로그 탄젠트 적분(log tangent integral)

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

• 수학사 연표

 

 

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 \[ \begin{align} L'(1)-\gamma \frac{\pi}{4}&=\int_0^{1}\frac{z-z^3}{1-z^4}\log \log\frac{1}{z} \,\frac{dz}{z}\\ &=\int_0^{1}\log \log\frac{1}{z} \,\frac{dz}{1+z^2}\\ &=\int_1^{\infty}\log \log u \,\frac{du}{1+u^2} \\ &=\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx \end{align} \] \[\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=L'(1)- \frac{\pi}{4}\gamma=\frac{\pi}{2}\ln(\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi})\]

• http://math.stackexchange.com/questions/285671/vardis-integral-int-pi-4-pi-2-ln-ln-tan-xdx

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

• http://arxiv.org/PS_cache/arxiv/pdf/0706/0706.0356v1.pdf

 

 

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

• 대한수학회 수학 학술 용어집
  • http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=
• 대한수학회 수학용어한글화 게시판

 

 

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