디리클레 베타함수
이 항목의 스프링노트 원문주소==
개요==
- 정의
\(\beta(s) = \sum_{n=0}^\infty \frac{(-1)^n} {(2n+1)^s} = \frac{1}{\Gamma(s)}\int_0^{\infty}\frac{t^{s-1}e^{-t}}{1 + e^{-2t}}\,dt=\frac{1}{2\Gamma(s)}\int_{0}^{\infty}\frac{1}{\cosh t}t^s \frac{\,dt}{t}\)
- \(\chi \colon(\mathbb{Z}/4\mathbb{Z})^\times \to \mathbb C^{*}\) , \(\chi(1)=1\), \(\chi(-1)=-1\) 인 경우의 디리클레 L-함수
\(L_{-4}(s) = \sum_{n\geq 1}\frac{\chi(n)}{n^s}, s>1\)
- 함수방정식
\(\Lambda(s)=(\frac{\pi}{4})^{-{(s+1)}/{2}}\Gamma(\frac{s+1}{2})\beta(s)\) 라 두면
\(\Lambda(s)=\Lambda(1-s)\) 를 만족
- 함수방정식에 대한 일반적인 정리는 디리클레 L-함수 참조
\(\beta(s) = \sum_{n=0}^\infty \frac{(-1)^n} {(2n+1)^s} = \frac{1}{\Gamma(s)}\int_0^{\infty}\frac{t^{s-1}e^{-t}}{1 + e^{-2t}}\,dt=\frac{1}{2\Gamma(s)}\int_{0}^{\infty}\frac{1}{\cosh t}t^s \frac{\,dt}{t}\)
\(L_{-4}(s) = \sum_{n\geq 1}\frac{\chi(n)}{n^s}, s>1\)
\(\Lambda(s)=(\frac{\pi}{4})^{-{(s+1)}/{2}}\Gamma(\frac{s+1}{2})\beta(s)\) 라 두면
\(\Lambda(s)=\Lambda(1-s)\) 를 만족
Special values==
- 아래에서 \(E_n\)은 오일러수를 뜻함.
\(E_0=1\),\(E_2 = â1\),\(E_4 = 5\),\(E_6 = â61\),\(E_8 = 1,385\),\(E_{10} = â50,521\),\(E_{12} = 2,702,765\),\(E_{14} = â199,360,981\),\(E_{16} = 19,391,512,145\),\(E_{18} = â2,404,879,675,441\)
- \(k\geq 0 \) 인 정수일 때,
\(\beta(2k+1)={{{({-1})^k}{E_{2k}}{\pi^{2k+1}} \over {4^{k+1}}(2k!)}}\)
- \(k\geq 0 \)인 정수일 때,
\(\beta(-k)={{E_{k}} \over {2}}\)
\(\beta(0)= \frac{1}{2}\)
\(\beta(1)\;=\;\tan^{-1}(1)\;=\;\frac{\pi}{4}\)
\(\beta(3)\;=\;\frac{\pi^3}{32}\)
\(\beta(5)\;=\;\frac{5\pi^5}{1536}\)
\(\beta(7)\;=\;\frac{61\pi^7}{184320}\)
\(G = \beta(2) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^2} = \frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} + \cdots \!=0.915965594\cdots\)
- 카탈란상수로 많은 정적분에 등장함
\(E_0=1\),\(E_2 = â1\),\(E_4 = 5\),\(E_6 = â61\),\(E_8 = 1,385\),\(E_{10} = â50,521\),\(E_{12} = 2,702,765\),\(E_{14} = â199,360,981\),\(E_{16} = 19,391,512,145\),\(E_{18} = â2,404,879,675,441\)
\(\beta(2k+1)={{{({-1})^k}{E_{2k}}{\pi^{2k+1}} \over {4^{k+1}}(2k!)}}\)
\(\beta(-k)={{E_{k}} \over {2}}\)
\(\beta(0)= \frac{1}{2}\)
\(\beta(1)\;=\;\tan^{-1}(1)\;=\;\frac{\pi}{4}\)
