"르장드르 카이 함수"의 두 판 사이의 차이

2009년 10월 8일 (목) 10:12 판

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

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

\(\chi_2(i) = iG\), \(G\)는 카탈란 상수

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

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

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

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

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

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

@@ 44번째 줄: / 44번째 줄: @@
 <h5>관련된 다른 주제들</h5>
+* [[다이로그 함수(dilogarithm)|Dilogarithm]]
 * [[로그 탄젠트 적분(log tangent integral)|적분쇼]]
 * [[후르비츠 제타함수(Hurwitz zeta function)]]
-* [[다이로그 함수(dilogarithm)|Dilogarithm]]
 * [[황금비]]
@@ 77번째 줄: / 77번째 줄: @@
 <h5>관련논문</h5>
-*
+* [http://www.mathnet.or.kr/mathnet/kms_content.php?no=378689 Some Identities Involving the Legendre's Chi-Function]<br>
+** 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]<br>
+** Djurdje Cvijović, Jacek Klinowski, Mathematics of Computation archive Volume 68 ,  Issue 228  (October 1999)
 * http://www.jstor.org/action/doBasicSearch?Query=