"로그 탄젠트 적분(log tangent integral)"의 두 판 사이의 차이
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Pythagoras0 (토론 | 기여) |
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| (사용자 2명의 중간 판 70개는 보이지 않습니다) | |||
| 1번째 줄: | 1번째 줄: | ||
| − | + | ==개요== | |
| − | + | * [[로그 사인 적분 (log sine integrals)]]과 밀접하게 관련되어 있음 | |
| + | * 다음과 같은 정적분값의 계산:<math>\int_{\pi/4}^{\pi/2} \ln \tan x\, dx=G</math>, <math>G</math>는 [[카탈란 상수]]:<math>\int_{0}^{\pi/4} \ln \tan x\,dx=-G</math> | ||
| + | :<math>\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)</math>:<math>\int_{0}^{\pi} (\ln \tan \frac{x}{4})^2\,dx=\frac{\pi^3}{4}</math>:<math>\int_0^{\infty}\frac{(\ln x)^2}{1+x^2} dx =\int_{0}^{\pi/2}(\ln \tan x)^2\,dx = \frac{ \pi^3}{8}</math> | ||
| + | * ''''''[Vardi1988] '''참조''' | ||
| + | * 적분:<math>\int_{0}^{\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}\,dx=\pi\ln2</math> 에 대해서는 [[로그함수와 유리함수가 있는 정적분]] | ||
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| − | + | ==증명== | |
| − | + | (보조정리) | |
| − | + | <math>\Gamma(s)\beta(s)=\int_{\pi/4}^{\pi/2} \ln^{s-1}\tan x\, dx</math> | |
| − | <math> | + | 여기서 <math>\Gamma(s)</math>는 [[감마함수]],<math>\beta(s)</math>는 [[디리클레 베타함수]]. |
| − | + | ||
| − | + | (증명) | |
| − | <math>F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}</math> | + | <math>F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}</math> 라 하자. |
| − | <math>\Gamma(s)F(s)=\int_0^{\infty}(\sum_{n=1}^{\infty}f(n)e^{-nt})t^{s-1}\,dt</math><math>z=e^{-t}</math> | + | <math>\Gamma(s)F(s)=\int_0^{\infty}(\sum_{n=1}^{\infty}f(n)e^{-nt})t^{s-1}\,dt</math> |
| + | |||
| + | <math>z=e^{-t}</math> 로 치환하면, | ||
<math>\Gamma(s)F(s)=\int_0^{1}(\sum_{n=1}^{\infty}f(n)z^n)(\log\frac{1}{z})^{s-1}\,\frac{dz}{z}</math> | <math>\Gamma(s)F(s)=\int_0^{1}(\sum_{n=1}^{\infty}f(n)z^n)(\log\frac{1}{z})^{s-1}\,\frac{dz}{z}</math> | ||
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| − | + | 만약 <math>f(n+q)=f(n)</math> 을 만족하면 (가령 디리클레 캐릭터의 경우) | |
| − | <math>p(z)=\sum_{n=1}^{q-1}f(n)z^n</math>라면, | + | <math>p(z)=\sum_{n=1}^{q-1}f(n)z^n</math>라면, <math>\sum_{n=1}^{\infty}f(n)z^n=\frac{p(z)}{1-z^q}</math> 로 쓸 수 있다. |
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| − | 이를 이용하면, | + | 이를 이용하면, |
| − | <math>\Gamma(s)F(s)=\int_0^{1}\frac{p(z)(\log\frac{1}{z})^{s-1}}{1-z^q}\,\frac{dz}{z}</math> | + | <math>\Gamma(s)F(s)=\int_0^{1}\frac{p(z)(\log\frac{1}{z})^{s-1}}{1-z^q}\,\frac{dz}{z}</math> 를 얻는다. |
| − | + | <math>f</math>가 <math>f(3)=-1</math>인 주기가 4인 디리클레 캐릭터라면, <math>q=4</math>, <math>p(z)=z-z^3</math> | |
| − | + | 따라서 | |
| − | <math>s=1 | + | <math>\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</math> ■ |
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| − | + | ||
| − | + | (따름정리1) | |
| − | <math>\ | + | <math>\int_{\pi/4}^{\pi/2} \ln \tan x\, dx=G</math>, G는 [[카탈란 상수]]. |
| − | + | (증명) | |
| − | + | 위에서 얻은 보조정리에 <math>s=2</math>를 적용하면, | |
| − | + | <math>\int_{\pi/4}^{\pi/2} \ln^{2-1}\tan x\, dx=\Gamma(2)\beta(2)=G</math> ■ | |
| − | + | ||
| − | + | ||
| − | + | (따름정리2) | |
| − | <math>\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx | + | <math>\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)</math> |
| − | + | ||
| − | + | (증명) | |
| − | + | <math>\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\frac{d}{ds}(\Gamma(s)\beta(s))|_{s=1}</math>임을 보이자. | |
| − | < | + | <math>\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</math> |
| − | + | <math>s=1</math> 일때, | |
| − | + | <math>\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</math> | |
| − | + | 이제 [[다이감마 함수(digamma function)]]와 [[디리클레 베타함수]]에서 얻은 결과를 사용하자. | |
