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

수학노트
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(사용자 2명의 중간 판 21개는 보이지 않습니다)
1번째 줄: 1번째 줄:
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<h5 style="LINE-HEIGHT: 3.42em; MARGIN: 0px; FONT-FAMILY: 'malgun gothic', dotum, gulim, sans-serif; BACKGROUND-POSITION: 0px 100%; COLOR: rgb(34,61,103); FONT-SIZE: 1.16em;">이 항목의 스프링노트 원문주소</h5>
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==개요==
  
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* [[로그 탄젠트 적분(log tangent integral)]]<br>
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* [[로그 사인 적분 (log sine integrals)]]과 밀접하게 관련되어 있음
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*  다음과 같은 정적분값의 계산:<math>\int_{\pi/4}^{\pi/2} \ln \tan x\, dx=G</math>, <math>G</math>는 [[카탈란 상수]]:<math>\int_{0}^{\pi/4} \ln \tan x\,dx=-G</math>
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:<math>\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\frac{\pi}{2}\ln \left(\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}\right)</math>:<math>\int_{0}^{\pi} (\ln \tan \frac{x}{4})^2\,dx=\frac{\pi^3}{4}</math>:<math>\int_0^{\infty}\frac{(\ln x)^2}{1+x^2} dx =\int_{0}^{\pi/2}(\ln \tan x)^2\,dx = \frac{ \pi^3}{8}</math>
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* ''''''[Vardi1988] '''참조'''
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*  적분:<math>\int_{0}^{\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}\,dx=\pi\ln2</math> 에 대해서는 [[로그함수와 유리함수가 있는 정적분]]
  
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<h5 style="LINE-HEIGHT: 3.42em; MARGIN: 0px; FONT-FAMILY: 'malgun gothic', dotum, gulim, sans-serif; BACKGROUND-POSITION: 0px 100%; COLOR: rgb(34,61,103); FONT-SIZE: 1.16em;">개요</h5>
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==증명==
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* [[로그 사인 적분 (log sine integrals)]]과 밀접하게 관련되어 있음<br>
 
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*  다음과 같은 정적분값의 계산<br><math>\int_{\pi/4}^{\pi/2} \ln \tan x\, dx=G</math>, <math>G</math>는 [[카탈란 상수]]<br><math>\int_{0}^{\pi/4} \ln \tan x\,dx=-G</math><br><math>\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}</math><br><math>\int_{0}^{\pi} \ln^2 \tan \frac{x}{4}\,dx=\frac{\pi^3}{4}</math><br><math>\int_0^{\infty}\frac{\ln^2 x}{1+x^2} dx =\int_{0}^{\pi/2}\ln^2 \tan x\,dx = \frac{ \pi^3}{8}</math><br>
 
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<h5 style="LINE-HEIGHT: 2em; MARGIN: 0px;">증명</h5>
 
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''''''[Vardi1988] '''참조 '''
 
  
 
(보조정리)
 
(보조정리)
24번째 줄: 17번째 줄:
 
<math>\Gamma(s)\beta(s)=\int_{\pi/4}^{\pi/2}  \ln^{s-1}\tan x\, dx</math>
 
<math>\Gamma(s)\beta(s)=\int_{\pi/4}^{\pi/2}  \ln^{s-1}\tan x\, dx</math>
  
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여기서 <math>\Gamma(s)</math>는 [[감마함수]],<math>\beta(s)</math>는 [[디리클레 베타함수]].
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여기서 <math>\Gamma(s)</math>는 [[감마함수]],<math>\beta(s)</math>는 [[디리클레 베타함수]].
  
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(증명)
 
(증명)
  
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<math>F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}</math> 라 하자.
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<math>F(s)=\sum_{n=1}^{\infty}\frac{f(n)}{n^s}</math> 라 하자.
  
 
<math>\Gamma(s)F(s)=\int_0^{\infty}(\sum_{n=1}^{\infty}f(n)e^{-nt})t^{s-1}\,dt</math>
 
<math>\Gamma(s)F(s)=\int_0^{\infty}(\sum_{n=1}^{\infty}f(n)e^{-nt})t^{s-1}\,dt</math>
  
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<math>z=e^{-t}</math> 로 치환하면,
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<math>z=e^{-t}</math> 로 치환하면,
  
 
<math>\Gamma(s)F(s)=\int_0^{1}(\sum_{n=1}^{\infty}f(n)z^n)(\log\frac{1}{z})^{s-1}\,\frac{dz}{z}</math>
 
