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

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<h5 style="margin: 0px; line-height: 2em;">메모</h5>
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<math>\int_{0}^{\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}=\pi\ln2</math>
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* [http://www.math.tulane.edu/%7Evhm/Table.html http://www.math.tulane.edu/~vhm/Table.html]<br>
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* [http://www.math.tulane.edu/%7Evhm/web_html/directory.html Directory of notes on the integrals in GR]<br>
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* http://arxiv.org/find/math/1/au:+Moll_V/0/1/0/all/0/1<br>
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* [http://arxiv.org/abs/0704.3872v2 The integrals in Gradshteyn and Ryzhik. Part 1: A family of logarithmic integrals.]<br>
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**  Victor H. Moll<br>
  
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[http://cjackal.tistory.com/109 http://cjackal.tistory.com/10][http://cjackal.tistory.com/109 9]
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http://www.artofproblemsolving.com/Forum/viewtopic.php?t=340081
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<math>\int_{0}^{\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}\,dx=\pi\ln2</math>
  
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* [http://cjackal.tistory.com/109 http://cjackal.tistory.com/10][http://cjackal.tistory.com/109 9]<br>
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* http://www.artofproblemsolving.com/Forum/viewtopic.php?t=340081<br>
  
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<h5 style="margin: 0px; line-height: 3.428em; color: rgb(34, 61, 103); font-family: 'malgun gothic',dotum,gulim,sans-serif; font-size: 1.166em; background-position: 0px 100%;">Gradshteyn and Ryzhik</h5>
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<math>\int_{0}^{\infty}\ln(1+e^{-x})}\,dx=\frac{\pi^2}{12}</math>
  
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* [http://www.math.tulane.edu/%7Evhm/Table.html http://www.math.tulane.edu/~vhm/Table.html]<br>
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[[다이로그 함수(dilogarithm)|다이로그 함수(dilogarithm )]]
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* [http://www.math.tulane.edu/%7Evhm/web_html/directory.html Directory of notes on the integrals in GR]<br>
 
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* http://arxiv.org/find/math/1/au:+Moll_V/0/1/0/all/0/1<br>
 
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* [http://arxiv.org/abs/0704.3872v2 The integrals in Gradshteyn and Ryzhik. Part 1: A family of logarithmic integrals.]<br>
 
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**  Victor H. Moll<br>
 
  
 
 
 
 
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* [[르장드르 카이 함수]]<br>
 
* [[르장드르 카이 함수]]<br>
 
* [[모듈라 군, j-invariant and the singular moduli]]<br>
 
* [[모듈라 군, j-invariant and the singular moduli]]<br>
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* [[카탈란 상수]]<br>
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2010년 6월 17일 (목) 17:08 판

쇼1

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

 

 

증명

[Vardi1988] 참조 

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

이것이 참이라면, Digamma 함수와 디리클레 베타함수에서 얻은 결과를 사용하여 계산할 수 있다.

 

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

\(\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}\), \(\psi(1) = -\gamma\,\!\)

 

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

 

 

\(F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}\) 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, 

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

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

따라서

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

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


그러므로

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

임이 증명된다.

(증명끝)

 

 

쇼2

\(\int_0^{\pi}\frac{x\sin x}{1+\cos^2 x}dx=\frac{\pi^2}{4}\)

\(\int_0^{\pi}\frac{x\cos x}{1+\sin^2 x}dx=\ln^2(\sqrt{2}+1)-\frac{\pi^2}{4}\)

 

 

 

 

Gradshteyn and Ryzhik

 

 

 

 

 

메모

\(\int_{0}^{\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}\,dx=\pi\ln2\)

\(\int_{0}^{\infty}\ln(1+e^{-x})}\,dx=\frac{\pi^2}{12}\)

다이로그 함수(dilogarithm )

 

 

[[란덴변환(Landen's transformation)|]]

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