Originally Posted by
Opalg For the function
, if there is a simple pole at
then the residue there is
.
Oh yea, I forgot about that. So let's say the upper right pole is about . Then using Mathematica's , I get for the residue there (last line):
Code:
In[7]:= f[z_] = 1.3*Sin[1.7*z] + 0.6*Sin[4*z];
urroot = FindRoot[f[z] == 0, {z, 1 + 0.25*I}]
r1 = 1/D[f[z], z] /. urroot
Out[8]= {z -> 1.2794588067848602 +
0.3743082054455279*I}
Out[9]= 0.05685562855624362 - 0.2787028735656702*I
Suppose could do that 4 more times and get a good approximation to the integral via residues. Just news to me, that's all.