"르장드르 카이 함수"의 두 판 사이의 차이

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==이 항목의 스프링노트 원문주소==
 
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* [[르장드르 카이 함수]]
 
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==개요==
 
==개요==
  
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==정의==
 
==정의==
21번째 줄: 11번째 줄:
 
<math>\chi_\nu(z) = \frac{1}{2}\left[\operatorname{Li}_\nu(z) - \operatorname{Li}_\nu(-z)\right]</math>
 
<math>\chi_\nu(z) = \frac{1}{2}\left[\operatorname{Li}_\nu(z) - \operatorname{Li}_\nu(-z)\right]</math>
  
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* [[다이로그 함수(dilogarithm)|Dilogarithm 함수]]<br>
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* [[다이로그 함수(dilogarithm)|Dilogarithm 함수]]
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*  polylogarithm 함수<br>
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*  polylogarithm 함수
  
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<math>\chi_2(z) =\frac{1}{2}\int_0^z{{\log (\frac{1+t}{1-t})\frac{dt}{t}</math>
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<math>\chi_2(z) =\frac{1}{2}\int_0^z \log (\frac{1+t}{1-t})\frac{dt}{t}</math>
  
 
(증명)
 
(증명)
  
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<math>\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}</math>  ■
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<math>\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}</math> ■
  
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==성질==
 
==성질==
44번째 줄: 34번째 줄:
 
<math>\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}</math>
 
<math>\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}</math>
  
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양변을 적분하면, 즉 <math>\int_0^z \cdots {dz}</math> 을 씌우면,
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양변을 적분하면, 즉 <math>\int_0^z \cdots {dz}</math> 을 씌우면,
  
 
<math>\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}</math>
 
<math>\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}</math>
  
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<math>\chi_2(z) =\frac{1}{2}\int_0^z{{\log (\frac{1+t}{1-t})\frac{dt}{t}</math>와 <math>\chi_2(1) = \frac{\pi^2}{8}</math>를 이용하면, 
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<math>\chi_2(z) =\frac{1}{2}\int_0^z{{\log (\frac{1+t}{1-t})\frac{dt}{t}</math>와 <math>\chi_2(1) = \frac{\pi^2}{8}</math>를 이용하면,  
  
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<math>\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})</math> 를 얻는다. ♥
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<math>\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})</math> 를 얻는다. ♥
  
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==dilogarithm 항등식과의 관계==
 
==dilogarithm 항등식과의 관계==
  
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* [[다이로그 함수(dilogarithm)|dilogarithm 함수]]의 곱셈공식<br><math>\mbox{Li}_2(x^2)=2(\mbox{Li}_2(x)+\mbox{Li}_2(-x))</math><br>
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* [[다이로그 함수(dilogarithm)|dilogarithm 함수]]의 곱셈공식:<math>\mbox{Li}_2(x^2)=2(\mbox{Li}_2(x)+\mbox{Li}_2(-x))</math>
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* <math>\nu=2</math>인 경우의 르장드르 카이 함수는 다음과 같이 쓸 수 있다<br><math>\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)</math><br>
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* <math>\nu=2</math>인 경우의 르장드르 카이 함수는 다음과 같이 쓸 수 있다:<math>\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)</math>
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*  special value의 계산으로부터 [[다이로그 항등식 (dilogarithm identities)|dilogarithm 항등식]] 을 얻을 수 있다<br>
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*  special value의 계산으로부터 [[다이로그 항등식 (dilogarithm identities)|dilogarithm 항등식]] 을 얻을 수 있다
  
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==special values==
 
==special values==
  
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<math>\chi_2(i) = iG</math>, <math>G</math>는 [[카탈란 상수]]
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<math>\chi_2(i) = iG</math>, <math>G</math>는 [[카탈란 상수]]
  
 
<math>\chi_2(\sqrt2 -1) = \frac{\pi^2}{16}-\frac{\ln^2(\sqrt{2}+1)}{4}</math>
 
<math>\chi_2(\sqrt2 -1) = \frac{\pi^2}{16}-\frac{\ln^2(\sqrt{2}+1)}{4}</math>
80번째 줄: 70번째 줄:
 
<math>\chi_2(1) = \frac{\pi^2}{8}</math>
 
<math>\chi_2(1) = \frac{\pi^2}{8}</math>
  
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==special value의 계산==
 
==special value의 계산==
88번째 줄: 78번째 줄:
 