<math>\Gamma(s)F(s)=\int_0^{1}(\sum_{n=1}^{\infty}f(n)z^n)(\log\frac{1}{z})^{s-1}\,\frac{dz}{z}</math>
  
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만약 <math>f(n+q)=f(n)</math> 을 만족하면 (가령 디리클레 캐릭터의 경우)
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만약 <math>f(n+q)=f(n)</math> 을 만족하면 (가령 디리클레 캐릭터의 경우)
  
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<math>p(z)=\sum_{n=1}^{q-1}f(n)z^n</math>라면,  <math>\sum_{n=1}^{\infty}f(n)z^n=\frac{p(z)}{1-z^q}</math> 로 쓸 수 있다.
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<math>p(z)=\sum_{n=1}^{q-1}f(n)z^n</math>라면, <math>\sum_{n=1}^{\infty}f(n)z^n=\frac{p(z)}{1-z^q}</math> 로 쓸 수 있다.
  
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이를 이용하면, 
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이를 이용하면,  
  
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<math>\Gamma(s)F(s)=\int_0^{1}\frac{p(z)(\log\frac{1}{z})^{s-1}}{1-z^q}\,\frac{dz}{z}</math> 를 얻는다.
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<math>\Gamma(s)F(s)=\int_0^{1}\frac{p(z)(\log\frac{1}{z})^{s-1}}{1-z^q}\,\frac{dz}{z}</math> 를 얻는다.
  
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<math>f</math>가 <math>f(3)=-1</math>인 주기가 4인 디리클레 캐릭터라면, <math>q=4</math>, <math>p(z)=z-z^3</math>
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<math>f</math>가 <math>f(3)=-1</math>인 주기가 4인 디리클레 캐릭터라면, <math>q=4</math>, <math>p(z)=z-z^3</math>
  
 
따라서
 
따라서
  
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<math>\Gamma(s)\beta(s)=\int_0^{1}\frac{(\log\frac{1}{z})^{s-1}}{1+z^2}  \,dz=\int_1^{\infty}\frac{(\log u)^{s-1}}{1+u^2}  \,du=\int_{\pi/4}^{\pi/2}  \ln^{s-1}\tan x\, dx</math> ■
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<math>\Gamma(s)\beta(s)=\int_0^{1}\frac{(\log\frac{1}{z})^{s-1}}{1+z^2}  \,dz=\int_1^{\infty}\frac{(\log u)^{s-1}}{1+u^2}  \,du=\int_{\pi/4}^{\pi/2}  \ln^{s-1}\tan x\, dx</math> ■
  
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(따름정리1)
 
(따름정리1)
  
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<math>\int_{\pi/4}^{\pi/2} \ln \tan x\, dx=G</math>, G는 [[카탈란 상수]].
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<math>\int_{\pi/4}^{\pi/2} \ln \tan x\, dx=G</math>, G는 [[카탈란 상수]].
  
 
(증명)
 
(증명)
  
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위에서 얻은 보조정리에 <math>s=2</math>를 적용하면, 
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위에서 얻은 보조정리에 <math>s=2</math>를 적용하면,  
  
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<math>\int_{\pi/4}^{\pi/2} \ln^{2-1}\tan x\, dx=\Gamma(2)\beta(2)=G</math> ■
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<math>\int_{\pi/4}^{\pi/2} \ln^{2-1}\tan x\, dx=\Gamma(2)\beta(2)=G</math> ■
  
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(따름정리2)
 
(따름정리2)
  
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<math>\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}</math>
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<math>\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\frac{\pi}{2}\ln \left(\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}\right)</math>
  
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(증명)
 
(증명)
86번째 줄: 79번째 줄:
 
<math>\frac{d}{ds}(\Gamma(s)\beta(s))=\frac{d}{ds}\int_1^{\infty}\frac{(\log u)^{s-1}}{1+u^2} \,du=\int_1^{\infty}\frac{(\log u)^{s-1}}{1+u^2}\log \log u \,du</math>
 
<math>\frac{d}{ds}(\Gamma(s)\beta(s))=\frac{d}{ds}\int_1^{\infty}\frac{(\log u)^{s-1}}{1+u^2} \,du=\int_1^{\infty}\frac{(\log u)^{s-1}}{1+u^2}\log \log u \,du</math>
  
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<math>s=1</math> 일때,
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<math>s=1</math> 일때,
  
 
<math>\Gamma'(1)\beta(1)+\Gamma(1)\beta'(1)=\int_1^{\infty}\log \log u \,\frac{du}{1+u^2}=\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx</math>
 
<math>\Gamma'(1)\beta(1)+\Gamma(1)\beta'(1)=\int_1^{\infty}\log \log u \,\frac{du}{1+u^2}=\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx</math>
  
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이제 [[다이감마 함수(digamma function)|Digamma 함수]]와 [[디리클레 베타함수]]에서 얻은 결과를 사용하자. 
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이제 [[다이감마 함수(digamma function)]]와 [[디리클레 베타함수]]에서 얻은 결과를 사용하자.  
  
