Is Cauchy-Riemann a necessary or sufficient condition?

When integrating using the method of residues, one needs to check if the integrand satisfies the Cauchy-Riemann equations first.

However, it is possible to have an integral that doesn't satisfy Cauchy-Riemann and yet the method of residues still works! To see an example of such an integral, see the Appendix of this document:

http://www.eleceng.adelaide.edu.au/personal/dabbott/conference/UPN_abbott1996b.pdf

The questions are:

a) Is this type of exception generally known?

b) Can you explain what is going on and why contour integration by residues still works?

c) Is it possible to define a class of integrals that are solvable by residues, and yet do not satisfy Cauchy-Riemann?