"로그 탄젠트 적분(log tangent integral)"의 두 판 사이의 차이

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http://www.math.tulane.edu/~vhm/Table.html
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* [[등차수열의 소수분포에 관한 디리클레 정리]]<br>
 
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* [[감마함수]]<br>
 
* [[감마함수]]<br>
  
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* [http://www.springerlink.com/content/p2k0106727416271/?p=03915f5244d74523b6d36406299c80d5&pi=6 A class of logarithmic integrals]<br>
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** Luis A. Medina1  and Victor H. Moll, The Ramanujan Journal, Volume 20, Number 1 / 2009년 10월
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* [http://www.springerlink.com/content/wql8d40h20jljxp2/ On Some Integrals Involving the Hurwitz Zeta Function: Part 1]<br>
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** Olivier Espinosa  and Victor H. Moll
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* [http://www.springerlink.com/content/t285842772wv0767/ On Some Integrals Involving the Hurwitz Zeta Function: Part 2]<br>
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** Olivier Espinosa  and Victor H. Moll
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* [http://doi.acm.org/10.1145/258726.258736 A class of logarithmic integrals]<br>
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** Victor Adamchik, 1997
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* [http://www.jstor.org/stable/2323562 Integrals, an Introduction to Analytic Number Theory]<br>
 
* [http://www.jstor.org/stable/2323562 Integrals, an Introduction to Analytic Number Theory]<br>
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* [http://www.amazon.com/Irresistible-Integrals-Symbolics-Experiments-Evaluation/dp/0521796369 Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals]<br>
 
* [http://www.amazon.com/Irresistible-Integrals-Symbolics-Experiments-Evaluation/dp/0521796369 Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals]<br>
 
** George Boros and Victor Moll
 
** George Boros and Victor Moll
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2009년 10월 7일 (수) 18:27 판

간단한 소개

 

\(\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\frac{\pi}{2}\ln{\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}\)

 

 

증명

 

\(\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\frac{d}{ds}\Gamma(s)\beta(s)|_{s=1}\) 임을 먼저 보이자

여기서 \(\Gamma(s)\)는 감마함수,\(\beta(s)\)는 디리클레 베타함수.

 

\(F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}\) 라 하자.

\(\Gamma(s)F(s)=\int_0^{\infty}(\sum_{n=1}^{\infty}f(n)e^{-nt})t^{s-1}\,dt\)\(z=e^{-t}\) 로 치환하면,

\(\Gamma(s)F(s)=\int_0^{1}(\sum_{n=1}^{\infty}f(n)z^n)(\log\frac{1}{z})^{s-1}\,\frac{dz}{z}\)

 

만약 \(f(n+q)=f(n)\) 을 만족하면 (가령 디리클레 캐릭터의 경우)

\(p(z)=\sum_{n=1}^{q-1}f(n)z^n\)라면,  \(\sum_{n=1}^{\infty}f(n)z^n=\frac{p(z)}{1-z^q}\) 로 쓸 수 있다.

 

이를 이용하면, 

\(\Gamma(s)F(s)=\int_0^{1}\frac{p(z)(\log\frac{1}{z})^{s-1}}{1-z^q}\,\frac{dz}{z}\) 를 얻는다.

 

\(\frac{d}{ds}\Gamma(s)F(s)=\int_0^{1}\frac{p(z)(\log\frac{1}{z})^{s-1}}{1-z^q}\log \log\frac{1}{z} \,\frac{dz}{z}\)

\(s=1\) 에서 \(F(s)\)가  미분가능하다면, 

\(F'(1)-\gamma F(1)=\int_0^{1}\frac{p(z)}{1-z^q}\log \log\frac{1}{z} \,\frac{dz}{z}\)

\(f\)가 \(f(3)=-1\)인 주기가 4인 디리클레 캐릭터라고 하면, \(p(z)=z-z^3\)

\(\beta(s) = \sum_{n\geq 1}\frac{f(n)}{n^s}\)

\(\beta'(1)-\gamma \frac{\pi}{4}=\int_0^{1}\frac{z-z^3}{1-z^4}\log \log\frac{1}{z} \,\frac{dz}{z}=\int_0^{1}\log \log\frac{1}{z} \,\frac{dz}{1+z^2}=\int_1^{\infty}\log \log u \,\frac{du}{1+u^2}\)

\(=\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx\)

 

이제 디리클레 베타함수에서 얻은 결과를 사용하자.

\(\beta'(1)=\frac{\pi}{4}\gamma+\frac{\pi}{2}\ln(\frac{\Gamma(3/4)}{\Gamma(1/4)}\sqrt{2\pi})\)

 

따라서

\(\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\beta'(1)- \frac{\pi}{4}\gamma=\frac{\pi}{2}\ln{\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}\)

(증명끝)

 

 

Gradshteyn and Ryzhik

http://www.math.tulane.edu/~vhm/Table.html

 

 

The integrals in Gradshteyn and Ryzhik. Part 1: A family of logarithmic integrals.

[1]Victor H. Moll

 

 

란덴변환(Landen's transformation)

 

 

재미있는 사실

 

 

역사

 

 

관련된 다른 주제들

 

수학용어번역

 

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