"후르비츠 제타함수(Hurwitz zeta function)"의 두 판 사이의 차이

2009년 8월 24일 (월) 17:18 판

\(\zeta(s,a) = \sum_{n=0}^\infty \frac{1}{(n+a)^{s}}\)

\(\frac{\partial }{\partial s}\zeta(s,a)|_{s=0} =\log \frac{\Gamma(a)}{\sqrt{2\pi}}\)

\(\zeta(s,a) = \sum_{n=0}^\infty \frac{1}{(n+a)^{s}}\)

\(\zeta(s,a+1) =\zeta(s,a)-\frac{1}{a^s}\)

\(\zeta'(s,a+1) =\zeta'(s,a)+a^{-s}\log a\)

\(\zeta'(0,a+1) =\zeta'(0,a)+}\log a\)

\(G(a)=e^{\zeta'(0,a)}\)라고 두면, \(G(a+1)=aG(a)\).

\(a>0\) 일때, \(\frac{d^2}{da^2}\log G(a)=\sum_{n=1}^{\infty}\frac{1}{(n+a)^2}> 0\)

또한 \(G(a)\)는 \(a>0\)에서 해석함수이다.

감마함수의 성질로부터 \(G(a)=G(1)\Gamma(a)\)

\(G(1)=\zeta'(0)\) 이므로, \(\zeta'(0)=-\log{\sqrt{2\pi}}\) 로부터 (모든 자연수의 곱과 리만제타함수 참조)

\(\zeta'(s,a) =\log \frac{\Gamma(a)}{\sqrt{2\pi}}\) 임이 증명된다.

@@ 7번째 줄: / 7번째 줄: @@
-<math>\zeta'(s,a) =\log \frac{\Gamma(a)}{\sqrt{2\pi}}</math>
+<math>\frac{\partial }{\partial s}\zeta(s,a)|_{s=0} =\log \frac{\Gamma(a)}{\sqrt{2\pi}}</math>
 <h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px;">증명</h5>
@@ 79번째 줄: / 79번째 줄: @@
 <h5 style="line-height: 2em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px;">관련논문</h5>
-On the Hurwitz zeta-functionBruce C. Berndt, Source: Rocky Mountain J. Math. Volume 2, Number 1 (1972), 151-158.
+* [http://www.springerlink.com/content/wql8d40h20jljxp2/ On Some Integrals Involving the Hurwitz Zeta Function: Part 1]
+* [http://www.springerlink.com/content/t285842772wv0767/ On Some Integrals Involving the Hurwitz Zeta Function: Part 2]<br>
+** Olivier Espinosa  and Victor H. Moll
+* [http://projecteuclid.org/euclid.rmjm/1250131668 On the Hurwitz zeta-function]<br>
+**  Bruce C. Berndt, Source: Rocky Mountain J. Math. Volume 2, Number 1 (1972), 151-158.<br>