07/21/2010, 03:24 AM
(07/01/2010, 03:31 PM)sheldonison Wrote: Kouznetsov's method requires three consecutive exponents, (he uses 0, 1, e), and their vertical (along increasing imaginary) paths to the fixed point.
As I think a second time about it, why does he need paths to the fixed point?
He integrates along the paths Re(z)=1, Re(z)=-1.
He merely forces the value of the superfunction to be the fixed point for imaginary part going to infinity. How the path behaves while going to the fixed point is not essential for his computation, isnt it?
