integrals
- Victor Adamchik, 1997
- [Vardi1988]Integrals, an Introduction to Analytic Number Theory
- Ilan Vardi, The American Mathematical Monthly, Vol. 95, No. 4 (Apr., 1988), pp. 308-315
관련도서 및 추천도서
- Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals
- George Boros and Victor Moll
- 도서내검색
- 도서검색
관련기사
- 네이버 뉴스 검색 (키워드 수정)
블로그
- 오늘의 계산 12
- 수학 잡담/오늘의 계산, 2008/08/10
- 구글 블로그 검색 http://blogsearch.google.com/blogsearch?q=
- 네이버 오늘의과학