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THE FOURIER TRANSFORM AND THE MELLIN TRANSFORM For suitable functions f : R −→ C the Fourier transform of f is the integral g : R −→ C, g(y) = f (x)e2πixy dx, R and for suitable functions g : R −→ C the inverse Fourier transform of g is the integral f : R −→ C, f (x) = g(y)e−2πiyx dy. R The Fourier inversion formula says that if the functions f and g are well enough behaved then g is the Fourier transform of f if and only if f is the inverse Fourier transform of g. The exponential map is a topological isomorphism exp : (R, +) −→ (R+ , ·) The Mellin transform, inverse Mellin transform, and Mellin inversion formula are essentially the Fourier ideas passed through the isomorphism. Speciﬁcally, given a suitable function on the positive real axis, f : R+ −→ C, we can make a corresponding function on the real line, f : R −→ C, f = f ◦ exp . The Fourier transform of f is g : R −→ C where g(y) = f (x)e2πixy dx R d(ex ) = f (ex )(ex )2πiy R ex dt = f (t)t2πiy R+ t dt = f (t)ts where s = 2πiy. R+ t If we assume that our function f (t) decreases at least as a polynomial in t as t → 0+ and that f decreases quickly as t → ∞ then in fact the integral converges for some complex right half plane of s-values {Re(s) > σo } where σ0 < 0. Thus we are led to deﬁne the Mellin transform of f , dt g : {Re(s) > σo } −→ C, g(s) = f (t)ts . R + t 1 2 THE FOURIER TRANSFORM AND THE MELLIN TRANSFORM The next question is how to recover f from g. Since g is simply the Fourier transform of f up to a coordinate change, f must be essentially the inverse Fourier transform of g. More speciﬁcally, the fact that f is exactly the inverse Fourier transform of g, f (x) = g(y)e−2πiyx dy, R rewrites as 1 f (ex ) = g(s)(ex )−s ds. 2πi s=2πiy That is, 1 f (t) = g(s)t−s ds. 2πi s=2πiy Contour integration shows that the vertical line of integration can be shifted hor- izontally within the right half plane of convergence with no eﬀect on the integral. Thus the deﬁnition of the inverse Mellin transform of g is inevitably 1 f : R+ −→ C, f (t) = g(s)t−s ds for any suitable σ. 2πi s=σ+2πiy Here it is understood that the integral proceed up the vertical line. Naturally, the Mellin inversion formula says that if the functions f and g are well enough behaved then g is the Mellin transform of f if and only if f is the inverse Mellin transform of g. For practice with Mellin inversion, it is an exercise to evaluate the integral σ+i∞ Γ(s)x−s ds, σ > 0. s=σ−i∞ Often what this writeup calls the Fourier transform is called the inverse Fourier transform and conversely. It is a small exercise to show that which convention is adopted has no eﬀect on the resulting deﬁnition of the Mellin transform and the inverse Mellin transform.

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posted: | 12/13/2011 |

language: | English |

pages: | 2 |

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