| − | + | <math>\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}</math>, <math>\psi(1) = -\gamma\,\!</math>. 따라서 <math>\Gamma(1)=-\gamma</math>. | |
| − | + | <math>\beta'(1)=\frac{\pi}{4}\gamma+\frac{\pi}{2}\ln(\frac{\Gamma(3/4)}{\Gamma(1/4)}\sqrt{2\pi})</math>. | |
| − | + | ||
| − | + | 그러므로 | |
| − | + | <math>\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)</math> | |
| − | + | 임이 증명된다. ■ | |
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| − | + | ==역사== | |
| − | + | * [[수학사 연표]] | |
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| − | + | ==메모== | |
| + | <math>f</math>가 <math>f(3)=-1</math>인 주기가 4인 디리클레 캐릭터라고 하면, <math>p(z)=z-z^3</math> | ||
| + | 다음의 <math>L</math>-함수 | ||
| + | :<math>L(s) = \sum_{n\geq 1}\frac{f(n)}{n^s}</math> | ||
| + | 는 아래의 식을 만족한다 | ||
| + | :<math> | ||
| + | \begin{align} | ||
| + | L'(1)-\gamma \frac{\pi}{4}&=\int_0^{1}\frac{z-z^3}{1-z^4}\log \log\frac{1}{z} \,\frac{dz}{z}\\ | ||
| + | &=\int_0^{1}\log \log\frac{1}{z} \,\frac{dz}{1+z^2}\\ | ||
| + | &=\int_1^{\infty}\log \log u \,\frac{du}{1+u^2} \\ | ||
| + | &=\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx | ||
| + | \end{align} | ||
| + | </math> | ||
| + | :<math>\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=L'(1)- \frac{\pi}{4}\gamma=\frac{\pi}{2}\ln(\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi})</math> | ||
| − | + | * http://math.stackexchange.com/questions/285671/vardis-integral-int-pi-4-pi-2-ln-ln-tan-xdx | |
| + | <math>\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</math> | ||
| + | * http://arxiv.org/PS_cache/arxiv/pdf/0706/0706.0356v1.pdf | ||
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| − | + | ==개요== | |
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| − | + | * [[등차수열의 소수분포에 관한 디리클레 정리]] | |
| + | * [[L-함수, 제타함수와 디리클레 급수|디리클레 급수]] | ||
| + | * [[후르비츠 제타함수(Hurwitz zeta function)|Hurwitz 제타함수]] | ||
| + | * [[감마함수]] | ||
| + | * [[다이로그 함수(dilogarithm)|다이로그 함수(dilogarithm )]] | ||
| + | * [[란덴변환(Landen's transformation)]] | ||
| + | * [[르장드르 카이 함수]] | ||
| + | * [[모듈라 군, j-invariant and the singular moduli]] | ||
| + | * [[카탈란 상수]] | ||
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| − | + | ==수학용어번역== | |
| − | + | * [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집] | |
| + | ** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr= | ||
| + | * [http://kms.or.kr/home/kor/board/bulletin_list_subject.asp?bulletinid=%7BD6048897-56F9-43D7-8BB6-50B362D1243A%7D&boardname=%BC%F6%C7%D0%BF%EB%BE%EE%C5%E4%B7%D0%B9%E6&globalmenu=7&localmenu=4 대한수학회 수학용어한글화 게시판] | ||
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| − | + | ==매스매티카 파일 및 계산 리소스== | |
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| − | + | * https://docs.google.com/file/d/0B8XXo8Tve1cxNmRFU0Vyak14NGM/edit | |
| + | * [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)] | ||
| + | * [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] | ||
| + | * [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] | ||
| + | * [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] | ||
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| − | + | ==사전 형태의 자료== | |
| − | + | * http://ko.wikipedia.org/wiki/ | |
| + | * http://en.wikipedia.org/wiki/ | ||
| + | * http://www.wolframalpha.com/input/?i= | ||
| + | * [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions] | ||
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| − | + | ==관련논문== | |
| − | * | + | * [http://dx.doi.org/10.1007/s11040-010-9074-y Alternative Evaluation of a ln tan Integral Arising in Quantum Field Theory] |
| − | ** http:// | + | ** Mark W. Coffey, Mathematical Physics, Analysis and Geometry, Volume 13, Number 2, 2010 |