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<math>\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}</math>, <math>\psi(1) = -\gamma\,\!</math>. 따라서 <math>\Gamma(1)=-\gamma</math>.
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<math>\psi(x) =\frac{d}{dx} \ln{\Gamma(x)}= \frac{\Gamma'(x)}{\Gamma(x)}</math>, <math>\psi(1) = -\gamma\,\!</math>. 따라서 <math>\Gamma(1)=-\gamma</math>.
 
  
 
<math>\beta'(1)=\frac{\pi}{4}\gamma+\frac{\pi}{2}\ln(\frac{\Gamma(3/4)}{\Gamma(1/4)}\sqrt{2\pi})</math>.
 
<math>\beta'(1)=\frac{\pi}{4}\gamma+\frac{\pi}{2}\ln(\frac{\Gamma(3/4)}{\Gamma(1/4)}\sqrt{2\pi})</math>.
  
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그러므로
 
그러므로
  
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<math>\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\Gamma'(1)\beta(1)+\Gamma(1)\beta'(1)= -\frac{\pi}{4}\gamma+\beta'(1)=\frac{\pi}{2}\ln{\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}</math>
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<math>\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\Gamma'(1)\beta(1)+\Gamma(1)\beta'(1)= -\frac{\pi}{4}\gamma+\beta'(1)=\frac{\pi}{2}\ln \left(\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}\right)</math>
  
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임이 증명된다. ■
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임이 증명된다. ■
  
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<h5 style="LINE-HEIGHT: 2em; MARGIN: 0px;">메모</h5>
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==역사==
  
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<math>\int_{0}^{\infty}\frac{\ln(x^{2}+1)}{x^{2}+1}\,dx=\pi\ln2</math>
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* [[수학사 연표]]
  
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* [[로그 사인 적분 (log sine integrals)]]<br>
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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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==메모==
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<math>f</math>가 <math>f(3)=-1</math>인 주기가 4인 디리클레 캐릭터라고 하면, <math>p(z)=z-z^3</math>
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다음의 <math>L</math>-함수
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:<math>L(s) = \sum_{n\geq 1}\frac{f(n)}{n^s}</math>
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는 아래의 식을 만족한다
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:<math>
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\begin{align}
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L'(1)-\gamma \frac{\pi}{4}&=\int_0^{1}\frac{z-z^3}{1-z^4}\log \log\frac{1}{z} \,\frac{dz}{z}\\
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&=\int_0^{1}\log \log\frac{1}{z} \,\frac{dz}{1+z^2}\\
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&=\int_1^{\infty}\log \log u \,\frac{du}{1+u^2} \\
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&=\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx
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\end{align}
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</math>
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:<math>\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=L'(1)- \frac{\pi}{4}\gamma=\frac{\pi}{2}\ln(\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi})</math>
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* http://math.stackexchange.com/questions/285671/vardis-integral-int-pi-4-pi-2-ln-ln-tan-xdx
 
<math>\frac{24}{7\sqrt{7}}\int_{\pi/3}^{\pi/2}\ln|\frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}|\,dt=L_{-7}(2)=1.15192547054449\cdots</math>
 
<math>\frac{24}{7\sqrt{7}}\int_{\pi/3}^{\pi/2}\ln|\frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}|\,dt=L_{-7}(2)=1.15192547054449\cdots</math>
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* http://arxiv.org/PS_cache/arxiv/pdf/0706/0706.0356v1.pdf
  
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[[란덴변환(Landen's transformation)|]]
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<h5 style="LINE-HEIGHT: 3.42em; MARGIN: 0px; FONT-FAMILY: 'malgun gothic', dotum, gulim, sans-serif; BACKGROUND-POSITION: 0px 100%; COLOR: rgb(34,61,103); FONT-SIZE: 1.16em;">재미있는 사실</h5>
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==개요==
  
