로그 사인 적분 (log sine integrals)

개요

• 정의 \[\operatorname{Ls}_{a+b,a}(\theta):=-\int_{0}^{\theta}x^a\log^{b-1}|2\sin \frac{x}{2}|\,dx\] \[\operatorname{Ls}_{n}(\theta):=\operatorname{Ls}_{n,0}(\theta)=-\int_{0}^{\theta}\log^{n-1}(2\sin \frac{x}{2})\,dx\]
• 다음과 같은 적분에서 등장한다

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

• 클라우센 함수의 일반화로 볼 수 있다 \[\operatorname{Cl}_2(\theta)=-\int_0^{\theta} \ln |2\sin \frac{t}{2}| \,dt=\sum_{n=1}^{\infty}\frac{\sin (n\theta)}{n^2}\]

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

지수생성함수

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 \[\log\Gamma(1+x) =-\gamma x+\sum_{k=2}^{\infty}(-1)^k \frac{\zeta(k)}{k}x^k.\] ■

• \(I(x)\) 에 대하여 다음이 성립한다

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

좀 더 일반적으로, \[\frac{1}{\pi}\int_{0}^{\pi}(2 \cos \frac{1}{2}\theta)^{x}\cos y\theta \,d\theta=\frac{\Gamma(1+x)}{\Gamma(1+\frac{1}{2}x+y)\Gamma(1+\frac{1}{2}x-y)}\] 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

• 중심이항계수(central binomial coefficient)\[\sum_{n=1}^\infty \frac{1}{n^3\binom{2n}{n}}=4\int_{0}^{\frac{1}{2}}(\arcsin x)^2\frac{dx}{x}=-2\int_{0}^{\pi/3}x\log(2\sin \frac{x}{2})\,dx\]
• http://cjackal.tistory.com/109
• http://www.wolframalpha.com/input/?i=integrate+(log+(2sin(x/2)))^2+dx+from+0+to+pi/3
• http://mathworld.wolfram.com/RamanujanLog-TrigonometricIntegrals.html

역사

• 수학사 연표

관련된 항목들

• 로그 탄젠트 적분(log tangent integral)
• 다이로그 함수(dilogarithm )
• 폴리로그 함수(polylogarithm)
• 로바체프스키와 클라우센 함수
• 정수에서의 리만제타함수의 값
• 중심이항계수(central binomial coefficient)
• 오일러 베타적분
• 로그 탄젠트 적분

계산 리소스

• LsToLi: A Mathematica package for evaluating log-sine integrals
  • http://arminstraub.com/pub/log-sine-integrals
• Kalmykov, M. Yu, and A. Sheplyakov. 2004. “LSJK - a C++ Library for Arbitrary-precision Numeric Evaluation of the Generalized Log-sine Functions.” arXiv:hep-ph/0411100 (November 7). doi:10.1016/j.cpc.2005.04.013. http://arxiv.org/abs/hep-ph/0411100.

관련논문

• 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.
• Borwein, Jonathan M, and Armin Straub. 2011. Log-sine evaluations of Mahler measures. 1103.3893 (March 20). http://arxiv.org/abs/1103.3893.
• Mark W. Coffey On some log-cosine integrals related to ζ(3), ζ(4), and ζ(6), 2003
• Multiple Gamma and Related Functions
  • J. Choi, H. M. Srivastava, V.S. Adamchik , Applied Mathematics and Computation, 134 (2003), 515-533
• [Borwein1995]On an Intriguing Integral and Some Series Related to ζ(4)
  • David Borwein and Jonathan M. Borwein, Proceedings of the American Mathematical Society, Vol. 123, No. 4 (Apr., 1995), pp. 1191-1198
• Some wonderful formulas ... an introduction to polylogarithms
  • A.J. Van der Poorten, Queen's papers in Pure and Applied Mathematics, 54 (1979), 269-286
• On the value of a logarithmic-trigonometric integral
  • K. S. Kölbig, 1970
• [Lewin1958] L. Lewin On the Evaluation of log-sine Integrals The Mathematical Gazette, Vol. 42, No. 340 (May, 1958), pp. 125-128
• [Bowman1947]Note on the Integral, J. London Math. Soc. 1947 s1-22: 172-173