| − | ** http:// | + | * '''[BBBZ2010]'''[http://arxiv.org/abs/1005.0414 Experimental Mathematics and Mathematical Physics] |
| − | * | + | ** DH Bailey, JM Borwein, D Broadhurst, W Zudilin, 2010 |
| − | * | + | * [http://www.springerlink.com/content/p2k0106727416271/?p=03915f5244d74523b6d36406299c80d5&pi=6 A class of logarithmic integrals] |
| − | ** http:// | + | ** Luis A. Medina and Victor H. Moll, The Ramanujan Journal, Volume 20, Number 1 / 2009년 10월 |
| − | ** http:// | + | * [http://link.aip.org/link/?JMAPAQ/49/093508/1 Evaluation of a ln tan integral arising in quantum field theory] |
| + | ** Mark W. Coffey, J. Math. Phys. 49, 093508 (2008) | ||
| + | * '''[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] | ||
| + | ** J.M. Borwein, D.J. Broadhurst, 1998 | ||
| + | * [http://doi.acm.org/10.1145/258726.258736 A class of logarithmic integrals] | ||
| + | ** Victor Adamchik, 1997 | ||
| + | * '''[Vardi1988]'''[http://www.jstor.org/stable/2323562 Integrals, an Introduction to Analytic Number Theory] | ||
| + | ** Ilan Vardi, The American Mathematical Monthly, Vol. 95, No. 4 (Apr., 1988), pp. 308-315 | ||
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| − | + | ==관련도서== | |
| − | * | + | * [http://www.amazon.com/Irresistible-Integrals-Symbolics-Experiments-Evaluation/dp/0521796369 Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals] |
| − | + | ** George Boros and Victor Moll | |
| − | + | [[분류:적분]] | |
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2020년 12월 28일 (월) 02:17 기준 최신판
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\[\int_{0}^{\pi/4} \ln \tan x\,dx=-G\]
\[\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\frac{\pi}{2}\ln \left(\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}\right)\]\[\int_{0}^{\pi} (\ln \tan \frac{x}{4})^2\,dx=\frac{\pi^3}{4}\]\[\int_0^{\infty}\frac{(\ln x)^2}{1+x^2} dx =\int_{0}^{\pi/2}(\ln \tan x)^2\,dx = \frac{ \pi^3}{8}\]
- '[Vardi1988] 참조'
- 적분\[\int_{0}^{\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}\,dx=\pi\ln2\] 에 대해서는 로그함수와 유리함수가 있는 정적분
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
\(f\)가 \(f(3)=-1\)인 주기가 4인 디리클레 캐릭터라고 하면, \(p(z)=z-z^3\) 다음의 \(L\)-함수 \[L(s) = \sum_{n\geq 1}\frac{f(n)}{n^s}\] 는 아래의 식을 만족한다 \[ \begin{align} L'(1)-\gamma \frac{\pi}{4}&=\int_0^{1}\frac{z-z^3}{1-z^4}\log \log\frac{1}{z} \,\frac{dz}{z}\\ &=\int_0^{1}\log \log\frac{1}{z} \,\frac{dz}{1+z^2}\\ &=\int_1^{\infty}\log \log u \,\frac{du}{1+u^2} \\ &=\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx \end{align} \] \[\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=L'(1)- \frac{\pi}{4}\gamma=\frac{\pi}{2}\ln(\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi})\]
\(\frac{24}{7\sqrt{7}}\int_{\pi/3}^{\pi/2}\ln|\frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}|\,dt=L_{-7}(2)=1.15192547054449\cdots\)
개요
- 등차수열의 소수분포에 관한 디리클레 정리
- 디리클레 급수
- Hurwitz 제타함수
- 감마함수
- 다이로그 함수(dilogarithm )
- 란덴변환(Landen's transformation)
- 르장드르 카이 함수
- 모듈라 군, j-invariant and the singular moduli
- 카탈란 상수
수학용어번역
매스매티카 파일 및 계산 리소스
- https://docs.google.com/file/d/0B8XXo8Tve1cxNmRFU0Vyak14NGM/edit
- 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
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
- Alternative Evaluation of a ln tan Integral Arising in Quantum Field Theory
- Mark W. Coffey, Mathematical Physics, Analysis and Geometry, Volume 13, Number 2, 2010
- [BBBZ2010]Experimental Mathematics and Mathematical Physics
- DH Bailey, JM Borwein, D Broadhurst, W Zudilin, 2010
- A class of logarithmic integrals
- Luis A. Medina 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
- Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals
- George Boros and Victor Moll