르장드르 카이 함수

이 항목의 스프링노트 원문주소

 

 

 

개요

 

 

정의

\chi_\nu(z) = \sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)^\nu}

\chi_\nu(z) = \frac{1}{2}\left[\operatorname{Li}_\nu(z) - \operatorname{Li}_\nu(-z)\right]

\chi_2(z) =\frac{1}{2}\int_0^z{{\log (\frac{1+t}{1-t})\frac{dt}{t}

(증명)

\chi_2(z) = \frac{1}{2}\left[\operatorname{Li}_2(z) - \operatorname{Li}_2(-z)\right]=-\frac{1}{2}\int_0^z{{\log (1-t)}\over t} dt +\frac{1}{2}\int_0^z{{\log (1+t)}\over t} dt =\frac{1}{2}\int_0^z{{\log (\frac{1+t}{1-t})\frac{dt}{t}  ■

 

 

 

성질

\chi_2(\frac{1-z}{1+z})+\chi_2(z) =\frac{\pi^2}{8}+\frac{1}{2}\log z\log (\frac{1+z}{1-z})

(증명)

\frac{d}{dz}[\chi_2(\frac{1-z}{1+z})] = \frac{1}{2}{{\log (\frac{1+\frac{1-z}{1+z}}{1-\frac{1-z}{1+z}})(\frac{1+z}{1-z})(\frac{1-z}{1+z})'=\frac{\log z}{1-z^2}

양변을 적분하면, 즉 \int_0^z \cdots {dz} 을 씌우면,

\chi_2(\frac{1-z}{1+z})-\chi_2(1) =\int_0^z \frac{\log z}{1-z^2}\,dz=\frac{1}{2}\log z \log (\frac{1-z}{1+z})-\frac{1}{2}\int_0^z \log (\frac{1+z}{1-z})\frac{dz}{z}

\chi_2(z) =\frac{1}{2}\int_0^z{{\log (\frac{1+t}{1-t})\frac{dt}{t}와 \chi_2(1) = \frac{\pi^2}{8}를 이용하면, 

\chi_2(\frac{1-z}{1+z})+\chi_2(z) =\frac{\pi^2}{8}+\frac{1}{2}\log z\log (\frac{1+z}{1-z}) 를 얻는다. ♥

 

 

dilogarithm 항등식과의 관계

 

 

special values

\chi_2(i) = iG, G는 카탈란 상수

\chi_2(\sqrt2 -1) = \frac{\pi^2}{16}-\frac{\ln^2(\sqrt{2}+1)}{4}

\chi_2(\frac{\sqrt5 -1}{2}) = \frac{\pi^2}{12}-\frac{3}{4}\ln^2(\frac{\sqrt{5}+1}{2})}

\chi_2(\sqrt5 -2}) = \frac{\pi^2}{24}-\frac{3}{4}\ln^2(\frac{\sqrt{5}+1}{2})}

\chi_2(-1) = -\frac{\pi^2}{8}

\chi_2(1) = \frac{\pi^2}{8}

 

 

special value의 계산

\chi_2(\frac{\sqrt5 -1}{2}) = \frac{\pi^2}{12}-\frac{3}{4}\ln^2(\frac{\sqrt{5}+1}{2})}=\frac{\pi^2}{12}-\frac{3}{4}\ln^2(\frac{\sqrt{5}-1}{2})}

\mbox{Li}_{2}(\frac{1-\sqrt{5}}{2})=-\frac{\pi^2}{15}+\frac{1}{2}\log^2(\frac{1+\sqrt{5}}{2})

\chi_2(\sqrt2 -1) = \frac{\pi^2}{16}-\frac{\ln^2(\sqrt{2}+1)}{4}

위에서 증명한 다음 성질을 이용

\chi_2(\frac{1-z}{1+z})+\chi_2(z) =\frac{\pi^2}{8}+\frac{1}{2}\log z\log (\frac{1+z}{1-z})

z=\sqrt2 -1 로 두면, 원하는 결과를 얻는다. 

(* 또는 Dilogarithm 함수에서 얻은 다음 결과를 이용 

2[\mbox{Li}_2(1-\sqrt 2)-\mbox{Li}_2(\sqrt2 -1)]=\ln^2(\sqrt{2}-1)-\frac{\pi^2}{4}=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4} *)

\chi_2(\sqrt5 -2}) = \frac{\pi^2}{24}-\frac{3}{4}\ln^2(\frac{\sqrt{5}+1}{2})}

위에서 증명한 다음 성질을 이용

\chi_2(\frac{1-z}{1+z})+\chi_2(z) =\frac{\pi^2}{8}+\frac{1}{2}\log z\log (\frac{1+z}{1-z})

z=\frac{\sqrt5 -1}{2} 로 두면, \frac{1-z}{1+z}=z^3=\sqrt{5}-2, z^{-3}=\sqrt{5}+2

\chi_2(\sqrt{5}-2)+\chi_2(\frac{\sqrt5 -1}{2}) =\frac{\pi^2}{8}+\frac{1}{2}\log (\frac{\sqrt5 -1}{2})\log (\sqrt{5}+2)

앞에서 얻은 \chi_2(\frac{\sqrt5 -1}{2}) = \frac{\pi^2}{12}-\frac{3}{4}\ln^2(\frac{\sqrt{5}-1}{2})}를  이용하자. 

