"르장드르 카이 함수"의 두 판 사이의 차이
Pythagoras0 (토론 | 기여) |
Pythagoras0 (토론 | 기여) |
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==개요== | ==개요== | ||
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==정의== | ==정의== | ||
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(증명) | (증명) | ||
| − | <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> | + | <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== | ||
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<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> | ||
| − | 양변을 적분하면, | + | 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> | ||
| − | <math>\chi_2(z) =\frac{1}{2}\int_0^z{{\log (\frac{1+t}{1-t})\frac{dt}{t}</math> | + | <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, |
| − | <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> 를 얻는다. ♥ |
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==dilogarithm 항등식과의 관계== | ==dilogarithm 항등식과의 관계== | ||
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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> | * [[다이로그 함수(dilogarithm)|dilogarithm 함수]]의 곱셈공식:<math>\mbox{Li}_2(x^2)=2(\mbox{Li}_2(x)+\mbox{Li}_2(-x))</math> | ||
* <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> | * <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> | ||
| − | * special value의 | + | * special value의 계산으로부터 [[다이로그 항등식 (dilogarithm identities)|dilogarithm 항등식]] 을 얻을 수 있다 |
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==special values== | ==special values== | ||
| − | <math>\chi_2(i) = iG</math>, | + | <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> | ||
| 70번째 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== | ||
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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> | <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>z=\sqrt2 -1</math> | + | <math>z=\sqrt2 -1</math> 로 두면, 원하는 결과를 얻는다. |
| − | (* | + | (* 또는 [[다이로그 함수(dilogarithm)|Dilogarithm 함수]]에서 얻은 다음 결과를 이용 |
| − | <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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| 102번째 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> | ||
| − | <math>z=\frac{\sqrt5 -1}{2}</math> | + | <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> | ||
| − | 앞에서 | + | 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> | ||
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<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== | ||
| 123번째 줄: | 123번째 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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<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> | ||
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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> | <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) | ||
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<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> | ||
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<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> | ||
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<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, | ||
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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> | <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== | ||
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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> | <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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* [[수학사 연표]] | * [[수학사 연표]] | ||
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==관련된 항목들== | ==관련된 항목들== | ||
| 211번째 줄: | 211번째 줄: | ||
* [[황금비]] | * [[황금비]] | ||
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| − | ==사전 | + | ==사전 형태의 자료== |
* http://ko.wikipedia.org/wiki/ | * http://ko.wikipedia.org/wiki/ | ||
| 225번째 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== | ||
* [http://www.mathnet.or.kr/mathnet/kms_content.php?no=378689 Some Identities Involving the Legendre's Chi-Function] | * [http://www.mathnet.or.kr/mathnet/kms_content.php?no=378689 Some Identities Involving the Legendre's Chi-Function] | ||
| − | ** Junesang Choi, | + | ** Junesang Choi, Communications of the Korean Mathematical Society ( Vol.22 NO.2 / 2007 ) |
* [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] | * [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] | ||
| − | ** Djurdje Cvijović, Jacek Klinowski, | + | ** 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= | ||
2020년 12월 28일 (월) 02:19 판
개요
정의
\(\chi_\nu(z) = \sum_{k=0}^\infty \frac{z^{2k+1}}{(2k+1)^\nu}\)
\(\chi_\nu(z) = \frac{1}{2}\left[\operatorname{Li}_\nu(z) - \operatorname{Li}_\nu(-z)\right]\)
- Dilogarithm 함수
- polylogarithm 함수
\(\chi_2(z) =\frac{1}{2}\int_0^z \log (\frac{1+t}{1-t})\frac{dt}{t}\)
(증명)
\(\chi_2(z) = \frac{1}{2}\left[\operatorname{Li}_2(z) - \operatorname{Li}_2(-z)\right]=-\frac{1}{2}\int_0^z \frac{\log (1-t)}{t} dt +\frac{1}{2}\int_0^z \frac{\log (1+t)}{t} dt =\frac{1}{2}\int_0^z \log (\frac{1+t}{1-t})\frac{dt}{t}\) ■
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
- 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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- 수학 잡담/오늘의 계산, 2008/08/10