@chris_juravich Check this thread for hints on how to solve this puzzle which is basically the way this Mellin transform was calculated before Cauchy 3/3
@chris_juravich and in doing so, you're implicitly using the definition of a contour integral. This is the reason why the basic scaling property for the Mellin transform works only for positive real constants 2/3
@psicoalfanista and to prove that analytic functions have a Taylor series expansion, we use the Cauchy integral formula. Tomorrow I will give a series of hints for this problem 2/2
@charrlie3045 Elliptic integrals of the first and second kinds are invariant under a sign change of the modulus, that's why the answer is basically the same!