<math>\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})}</math>
 
<math>\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})}</math>
  
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* [[다이로그 함수(dilogarithm)|Dilogarithm 함수]]에서 얻어진 다음 두 결과를 이용<br><math>\mbox{Li}_{2}(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}-\log^2(\frac{1+\sqrt{5}}{2})</math><br>
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* [[다이로그 함수(dilogarithm)|Dilogarithm 함수]]에서 얻어진 다음 두 결과를 이용:<math>\mbox{Li}_{2}(\frac{-1+\sqrt{5}}{2})=\frac{\pi^2}{10}-\log^2(\frac{1+\sqrt{5}}{2})</math>
  
 
<math>\mbox{Li}_{2}(\frac{1-\sqrt{5}}{2})=-\frac{\pi^2}{15}+\frac{1}{2}\log^2(\frac{1+\sqrt{5}}{2})</math>
 
<math>\mbox{Li}_{2}(\frac{1-\sqrt{5}}{2})=-\frac{\pi^2}{15}+\frac{1}{2}\log^2(\frac{1+\sqrt{5}}{2})</math>
99번째 줄: 89번째 줄:
 
<math>\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})</math>
 
<math>\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})</math>
  
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<math>z=\sqrt2 -1</math> 로 두면, 원하는 결과를 얻는다. 
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<math>z=\sqrt2 -1</math> 로 두면, 원하는 결과를 얻는다.  
  
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(* 또는 [[다이로그 함수(dilogarithm)|Dilogarithm 함수]]에서 얻은 다음 결과를 이용 
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(* 또는 [[다이로그 함수(dilogarithm)|Dilogarithm 함수]]에서 얻은 다음 결과를 이용
  
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<math>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}</math> *)
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<math>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}</math> *)
 
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112번째 줄: 102번째 줄:
 
<math>\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})</math>
 
<math>\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})</math>
  
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<math>z=\frac{\sqrt5 -1}{2}</math> 로 두면, <math>\frac{1-z}{1+z}=z^3=\sqrt{5}-2</math>, <math>z^{-3}=\sqrt{5}+2</math>
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<math>z=\frac{\sqrt5 -1}{2}</math> 로 두면, <math>\frac{1-z}{1+z}=z^3=\sqrt{5}-2</math>, <math>z^{-3}=\sqrt{5}+2</math>
  
 
<math>\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)</math>
 
<math>\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)</math>
  
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앞에서 얻은 <math>\chi_2(\frac{\sqrt5 -1}{2}) = \frac{\pi^2}{12}-\frac{3}{4}\ln^2(\frac{\sqrt{5}-1}{2})}</math>를  이용하자. 
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앞에서 얻은 <math>\chi_2(\frac{\sqrt5 -1}{2}) = \frac{\pi^2}{12}-\frac{3}{4}\ln^2(\frac{\sqrt{5}-1}{2})}</math>를 이용하자.  
  
 
<math>\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})}</math>
 
<math>\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})}</math>
124번째 줄: 114번째 줄:
 
<math>=\frac{\pi^2}{24}-\frac{3}{4}\log^2(\frac{\sqrt{5}+1}{2})}</math>
 
<math>=\frac{\pi^2}{24}-\frac{3}{4}\log^2(\frac{\sqrt{5}+1}{2})}</math>
  
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==재미있는 사실==
 
==재미있는 사실==
  
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* [[디리클레 베타함수]]<br><math>\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}</math><br>
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* [[디리클레 베타함수]]:<math>\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}</math>
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* <math>\int_0^{\pi}\frac{x\cos x}{1+\sin^2 x}dx=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}</math><br>
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* <math>\int_0^{\pi}\frac{x\cos x}{1+\sin^2 x}dx=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}</math>
  
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<math>2[\mbox{Li}_2(1-\sqrt 2)-\mbox{Li}_2(\sqrt2 -1)]=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}</math>
 
<math>2[\mbox{Li}_2(1-\sqrt 2)-\mbox{Li}_2(\sqrt2 -1)]=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}</math>
143번째 줄: 133번째 줄:
 
<math>\int_0^{\pi}\frac{x\cos x}{1+\sin^2 x}dx=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}</math>
 
<math>\int_0^{\pi}\frac{x\cos x}{1+\sin^2 x}dx=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}</math>
  
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(증명)
 
(증명)
157번째 줄: 147번째 줄:
 
<math>=\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</math>
 
<math>=\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</math>
  
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<math>\mbox{Li}_2(\sqrt 2-1)=\mbox{Li}_2(1-(2-\sqrt 2))</math>
 
<math>\mbox{Li}_2(\sqrt 2-1)=\mbox{Li}_2(1-(2-\sqrt 2))</math>
171번째 줄: 161번째 줄:
 
<math>=-\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})</math>
 
<math>=-\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})</math>
  
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<math>\mbox{Li}_2(1-\sqrt 2)-\mbox{Li}_2(\sqrt2 -1)=</math>
 
