It works perfectly for polynomials, which follows from Mellin transforms. I'm thinking that it should work for functions that don't grow too fast (polynomially, less than exponential) as the imaginary part goes to \pm infinity. If that's satisfied then I'm pretty sure for a large class of functions the result should hold. It starts with pretty much the same mellin inverse integral, so I'm thinking it works on the same class of functions as the first continuum sum method since the mellin inversion transformation guarantees that the rest will converge. On exponentials it should work as well, I haven't tested, but I'm confident it does.
Functions that blow up at \pm imaginary infinity like \( e^{-x^2} \) don't work. I'll have to think about that one, maybe solving for \( e^{x^2} \) and then making a change of argument in the continuum sum so you count across imaginary values (that should work!)
I'm really confident about this way and I think it may be more effective. My next mission is to define contour summation.
Functions that blow up at \pm imaginary infinity like \( e^{-x^2} \) don't work. I'll have to think about that one, maybe solving for \( e^{x^2} \) and then making a change of argument in the continuum sum so you count across imaginary values (that should work!)
I'm really confident about this way and I think it may be more effective. My next mission is to define contour summation.