"르장드르 카이 함수"의 두 판 사이의 차이

2009년 10월 8일 (목) 10:07 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
\(\int_0^{\pi}\frac{x\cos x}{1+\sin^2 x}dx=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}\)

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+<math>\chi_2(i) = iG</math>, <math>G</math>는 [[카탈란 상수]]
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-<math>\chi_2(i) = iK</math>, <math>K</math>는 [[카탈란 상수]]
 <math>\chi_2(\sqrt2 -1) = \frac{\pi^2}{16}-\frac{\ln^2(\sqrt{2}+1)}{4}</math>
@@ 53번째 줄: / 19번째 줄: @@
 <math>\chi_2(\sqrt5 -2}) = \frac{\pi^2}{24}-\frac{3}{4}\ln^2(\frac{\sqrt{5}+1}{2})}</math>
-<math>\chi_2(-1) = </math>
+<math>\chi_2(-1) = -\frac{\pi^2}{8}</math>
-<math>\chi_2(1) = </math>
+<math>\chi_2(1) = \frac{\pi^2}{8}</math>
@@ 64번째 줄: / 30번째 줄: @@
 * [[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>
+* [[로그 탄젠트 적분(log tangent integral)|적분쇼]]<br><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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 <h5>관련논문</h5>
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 * http://www.jstor.org/action/doBasicSearch?Query=