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* [[등차수열의 소수분포에 관한 디리클레 정리]]
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* [[L-함수, 제타함수와 디리클레 급수|디리클레 급수]]
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* [[후르비츠 제타함수(Hurwitz zeta function)|Hurwitz 제타함수]]
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* [[감마함수]]
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* [[다이로그 함수(dilogarithm)|다이로그 함수(dilogarithm )]]
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* [[란덴변환(Landen's transformation)]]
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* [[르장드르 카이 함수]]
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* [[모듈라 군, j-invariant and the singular moduli]]
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* [[카탈란 상수]]
  
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<h5 style="LINE-HEIGHT: 3.42em; MARGIN: 0px; FONT-FAMILY: 'malgun gothic', dotum, gulim, sans-serif; BACKGROUND-POSITION: 0px 100%; COLOR: rgb(34,61,103); FONT-SIZE: 1.16em;">역사</h5>
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* [[수학사연표 (역사)|수학사연표]]
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==수학용어번역==
  
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* [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집]
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** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=
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<h5 style="LINE-HEIGHT: 2em; MARGIN: 0px;">메모</h5>
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* [http://kms.or.kr/home/kor/board/bulletin_list_subject.asp?bulletinid=%7BD6048897-56F9-43D7-8BB6-50B362D1243A%7D&boardname=%BC%F6%C7%D0%BF%EB%BE%EE%C5%E4%B7%D0%B9%E6&globalmenu=7&localmenu=4 대한수학회 수학용어한글화 게시판]
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* http://arxiv.org/PS_cache/arxiv/pdf/0706/0706.0356v1.pdf<br>
 
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* [http://www.wolframalpha.com/input/?i=integrate_0%5E%28pi%29+x+cos+x+%2F%281%2Bsin%5E2+x%29 http://www.wolframalpha.com/input/?i=integrate_0^(pi)+x+cos+x+%2F(1%2Bsin^2+x)]<br>
 
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* [http://www.wolframalpha.com/input/?i=log%5E2+%281%2Bsqrt%282%29%29+-pi%5E2%2F4 http://www.wolframalpha.com/input/?i=log^2+(1%2Bsqrt(2))+-pi^2%2F4]<br>
 
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* [http://www.wolframalpha.com/input/?i=integrate+%28tan+x%2B1%29/sqrt%28tan%5E2+x+%2B1%29dx http://www.wolframalpha.com/input/?i=integrate+(tan+x%2B1)/sqrt(tan^2+x+%2B1)dx]<br>
 
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* [http://www.wolframalpha.com/input/?i=integrate+%28tan+x-1%29/sqrt%28tan%5E2+x+%2B1%29dx http://www.wolframalpha.com/input/?i=integrate+(tan+x-1)/sqrt(tan^2+x+%2B1)dx]<br>
 
  
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<h5 style="LINE-HEIGHT: 3.42em; MARGIN: 0px; FONT-FAMILY: 'malgun gothic', dotum, gulim, sans-serif; BACKGROUND-POSITION: 0px 100%; COLOR: rgb(34,61,103); FONT-SIZE: 1.16em;">관련된 다른 주제들</h5>
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==매스매티카 파일 및 계산 리소스==
  
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* [[등차수열의 소수분포에 관한 디리클레 정리]]<br>
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* https://docs.google.com/file/d/0B8XXo8Tve1cxNmRFU0Vyak14NGM/edit
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* [[L-함수, 제타함수와 디리클레 급수|디리클레 급수]]<br>
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* [http://www.wolframalpha.com/input/?i=integrate_0%5E%28pi%29+x+cos+x+%2F%281%2Bsin%5E2+x%29 http://www.wolframalpha.com/input/?i=integrate_0^(pi)+x+cos+x+%2F(1%2Bsin^2+x)]
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* [[후르비츠 제타함수(Hurwitz zeta function)|Hurwitz 제타함수]]<br>
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* [http://www.wolframalpha.com/input/?i=log%5E2+%281%2Bsqrt%282%29%29+-pi%5E2%2F4 http://www.wolframalpha.com/input/?i=log^2+(1%2Bsqrt(2))+-pi^2%2F4]
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* [[감마함수]]<br>
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* [http://www.wolframalpha.com/input/?i=integrate+%28tan+x%2B1%29/sqrt%28tan%5E2+x+%2B1%29dx http://www.wolframalpha.com/input/?i=integrate+(tan+x%2B1)/sqrt(tan^2+x+%2B1)dx]
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* [[다이로그 함수(dilogarithm)|다이로그 함수(dilogarithm )]]<br>
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* [http://www.wolframalpha.com/input/?i=integrate+%28tan+x-1%29/sqrt%28tan%5E2+x+%2B1%29dx http://www.wolframalpha.com/input/?i=integrate+(tan+x-1)/sqrt(tan^2+x+%2B1)dx]
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* [[란덴변환(Landen's transformation)]]<br>
 