<math>\mbox{Li}_2(1-\sqrt 2)-\mbox{Li}_2(\sqrt2 -1)=</math>
187번째 줄: 177번째 줄:
 
<math>=-\frac{\pi^2}{8}+\frac{1}{2}\ln^2(\sqrt{2}-1})</math>
 
<math>=-\frac{\pi^2}{8}+\frac{1}{2}\ln^2(\sqrt{2}-1})</math>
  
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따라서,
 
따라서,
193번째 줄: 183번째 줄:
 
<math>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}</math>
 
<math>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}</math>
  
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==메모==
 
==메모==
201번째 줄: 191번째 줄:
 
<math>\int_0^{\pi}\frac{x\cos x}{1+\sin^2 x}dx=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}</math>
 
<math>\int_0^{\pi}\frac{x\cos x}{1+\sin^2 x}dx=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}</math>
  
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==역사==
 
==역사==
  
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* [[수학사연표 (역사)|수학사연표]]
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* [[수학사 연표]]
  
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==관련된 다른 주제들==
 
  
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==관련된 항목들==
 
* [[다이로그 함수(dilogarithm)|Dilogarithm]]
 
* [[다이로그 함수(dilogarithm)|Dilogarithm]]
 
* [[로그 탄젠트 적분(log tangent integral)|적분쇼]]
 
* [[로그 탄젠트 적분(log tangent integral)|적분쇼]]
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* [[황금비]]
 
* [[황금비]]
  
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==수학용어번역==
 
  
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* http://www.google.com/dictionary?langpair=en|ko&q=
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* [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집]<br>
 
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** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=
 
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* [http://kms.or.kr/home/kor/board/bulletin_list_subject.asp?bulletinid=%7BD6048897-56F9-43D7-8BB6-50B362D1243A%7D&boardname=%BC%F6%C7%D0%BF%EB%BE%EE%C5%E4%B7%D0%B9%E6&globalmenu=7&localmenu=4 대한수학회 수학용어한글화 게시판]
 
  
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==사전 형태의 자료==
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==사전 형태의 자료==
  
 
* http://ko.wikipedia.org/wiki/
 
* http://ko.wikipedia.org/wiki/
244번째 줄: 225번째 줄:
 
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
 
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
  
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==관련논문==
 
==관련논문==
  
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* [http://www.mathnet.or.kr/mathnet/kms_content.php?no=378689 Some Identities Involving the Legendre's Chi-Function]<br>
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* [http://www.mathnet.or.kr/mathnet/kms_content.php?no=378689 Some Identities Involving the Legendre's Chi-Function]
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** Junesang Choi, Communications of the Korean Mathematical Society ( Vol.22 NO.2 / 2007 )
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** Junesang Choi, Communications of the Korean Mathematical Society ( Vol.22 NO.2 / 2007 )
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* [http://www.ams.org/mcom/1999-68-228/S0025-5718-99-01091-1/home.html Values of the Legendre chi and Hurwitz zeta functions at rational arguments]<br>
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* [http://www.ams.org/mcom/1999-68-228/S0025-5718-99-01091-1/home.html Values of the Legendre chi and Hurwitz zeta functions at rational arguments]
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** Djurdje Cvijović, Jacek Klinowski, Mathematics of Computation archive Volume 68 ,  Issue 228  (October 1999)
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** Djurdje Cvijović, Jacek Klinowski, Mathematics of Computation archive Volume 68 , Issue 228 (October 1999)
 
* http://www.jstor.org/action/doBasicSearch?Query=
 
* http://www.jstor.org/action/doBasicSearch?Query=
  
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==블로그==
  
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* '''[오늘의계산080810]'''[http://sos440.tistory.com/83 오늘의 계산 12]
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** 수학 잡담/오늘의 계산, 2008/08/10
  
  
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[[분류:다이로그]]
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[[분류:특수함수]]
  
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==메타데이터==
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===위키데이터===
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* ID :  [https://www.wikidata.org/wiki/Q1452853 Q1452853]
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===Spacy 패턴 목록===
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* [{'LOWER': 'legendre'}, {'LOWER': 'chi'}, {'LEMMA': 'function'}]
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==블로그==
 
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* '''[오늘의계산080810]'''[http://sos440.tistory.com/83 오늘의 계산 12]<br>
 
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** 수학 잡담/오늘의 계산, 2008/08/10
 
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* 구글 블로그 검색 http://blogsearch.google.com/blogsearch?q=
 
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네이버 ]
 

2021년 2월 17일 (수) 04:41 기준 최신판

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

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

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



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