\(\beta(3)\;=\;\frac{\pi^3}{32}\)
\(\beta(5)\;=\;\frac{5\pi^5}{1536}\)
\(\beta(7)\;=\;\frac{61\pi^7}{184320}\)
\(G = \beta(2) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^2} = \frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} + \cdots \!=0.915965594\cdots\)
- 카탈란상수로 많은 정적분에 등장함
증명== 정수에서의 리만제타함수의 값 에서 사용한 방식을 모방한다. \(\beta(5)\)의 경우를 예로 구해보자. \(\oint_{C_{R}}\frac{\pi/2\sec(\pi z/2)}{z^{5}}dz\) \(C_{R}\)는 원점을 중심으로 반지름이 \(R\) 인 원 이때 \(R\)이 커지면, 적분은 0으로 수렴한다. 유수정리를 사용하자. 정수 \(2k+1\)에 대하여 \(z\approx 2k+1\) 이면, \(\pi/2 \sec \pi z/2 \approx \frac{(-1)^{k+1}}{z-(2k+1)}\) \(\frac{\pi/2\sec(\pi z/2)}{z^{5}}\)의 정수 \(2k+1\)에서의 유수(residue)는 \((-1)^{k+1}\frac{1}{(2k+1)^{5}}\)로 주어진다. \(\sec x = 1 + \frac {x^2} {2} + \frac {5 x^4} {24} + \frac {61 x^6} {720} + \cdots=\sum_{n=0}^\infty \frac{(-1)^n E_{2n} x^{2n}}{(2n)!}\) 삼각함수와 쌍곡함수의 맥클로린 급수 참조 를 이용하면 0 에서의 유수는 \(\frac{\pi}{2}\times \frac{5}{24}\times \frac{\pi^4}{16}\)임을 알 수 있다. 그러므로 모든 유수의 합은 \(0=\frac{5\pi^5}{768}+\sum_{-\infty}^{\infty}\frac{(-1)^{k+1}}{(2k+1)^{5}}=\frac{5\pi^5}{768}+\sum_{k=0}^{\infty}\frac{(-1)^{k+1}}{(2k+1)^{5}}+\sum_{n=1}^{\infty}\frac{(-1)^{-n}}{(2n-1)^{5}}=\frac{5\pi^5}{768}+2\sum_{k=1}^{\infty}\frac{(-1)^{k}}{(2k+1)^{5}}\) 따라서 \(\beta(5)=\frac{5\pi^5}{1536}\) 일반적인 자연수 \(k\) 에 대하여도 마찬가지 방법으로 \(\beta(2k+1)={{{({-1})^k}{E_{2k}}{\pi^{2k+1}} \over {4^{k+1}}(2k!)}}\) 을 얻는다. 또한 함수방정식으로부터 \(\beta(0)=\frac{1}{2}\) 와 나머지 짝수인 음의 정수에서의 값을 구할 수 있음
special values for derivative \(\beta'(1)\)==
- \(\beta'(1)\) 의 값
- 후르비츠 제타함수(Hurwitz zeta function)를 사용하면, 함수를 다음과 같이 쓸 수 있음
\(\beta(s)=4^{-s}\{\zeta(s,1/4)-\zeta(s,3/4)\}\)
- 후르비츠 제타함수(Hurwitz zeta function) 의 에르미트 표현
\(\frac{\partial }{\partial s}\zeta(s,a)|_{s=0} =\log \frac{\Gamma(a)}{\sqrt{2\pi}}\)
- 미분은 다음과 주어짐
\(\beta'(s)=4^{-s}\{\zeta(s,1/4)-\zeta(s,3/4)\}(-\log 4)+4^{-s}\{\zeta'(s,1/4)-\zeta'(s,3/4)\}\)
\(\beta'(0)=\{\zeta(0,1/4)-\zeta(0,3/4)\}(-\log 4)+\{\zeta'(0,1/4)-\zeta'(0,3/4)\}=-\beta(0)\log4+\log\frac{\Gamma(1/4)}{\Gamma(3/4)}=\log\frac{\Gamma(1/4)}{2\Gamma(3/4)}\)
- 함수방정식
\(\Lambda(s)=(\frac{\pi}{4})^{-{(s+1)}/{2}}\Gamma(\frac{s+1}{2})\beta(s)\)
