I know that the Gaussian distribution is its own Fourier transform. There was a similar post about what function is its own Laplace transform that was answered. I was wondering if there exist a function such that $$ f(s) = \int_0^\infty f(x) x^{s-1}dx $$ I don't have any use for finding such a function but I am just curious about it.
2026-03-25 18:44:17.1774464257
Function that equals its own Mellin transform
59 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in CALCULUS
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