로그 사인 적분 (log sine integrals)

이 항목의 스프링노트 원문주소

• 로그 사인 적분 (log sine integrals)
  

 

 

개요

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

• 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)

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

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

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

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

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

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

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

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

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

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

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

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

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