\chi_2(\sqrt{5}-2)=\frac{\pi^2}{8}+\frac{1}{2}\log (\frac{\sqrt5 -1}{2})\log (\sqrt{5}+2)-\frac{\pi^2}{12}+\frac{3}{4}\log^2(\frac{\sqrt{5}-1}{2})}

=\frac{\pi^2}{24}-\frac{3}{2}\log^2(\frac{\sqrt5 -1}{2})+\frac{3}{4}\log^2(\frac{\sqrt{5}-1}{2})}=\frac{\pi^2}{24}-\frac{3}{4}\log^2(\frac{\sqrt{5}-1}{2})}

=\frac{\pi^2}{24}-\frac{3}{4}\log^2(\frac{\sqrt{5}+1}{2})}

 

 

재미있는 사실

 

 

2[\mbox{Li}_2(1-\sqrt 2)-\mbox{Li}_2(\sqrt2 -1)]=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}

이 결과는 다음 정적분과 같음.

\int_0^{\pi}\frac{x\cos x}{1+\sin^2 x}dx=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}

 

(증명)

[오늘의계산080810]

\mbox{Li}_2(1-\sqrt 2)=-\mbox{Li}_2(1-\frac{1}{\sqrt 2})-\frac{1}{2} \ln^2(\sqrt{2})

=-[-\mbox{Li}_2(\frac{1}{\sqrt 2})+\frac{\pi^2}{6}-\ln\frac{1}{\sqrt{2}}\ln(1-\frac{1}{\sqrt{2}})]-\frac{1}{8} \ln^2 2

=\mbox{Li}_2(\frac{1}{\sqrt 2})-\frac{\pi^2}{6}+\ln\frac{1}{\sqrt{2}}\ln(1-\frac{1}{\sqrt{2}})-\frac{1}{8} \ln^2 2

=\mbox{Li}_2(\frac{1}{\sqrt 2})-\frac{\pi^2}{6}-\frac{1}{2}\ln{2}\ln({\sqrt{2}}-1)+\frac{1}{8} \ln^2 2

 

 

\mbox{Li}_2(\sqrt 2-1)=\mbox{Li}_2(1-(2-\sqrt 2))

=-\mbox{Li}_2(1-\frac{1}{2-\sqrt 2})-\frac{1}{2}\ln^2(2-\sqrt{2})

=-\mbox{Li}_2(1-\frac{2+\sqrt{2}}{2})-\frac{1}{2}\ln^2(2-\sqrt{2})

=-\mbox{Li}_2(-\frac{1}{\sqrt{2}})-\frac{1}{2}(\frac{1}{2}\ln {2} + \ln(\sqrt{2}-1}))^2

=-\mbox{Li}_2(-\frac{1}{\sqrt{2}})-\frac{1}{8}\ln^2 {2}-\frac{1}{2}\ln 2\ln(\sqrt{2}-1})-\frac{1}{2}\ln^2(\sqrt{2}-1})

 

 

\mbox{Li}_2(1-\sqrt 2)-\mbox{Li}_2(\sqrt2 -1)=

=\mbox{Li}_2(\frac{1}{\sqrt 2})-\frac{\pi^2}{6}-\frac{1}{2}\ln{2}\ln({\sqrt{2}}-1)+\frac{1}{8} \ln^2 2-(-\mbox{Li}_2(-\frac{1}{\sqrt{2}})-\frac{1}{8}\ln^2 {2}-\frac{1}{2}\ln 2\ln(\sqrt{2}-1})-\frac{1}{2}\ln^2(\sqrt{2}-1}))

=\mbox{Li}_2(\frac{1}{\sqrt 2})+\mbox{Li}_2(-\frac{1}{\sqrt{2}})-\frac{\pi^2}{6}+\frac{1}{4} \ln^2 2+\frac{1}{2}\ln^2(\sqrt{2}-1})

=\frac{1}{2}\mbox{Li}_2\frac{1}{2}-\frac{\pi^2}{6}+\frac{1}{4} \ln^2 2 +\frac{1}{2}\ln^2(\sqrt{2}-1})

=\frac{1}{2}(\frac{\pi^2}{12}-\frac{1}{2}\ln^2 2)-\frac{\pi^2}{6}+\frac{1}{4} \ln^2 2+\frac{1}{2}\ln^2(\sqrt{2}-1})

=-\frac{\pi^2}{8}+\frac{1}{2}\ln^2(\sqrt{2}-1})

 

따라서,

2[\mbox{Li}_2(1-\sqrt 2)-\mbox{Li}_2(\sqrt2 -1)]=\ln^2(\sqrt{2}-1)-\frac{\pi^2}{4}=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}

 

메모

\int_0^{\pi}\frac{x\sin x}{1+\cos^2 x}dx=\frac{\pi^2}{4}

\int_0^{\pi}\frac{x\cos x}{1+\sin^2 x}dx=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}

 

 

역사

 

 

관련된 다른 주제들

 

 

수학용어번역

 

사전 형태의 자료

 

 

관련논문

 

 

관련도서 및 추천도서

 

 

관련기사

 

 

블로그