멜린-반스 적분

수학노트
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개요

  • 감마함수의 곱을 적분
  • 함수의 점근급수전개에 유용한 도구


멜린 변환

  • 함수 $g,h$의 멜린변환을 $\tilde{g},\tilde{h}$라 하면, 다음이 성립한다

$$ \int_0^{\infty}g(y)h(y)y^{z-1}\,dy=\frac{1}{2\pi i}\int_{\delta-i\infty}^{\delta+i\infty}\tilde{g}(w)\tilde{h}(z-w)\, dw $$

  • 멜린변환의 성질을 이용하면, 다음을 얻는다

$$ \int_0^{\infty}g(xy)h(y)y^{z-1}\,dy=\frac{1}{2\pi i}\int_{\delta-i\infty}^{\delta+i\infty}\tilde{g}(w)\tilde{h}(z-w)x^{-w}\, dw, \, \quad x>0 $$

  • $z=1$일 때,

$$ \int_0^{\infty}g(xy)h(y)\,dy=\frac{1}{2\pi i}\int_{\delta-i\infty}^{\delta+i\infty}\tilde{g}(w)\tilde{h}(1-w)\, dw $$

예1

  • 함수 $f(x)=\left(\frac{\sin x}{x}\right)^2$의 멜린변환 $\tilde{f}$는 다음과 같다

$$ \tilde{f}(z)=\int_0^{\infty}x^{z-1}f(x)\,dx=-\frac{\sqrt{\pi}}{4}\frac{\Gamma(\frac{z}{2}-1)}{\Gamma(\frac{3}{2}-\frac{z}{2})},\, \quad 0<\Re{z}<2 $$

  • 멜린역변환은 다음과 같다

$$ f(x)=\frac{1}{2\pi i}\int_{\delta-i\infty}^{\delta+i\infty}\tilde{f}(z)x^{-z}\, dz=-\frac{\sqrt{\pi}}{4}\frac{1}{2\pi i}\int_{\delta-i\infty}^{\delta+i\infty}\frac{\Gamma(\frac{z}{2}-1)}{\Gamma(\frac{3}{2}-\frac{z}{2})}x^{-z}\, dz,\, \quad 0<\delta<2 $$

  • 즉,

$$ \left(\frac{\sin x}{x}\right)^2=-\frac{\sqrt{\pi}}{4}\frac{1}{2\pi i}\int_{\delta-i\infty}^{\delta+i\infty}\frac{\Gamma(\frac{z}{2}-1)}{\Gamma(\frac{3}{2}-\frac{z}{2})}x^{-z}\, dz,\, \quad 0<\delta<2 $$

예2

  • $f$는 다음과 같다

$$ f(x)= \begin{cases} \frac{1}{\sqrt{1-x^2}}&x\in(0,1)\\ 0&x>0 \end{cases} $$

  • 멜린변환은 다음과 같다

$$ \tilde{f}(z)=\int_0^{\infty}x^{z-1}f(x)\,dx=\frac{\sqrt{\pi}}{2}\frac{\Gamma \left(\frac{z}{2}\right)}{\Gamma \left(\frac{z+1}{2}\right)},\, \quad \Re{z}>0 $$


예3

  • 적분으로 주어진 다음 함수 $f$를 생각하자

$$ f(x)=\int_{0}^{1}\left(\frac{\sin xy}{xy}\right)^2\frac{1}{\sqrt{1-y^2}}\, dy $$

  • $f$의 멜린-반스 적분표현은 다음과 같다

$$ f(x)=-\frac{\pi}{8}\frac{1}{2\pi i}\int_{\delta-i\infty}^{\delta+i\infty}\frac{\Gamma(\frac{z}{2}-1)\Gamma(\frac{1}{2}-\frac{z}{2})}{\Gamma(\frac{3}{2}-\frac{z}{2})\Gamma(1-\frac{z}{2})}x^{-z}\, dz,\, \quad 0<\delta<1 $$

  • 유수 정리를 이용하여 다음을 얻는다

$$ f(x)=\frac{\pi }{4}\sum_{n=0}^{\infty}\frac{(-1)^n \Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1) \Gamma \left(n+\frac{3}{2}\right) \Gamma (n+2)} x^{2 n} $$

메모

  • The origins of what we call the “Mellin-transform method” go far back: The idea of the Mellin inversion formula appeared in an 1876 memoir by Riemann, and the first accurate discussion was given by Mellin in 1896 and 1902. What we now call “Mellin–Barnes integrals”were first introduced by Pincherle in 1888, developed theoretically by Mellin by 1910, and used by Barnes in 1908 to discuss the asymptotic expansion of certain special functions. (Fikioris)


사전 형태의 자료

매스매티카 파일 및 계산 리소스

리뷰, 에세이, 강의노트

  • Fikioris, George. “Mellin-Transform Method for Integral Evaluation: Introduction and Applications to Electromagnetics.” Synthesis Lectures on Computational Electromagnetics 2, no. 1 (January 1, 2007): 1–67. doi:10.2200/S00076ED1V01Y200612CEM013.
  • Elizalde, E., K. Kirsten, and S. Zerbini. “Applications of the Mellin-Barnes Integral Representation.” Journal of Physics A: Mathematical and General 28, no. 3 (February 7, 1995): 617–29. doi:10.1088/0305-4470/28/3/016.
  • http://www2.math.umd.edu/~punshs/Mellin-Barnes.pdf