\(\Lambda'(s)=(\frac{\pi}{4})^{-{(s+1)}/{2}}(-\frac{1}{2}\log{\frac{\pi}{4})\Gamma(\frac{s+1}{2})\beta(s)+\frac{1}{2}(\frac{\pi}{4})^{-{(s+1)}/{2}}\Gamma'(\frac{s+1}{2})\beta(s)+(\frac{\pi}{4})^{-{(s+1)}/{2}}\Gamma(\frac{s+1}{2})\beta'(s)\)
\(\Lambda(s)=\Lambda(1-s)\)
\(\Lambda'(s)=-\Lambda'(1-s)\) 을 이용하면\(\beta'(0)\)과 \(\beta'(1)\)의 관계를 찾을 수 있다
\(\Lambda'(0)=(\frac{\pi}{4})^{-{1}/{2}}(-\frac{1}{2}\ln{\frac{\pi}{4})\Gamma(\frac{1}{2})\beta(0)+\frac{1}{2}(\frac{\pi}{4})^{-{1}/{2}}\Gamma'(\frac{1}{2})\beta(0)+(\frac{\pi}{4})^{-{1}/{2}}\Gamma(\frac{1}{2})\beta'(0)\)
\(=\frac{2}{\sqrt{\pi}}\ln(\frac{2}{\sqrt{\pi}})\frac{\sqrt{\pi}}{2}+\frac{1}{2}\frac{2}{\sqrt{\pi}}\Gamma'(\frac{1}{2})\frac{1}{2}+\frac{2}{\sqrt{\pi}}\sqrt{\pi}\beta'(0)\)
\(=\ln\frac{2}{\sqrt{\pi}}-\ln 2 -\frac{\gamma}{2}+2\beta'(0)\)
\(\Lambda'(1)=(\frac{\pi}{4})^{-{(1+1)}/{2}}(-\frac{1}{2}\ln{\frac{\pi}{4})\Gamma(\frac{1+1}{2})\beta(1)+\frac{1}{2}(\frac{\pi}{4})^{-{(1+1)}/{2}}\Gamma'(\frac{1+1}{2})\beta(1)+(\frac{\pi}{4})^{-{(1+1)}/{2}}\Gamma(\frac{1+1}{2})\beta'(1)\)
\(=(\frac{4}{\pi})\ln(\frac{2}{\sqrt{\pi}})\Gamma(1)\beta(1)+\frac{1}{2}\frac{4}{\pi}\Gamma'(1)\beta(1)+\frac{4}{\pi}\Gamma(1)\beta'(1)\)
\(=\ln\frac{2}{\sqrt{\pi}}-\frac{\gamma}{2}+\frac{4}{\pi}\beta'(1)\)
(* Digamma 함수 의 값을 이용하였음
\(\psi(1) = -\gamma\,\!\)
\(\Gamma'(1)=-\gamma\)
\(\psi\left(\frac{1}{2}\right) =\frac{\Gamma'(\frac{1}{2})}{\Gamma(\frac{1}{2})}= -2\ln{2} - \gamma\)
\(\Gamma'(1/2)=-\sqrt{\pi}(2\ln2+\gamma)\) *)
- \(\beta'(0)\)과 \(\beta'(1)\)의 관계
\(-\ln\frac{2}{\sqrt{\pi}}+\ln 2+\frac{\gamma}{2}-2\beta'(0)=\ln\frac{2}{\sqrt{\pi}}-\frac{\gamma}{2}+\frac{4}{\pi}\beta'(1)\)
\(\beta'(1)=\frac{\pi}{4}(-2\ln\frac{2}{\sqrt{\pi}}+\ln 2+\gamma-2\beta'(0))=\frac{\pi}{4}(-2\ln\frac{2}{\sqrt{\pi}}+\ln 2+\gamma-2\ln\frac{\Gamma(1/4)}{2\Gamma(3/4)})\)
\(=\frac{\pi}{4}(\ln 2+\ln \pi+ \gamma+2\ln\frac{\Gamma(3/4)}{\Gamma(1/4)})\)
따라서
\(\beta'(1)=L_{-4}'(1)=\frac{\pi}{4}(\gamma+\ln 2\pi)-\frac{\pi}{2}\ln(\frac{\Gamma(1/4)}{\Gamma(3/4)})\)
- 더 일반적인 경우에 대해서는 L-함수의 미분 참조
\(\beta(s)=4^{-s}\{\zeta(s,1/4)-\zeta(s,3/4)\}\)