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* [[르장드르 카이 함수]]<br>
 
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* [[모듈라 군, j-invariant and the singular moduli]]<br>
 
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* [[카탈란 상수]]<br>
 
  
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==사전 형태의 자료==
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* [http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=&fstr= 대한수학회 수학 학술 용어집]<br>
 
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** http://mathnet.kaist.ac.kr/mathnet/math_list.php?mode=list&ftype=eng_term&fstr=
 
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* [http://kms.or.kr/home/kor/board/bulletin_list_subject.asp?bulletinid=%7BD6048897-56F9-43D7-8BB6-50B362D1243A%7D&boardname=%BC%F6%C7%D0%BF%EB%BE%EE%C5%E4%B7%D0%B9%E6&globalmenu=7&localmenu=4 대한수학회 수학용어한글화 게시판]
 
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<h5 style="LINE-HEIGHT: 3.42em; MARGIN: 0px; FONT-FAMILY: 'malgun gothic', dotum, gulim, sans-serif; BACKGROUND-POSITION: 0px 100%; COLOR: rgb(34,61,103); FONT-SIZE: 1.16em;">사전 형태의 자료</h5>
 
  
 
* http://ko.wikipedia.org/wiki/
 
* http://ko.wikipedia.org/wiki/
188번째 줄: 179번째 줄:
 
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
 
* [http://dlmf.nist.gov/ NIST Digital Library of Mathematical Functions]
  
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==관련논문==
  
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* [http://dx.doi.org/10.1007/s11040-010-9074-y Alternative Evaluation of a ln tan Integral Arising in Quantum Field Theory]<br>
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* [http://dx.doi.org/10.1007/s11040-010-9074-y Alternative Evaluation of a ln tan Integral Arising in Quantum Field Theory]
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**  Mark W. Coffey, Mathematical Physics, Analysis and Geometry, Volume 13, Number 2, 2010<br>
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**  Mark W. Coffey, Mathematical Physics, Analysis and Geometry, Volume 13, Number 2, 2010
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*  <br>'''[BBBZ2010]'''[http://arxiv.org/abs/1005.0414 Experimental Mathematics and Mathematical Physics]<br>
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* '''[BBBZ2010]'''[http://arxiv.org/abs/1005.0414 Experimental Mathematics and Mathematical Physics]
 
** DH Bailey, JM Borwein, D Broadhurst, W Zudilin, 2010
 
** DH Bailey, JM Borwein, D Broadhurst, W Zudilin, 2010
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* [http://www.springerlink.com/content/p2k0106727416271/?p=03915f5244d74523b6d36406299c80d5&pi=6 A class of logarithmic integrals]<br>
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* [http://www.springerlink.com/content/p2k0106727416271/?p=03915f5244d74523b6d36406299c80d5&pi=6 A class of logarithmic integrals]
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** Luis A. Medina and Victor H. Moll, The Ramanujan Journal, Volume 20, Number 1 / 2009년 10월
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** Luis A. Medina and Victor H. Moll, The Ramanujan Journal, Volume 20, Number 1 / 2009년 10월
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* [http://link.aip.org/link/?JMAPAQ/49/093508/1 Evaluation of a ln tan integral arising in quantum field theory]<br>
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* [http://link.aip.org/link/?JMAPAQ/49/093508/1 Evaluation of a ln tan integral arising in quantum field theory]
 
** Mark W. Coffey, J. Math. Phys. 49, 093508 (2008)
 
** Mark W. Coffey, J. Math. Phys. 49, 093508 (2008)
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* '''[Borwein and Boradhurst 1998]'''[http://arxiv.org/abs/hep-th/9811173 Determinations of rational Dedekind-zeta invariants of hyperbolic manifolds and Feynman knots and links]<br>
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* '''[Borwein and Boradhurst 1998]'''[http://arxiv.org/abs/hep-th/9811173 Determinations of rational Dedekind-zeta invariants of hyperbolic manifolds and Feynman knots and links]
 
** J.M. Borwein, D.J. Broadhurst, 1998
 
** J.M. Borwein, D.J. Broadhurst, 1998
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* [http://doi.acm.org/10.1145/258726.258736 A class of logarithmic integrals]<br>
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* [http://doi.acm.org/10.1145/258726.258736 A class of logarithmic integrals]
 
