르장드르 카이 함수

수학노트
Pythagoras0 (토론 | 기여)님의 2020년 12월 28일 (월) 06:37 판 (→‎메타데이터: 새 문단)
둘러보기로 가기 검색하러 가기

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)

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




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

  • dilogarithm 함수의 곱셈공식\[\mbox{Li}_2(x^2)=2(\mbox{Li}_2(x)+\mbox{Li}_2(-x))\]
  • \(\nu=2\)인 경우의 르장드르 카이 함수는 다음과 같이 쓸 수 있다\[\chi_\nu(z) = \frac{1}{2}\left[\operatorname{Li}_\nu(z) - \operatorname{Li}_\nu(-z)\right]=\operatorname{Li}_2(z) - \frac{1}{4}\operatorname{Li}_2(z^2)\]
  • special value의 계산으로부터 dilogarithm 항등식 을 얻을 수 있다



special values

\(\chi_2(i) = iG\), \(G\)는 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

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

  • Dilogarithm 함수에서 얻어진 다음 두 결과를 이용\[\mbox{Li}_{2}(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}-\log^2(\frac{1+\sqrt{5}}{2})\]

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

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

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

  • 디리클레 베타함수\[\beta(s) = \sum_{n=0}^\infty \frac{(-1)^n} {(2n+1)^s} = \frac{1}{\Gamma(s)}\int_0^{\infty}\frac{t^{s-1}e^{-t}}{1 + e^{-2t}}\,dt=\frac{1}{2\Gamma(s)}\int_{0}^{\infty}\frac{1}{\cosh t}t^s \frac{\,dt}{t}\]
  • \(\int_0^{\pi}\frac{x\cos x}{1+\sin^2 x}dx=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}\)



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

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 항목들




사전 형태의 자료



관련논문


블로그

메타데이터

위키데이터

원본 주소 "https://wiki.mathnt.net/index.php?title=르장드르_카이_함수&oldid=48768"