"로그감마 함수"의 두 판 사이의 차이

2010년 6월 26일 (토) 16:08 판

Lerch의 공식 : 후르비츠 제타함수(Hurwitz zeta function)의 미분
\(\frac{\partial }{\partial s}\zeta(s,a)|_{s=0} =\log \frac{\Gamma(a)}{\sqrt{2\pi}}\)

Binet's second expression
\(\operatorname{Re} z > 0 \) 일 때, \(\log \Gamma(z)=(z-\frac{1}{2})\log z -z+\frac{1}{2}\log 2\pi+ 2\int_0^{\infty}\frac{\tan^{-1}(t/z)}{e^{2\pi t} -1}dt\)
http://dlmf.nist.gov/5/9/ 참고

쿰머 (1847)
\(\begin{eqnarray}\log\Gamma(x)=\log\sqrt{2\pi}-\frac{1}{2}\log(2\sin\pi x)+\frac{1}{2}(\gamma+2\log\sqrt{2\pi})(1-2x)+\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k}\sin 2\pi kx \nonumber \\ =(\frac{1}{2}-x)(\gamma+\log 2)+(1-x)\log \pi -\frac{1}{2}\log(\sin\pi x)+\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{\log k}{k}\sin 2\pi kx \nonumber \end{eqnarray} \)

\(\int_{0}^{1}\log\Gamma(x)\,dx=\log\sqrt{2\pi}\)

\(\int_{0}^{\frac{1}{2}}\log\Gamma(x+1)\,dx=-\frac{1}{2}-\frac{7}{24}\log 2+\frac{1}{4}\log \pi+\frac{3}{2}\log A\)

@@ 39번째 줄: / 39번째 줄: @@
 <h5 style="line-height: 3.428em; margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0px; color: rgb(34, 61, 103); font-family: 'malgun gothic', dotum, gulim, sans-serif; font-size: 1.166em; background-image: ; background-color: initial; background-position: 0px 100%;">테일러 급수</h5>
-* [[로그감마 함수]]의 테일러 급수<br><math>\log\Gamma(1+x) =-\gamma x+\sum_{k=2}^{\infty}(-1)^k \frac{\zeta(k)}{k}x^k</math><br>
+* [[로그감마 함수]]의 테일러 급수 (http://www.wolframalpha.com/input/?i=taylor+series+of+log+gamma(1%2Bx)+at+x%3D0)<br><math>\log\Gamma(1+x) =-\gamma x+\sum_{k=2}^{\infty}(-1)^k \frac{\zeta(k)}{k}x^k</math><br>
+* [[정수에서의 리만제타함수의 값]]<br>
@@ 48번째 줄: / 49번째 줄: @@
 <math>\int_{0}^{1}\log\Gamma(x)\,dx=\log\sqrt{2\pi}</math>
+<math>\int_{0}^{\frac{1}{2}}\log\Gamma(x+1)\,dx=-\frac{1}{2}-\frac{7}{24}\log 2+\frac{1}{4}\log \pi+\frac{3}{2}\log A</math>