... numerator in the vicinity of zerois O(zν+1), and the order of its denominator is O(zν), the integral along the left part of contour vanishes. Now, consider the integrals along the remaining three ... therefore N = 1. Hence, beginning with some k,Equation2.6 has exactly one positive root corresponding to the imaginary root of Equation 2.5.Now, find the asymptotic of the imaginary roots of ... verified that in the vicinity of zero, the function g(z) is of order O (zν).By virtue of this asymptotic and because g(z) is an odd function, the integral along the left-hand side of the contour...