** Victor Adamchik, 1997
 
** Victor Adamchik, 1997
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* '''[Vardi1988]'''[http://www.jstor.org/stable/2323562 Integrals, an Introduction to Analytic Number Theory]<br>
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* '''[Vardi1988]'''[http://www.jstor.org/stable/2323562 Integrals, an Introduction to Analytic Number Theory]
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** Ilan Vardi, The American Mathematical Monthly, Vol. 95, No. 4 (Apr., 1988), pp. 308-315
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** Ilan Vardi, The American Mathematical Monthly, Vol. 95, No. 4 (Apr., 1988), pp. 308-315
  
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<h5 style="LINE-HEIGHT: 3.42em; MARGIN: 0px; FONT-FAMILY: 'malgun gothic', dotum, gulim, sans-serif; BACKGROUND-POSITION: 0px 100%; COLOR: rgb(34,61,103); FONT-SIZE: 1.16em;">관련도서 및 추천도서</h5>
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==관련도서==
  
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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>
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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]
 
** George Boros and Victor Moll
 
** George Boros and Victor Moll
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2020년 12월 28일 (월) 02:17 기준 최신판

개요

\[\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=\frac{\pi}{2}\ln \left(\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi}\right)\]\[\int_{0}^{\pi} (\ln \tan \frac{x}{4})^2\,dx=\frac{\pi^3}{4}\]\[\int_0^{\infty}\frac{(\ln x)^2}{1+x^2} dx =\int_{0}^{\pi/2}(\ln \tan x)^2\,dx = \frac{ \pi^3}{8}\]



증명

(보조정리)

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

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


(증명)

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

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

따라서

\(\Gamma(s)\beta(s)=\int_0^{1}\frac{(\log\frac{1}{z})^{s-1}}{1+z^2} \,dz=\int_1^{\infty}\frac{(\log u)^{s-1}}{1+u^2} \,du=\int_{\pi/4}^{\pi/2} \ln^{s-1}\tan x\, dx\) ■



(따름정리1)

\(\int_{\pi/4}^{\pi/2} \ln \tan x\, dx=G\), G는 카탈란 상수.

(증명)

위에서 얻은 보조정리에 \(s=2\)를 적용하면,

\(\int_{\pi/4}^{\pi/2} \ln^{2-1}\tan x\, dx=\Gamma(2)\beta(2)=G\) ■



(따름정리2)

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


(증명)

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

\(\frac{d}{ds}(\Gamma(s)\beta(s))=\frac{d}{ds}\int_1^{\infty}\frac{(\log u)^{s-1}}{1+u^2} \,du=\int_1^{\infty}\frac{(\log u)^{s-1}}{1+u^2}\log \log u \,du\)

\(s=1\) 일때,

\(\Gamma'(1)\beta(1)+\Gamma(1)\beta'(1)=\int_1^{\infty}\log \log u \,\frac{du}{1+u^2}=\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx\)

이제 다이감마 함수(digamma function)와 디리클레 베타함수에서 얻은 결과를 사용하자.

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

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


그러므로

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

임이 증명된다. ■



역사



메모

\(f\)가 \(f(3)=-1\)인 주기가 4인 디리클레 캐릭터라고 하면, \(p(z)=z-z^3\) 다음의 \(L\)-함수 \[L(s) = \sum_{n\geq 1}\frac{f(n)}{n^s}\] 는 아래의 식을 만족한다 \[ \begin{align} L'(1)-\gamma \frac{\pi}{4}&=\int_0^{1}\frac{z-z^3}{1-z^4}\log \log\frac{1}{z} \,\frac{dz}{z}\\ &=\int_0^{1}\log \log\frac{1}{z} \,\frac{dz}{1+z^2}\\ &=\int_1^{\infty}\log \log u \,\frac{du}{1+u^2} \\ &=\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx \end{align} \] \[\int_{\pi/4}^{\pi/2} \ln \ln \tan x\, dx=L'(1)- \frac{\pi}{4}\gamma=\frac{\pi}{2}\ln(\frac{\Gamma(\frac{3}{4})}{\Gamma(\frac{1}{4})}\sqrt{2\pi})\]

\(\frac{24}{7\sqrt{7}}\int_{\pi/3}^{\pi/2}\ln|\frac{\tan t+\sqrt{7}}{\tan t-\sqrt{7}}|\,dt=L_{-7}(2)=1.15192547054449\cdots\)




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