"오일러 베타적분(베타함수)"의 두 판 사이의 차이

수학노트
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* S. Ole Warnaar, [http://www.maths.uq.edu.au/%7Euqowarna/talks/Wien.pdf Beta Integrals]
 
* S. Ole Warnaar, [http://www.maths.uq.edu.au/%7Euqowarna/talks/Wien.pdf Beta Integrals]
 
* Askey, Richard. “Beta Integrals and the Associated Orthogonal Polynomials.” In Number Theory, Madras 1987, edited by Krishnaswami Alladi, 84–121. Lecture Notes in Mathematics 1395. Springer Berlin Heidelberg, 1989. http://link.springer.com/chapter/10.1007/BFb0086401.
 
* Askey, Richard. “Beta Integrals and the Associated Orthogonal Polynomials.” In Number Theory, Madras 1987, edited by Krishnaswami Alladi, 84–121. Lecture Notes in Mathematics 1395. Springer Berlin Heidelberg, 1989. http://link.springer.com/chapter/10.1007/BFb0086401.
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* Askey, Richard. ‘Ramanujan’s Extensions of the Gamma and Beta Functions’. The American Mathematical Monthly 87, no. 5 (1 May 1980): 346–59. doi:10.2307/2321202.
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[[분류:적분]]
 
[[분류:적분]]

2014년 11월 23일 (일) 18:41 판

개요



삼각함수의 적분과의 관계

\(B(x,y) = 2\int_0^{\pi/2}(\sin\theta)^{2x-1}(\cos\theta)^{2y-1}\,d\theta\)

\(\int_0^{\frac{\pi}{2}}\sin^{p}\theta{d\theta}= \frac{1}{2}B(\frac{p+1}{2},\frac{1}{2})=\frac{\sqrt{\pi}\Gamma(\frac{p}{2}+\frac{1}{2})}{2\Gamma(\frac{p}{2}+1)}\)

\(\int_0^{\frac{\pi}{2}}\cos^{p}\theta{d\theta}= \frac{1}{2}B(\frac{p+1}{2},\frac{1}{2})=\frac{\sqrt{\pi}\Gamma(\frac{p}{2}+\frac{1}{2})}{2\Gamma(\frac{p}{2}+1)}\)

\(\int_0^{\frac{\pi}{2}}\sin^{2n}\theta{d\theta}= \frac{\sqrt{\pi}\Gamma(n+\frac{1}{2})}{2\Gamma(n+1)}=\frac{\pi}{2}\frac{(\frac{1}{2})_n}{(1)_n}\)

(증명)

\(B(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt\) 에서\(t=\sin^{2} \theta\) 로 치환 ■





베타적분과 감마함수

  • 감마함수를 이용하여, 다음과 같이 표현할 수 있다\[B(x,y)=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}\]

(증명)

가우시안 적분의 아이디어와 비슷하다.

\(\Gamma(x)\Gamma(y) = \int_0^\infty\ e^{-u} u^{x-1}\,du \int_0^\infty\ e^{-v} v^{y-1}\,dv\)

\(u = a^2\)와 \(v = b^2\) 로 치환하면,

\(\Gamma(x)\Gamma(y) = 4\int_0^\infty\ e^{-a^2} a^{2x-1}\,da \int_0^\infty\ e^{-b^2} b^{2y-1}\,db\)

\(= 4\int_{0}^\infty\ \int_{0}^\infty\ e^{-(a^2+b^2)} a^{2x-1} b^{2y-1} \,da \,db\)

\(=4\int_0^{\frac{\pi}{2}}\int_0^\infty\ e^{-r^2} (r\cos\theta)^{2x-1} (r\sin\theta)^{2y-1} r \, dr \,d\theta\)

\(= 4\int_0^\infty\ e^{-r^2} r^{2x+2y-2} r\, dr \int_0^{\frac{\pi}{2}}(\cos\theta)^{2x-1} (\sin\theta)^{2y-1}\, d\theta\)

\(= 2\int_0^\infty\ e^{-r^2} r^{2(x+y-1)} \, d(r^2) \int_0^{\pi/2}\ (\cos\theta)^{2x-1} (\sin\theta)^{2y-1} \,d\theta\)

\(= \Gamma(x+y)B(x,y)\) ■



성질

  • \(x+y+z=1\) 이면, \(\frac{\pi B(y,z)}{\sin \pi x}=\Gamma(x)\Gamma(y)\Gamma(z)\) ;(증명)

\[\Gamma(1-x)\Gamma(x) = {\pi \over \sin{\pi x}} \,\!\]



무리함수의 적분과 감마함수

\(n>0\)에 대하여,

\(\int_0^1\frac{dx}{\sqrt{1-x^n}}=\frac{1}{n}B(\frac{1}{2},\frac{1}{n})\)

이 성립한다


(증명)

\(t=x^n\) 으로 치환하면, \(dt=nx^{n-1}\,dx=nt^{\frac{n-1}{n}}\,dx\).

\(\int_0^1\frac{dx}{\sqrt{1-x^n}}=\frac{1}{n}\int_0^1\frac{t^{-\frac{n-1}{n}}}{\sqrt{1-t}}dt=\frac{1}{n}\int_0^1{t^{\frac{1}{n}-1}}(1-t)^{\frac{1}{2}-1}dt=\frac{1}{n}B(\frac{1}{2},\frac{1}{n})\). ■



타원적분과의 관계

(증명)\[\int_0^1\frac{dx}{\sqrt{1-x^3}}=\frac{1}{3}B(\frac{1}{2},\frac{1}{3})=\frac{1}{6}B(\frac{1}{3},\frac{1}{6})\] ■




베타적분과 초월수

정리

\(a,b,a+b \in \mathbb{Q}\backslash\mathbb{Z}\) 라 하자. \(B(a,b)\) 는 초월수이다. 즉 \[B(a,b) = \frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}= \int_0^1t^{a-1}(1-t)^{b-1}\,dt\] 는 초월수이다.

역사



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사전 형태의 자료



리뷰, 에세이, 강의노트

  • Neretin, Yury A. ‘Matrix Beta-Integrals: An Overview’. arXiv:1411.2110 [math], 8 November 2014. http://arxiv.org/abs/1411.2110.
  • S. Ole Warnaar, Beta Integrals
  • Askey, Richard. “Beta Integrals and the Associated Orthogonal Polynomials.” In Number Theory, Madras 1987, edited by Krishnaswami Alladi, 84–121. Lecture Notes in Mathematics 1395. Springer Berlin Heidelberg, 1989. http://link.springer.com/chapter/10.1007/BFb0086401.
  • Askey, Richard. ‘Ramanujan’s Extensions of the Gamma and Beta Functions’. The American Mathematical Monthly 87, no. 5 (1 May 1980): 346–59. doi:10.2307/2321202.