Function that equals its own Mellin transform

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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.