르장드르 카이 함수
이 항목의 스프링노트 원문주소
간단한 소개
\(\chi_\nu(z) = \sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)^\nu}\)
\(\chi_\nu(z) = \frac{1}{2}\left[\operatorname{Li}_\nu(z) - \operatorname{Li}_\nu(-z)\right]\)
- Dilogarithm 함수
- polylogarithm 함수
\(\chi_2(z) =\frac{1}{2}\int_0^z{{\log (\frac{1+t}{1-t})\frac{dt}{t}\)
(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. ♥
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, \(\frac{1-z}{1+z}=\sqrt{5}-2\)
\(\chi_2(\sqrt{5}-2)+\chi_2(\frac{\sqrt5 -1}{2}) =\frac{\pi^2}{8}+\frac{1}{2}\log (\frac{\sqrt5 -1}{2})\log (\sqrt{5}+2)\)
앞에서 얻은 \(\chi_2(\frac{\sqrt5 -1}{2}) = \frac{\pi^2}{12}-\frac{3}{4}\ln^2(\frac{\sqrt{5}+1}{2})}\) 를 이용하자.
\(z=\frac{\sqrt5 -1}{2}\)
재미있는 사실
- 디리클레 베타함수
\(\beta(s) = \sum_{n=0}^\infty \frac{(-1)^n} {(2n+1)^s} = \frac{1}{\Gamma(s)}\int_0^{\infty}\frac{t^{s-1}e^{-t}}{1 + e^{-2t}}\,dt=\frac{1}{2\Gamma(s)}\int_{0}^{\infty}\frac{1}{\cosh t}t^s \frac{\,dt}{t}\) - 적분쇼
\(\int_0^{\pi}\frac{x\cos x}{1+\sin^2 x}dx=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}\)
역사
관련된 다른 주제들
수학용어번역
사전 형태의 자료
- http://ko.wikipedia.org/wiki/
- http://en.wikipedia.org/wiki/Legendre_chi_function
- http://en.wikipedia.org/wiki/
- http://mathworld.wolfram.com/LegendresChi-Function.html
- http://www.wolframalpha.com/input/?i=
- NIST Digital Library of Mathematical Functions
관련논문
- Some Identities Involving the Legendre's Chi-Function
- Junesang Choi, Communications of the Korean Mathematical Society ( Vol.22 NO.2 / 2007 )
- Values of the Legendre chi and Hurwitz zeta functions at rational arguments
- Djurdje Cvijović, Jacek Klinowski, Mathematics of Computation archive Volume 68 , Issue 228 (October 1999)
- http://www.jstor.org/action/doBasicSearch?Query=
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