\(\frac{\partial }{\partial s}\zeta(s,a)|_{s=0} =\log \frac{\Gamma(a)}{\sqrt{2\pi}}\)
\(\beta'(s)=4^{-s}\{\zeta(s,1/4)-\zeta(s,3/4)\}(-\log 4)+4^{-s}\{\zeta'(s,1/4)-\zeta'(s,3/4)\}\)
\(\beta'(0)=\{\zeta(0,1/4)-\zeta(0,3/4)\}(-\log 4)+\{\zeta'(0,1/4)-\zeta'(0,3/4)\}=-\beta(0)\log4+\log\frac{\Gamma(1/4)}{\Gamma(3/4)}=\log\frac{\Gamma(1/4)}{2\Gamma(3/4)}\)
\(\Lambda(s)=(\frac{\pi}{4})^{-{(s+1)}/{2}}\Gamma(\frac{s+1}{2})\beta(s)\)
\(\Lambda'(s)=(\frac{\pi}{4})^{-{(s+1)}/{2}}(-\frac{1}{2}\log{\frac{\pi}{4})\Gamma(\frac{s+1}{2})\beta(s)+\frac{1}{2}(\frac{\pi}{4})^{-{(s+1)}/{2}}\Gamma'(\frac{s+1}{2})\beta(s)+(\frac{\pi}{4})^{-{(s+1)}/{2}}\Gamma(\frac{s+1}{2})\beta'(s)\)
\(\Lambda(s)=\Lambda(1-s)\)
\(\Lambda'(s)=-\Lambda'(1-s)\) 을 이용하면\(\beta'(0)\)과 \(\beta'(1)\)의 관계를 찾을 수 있다
\(\Lambda'(0)=(\frac{\pi}{4})^{-{1}/{2}}(-\frac{1}{2}\ln{\frac{\pi}{4})\Gamma(\frac{1}{2})\beta(0)+\frac{1}{2}(\frac{\pi}{4})^{-{1}/{2}}\Gamma'(\frac{1}{2})\beta(0)+(\frac{\pi}{4})^{-{1}/{2}}\Gamma(\frac{1}{2})\beta'(0)\)
\(=\frac{2}{\sqrt{\pi}}\ln(\frac{2}{\sqrt{\pi}})\frac{\sqrt{\pi}}{2}+\frac{1}{2}\frac{2}{\sqrt{\pi}}\Gamma'(\frac{1}{2})\frac{1}{2}+\frac{2}{\sqrt{\pi}}\sqrt{\pi}\beta'(0)\)
\(=\ln\frac{2}{\sqrt{\pi}}-\ln 2 -\frac{\gamma}{2}+2\beta'(0)\)
\(\Lambda'(1)=(\frac{\pi}{4})^{-{(1+1)}/{2}}(-\frac{1}{2}\ln{\frac{\pi}{4})\Gamma(\frac{1+1}{2})\beta(1)+\frac{1}{2}(\frac{\pi}{4})^{-{(1+1)}/{2}}\Gamma'(\frac{1+1}{2})\beta(1)+(\frac{\pi}{4})^{-{(1+1)}/{2}}\Gamma(\frac{1+1}{2})\beta'(1)\)
\(=(\frac{4}{\pi})\ln(\frac{2}{\sqrt{\pi}})\Gamma(1)\beta(1)+\frac{1}{2}\frac{4}{\pi}\Gamma'(1)\beta(1)+\frac{4}{\pi}\Gamma(1)\beta'(1)\)
\(=\ln\frac{2}{\sqrt{\pi}}-\frac{\gamma}{2}+\frac{4}{\pi}\beta'(1)\)
\(-\ln\frac{2}{\sqrt{\pi}}+\ln 2+\frac{\gamma}{2}-2\beta'(0)=\ln\frac{2}{\sqrt{\pi}}-\frac{\gamma}{2}+\frac{4}{\pi}\beta'(1)\)
\(\beta'(1)=\frac{\pi}{4}(-2\ln\frac{2}{\sqrt{\pi}}+\ln 2+\gamma-2\beta'(0))=\frac{\pi}{4}(-2\ln\frac{2}{\sqrt{\pi}}+\ln 2+\gamma-2\ln\frac{\Gamma(1/4)}{2\Gamma(3/4)})\)
\(=\frac{\pi}{4}(\ln 2+\ln \pi+ \gamma+2\ln\frac{\Gamma(3/4)}{\Gamma(1/4)})\)