Principal Value and Principal Branch of an Integrand
Use parametric representation in exercise 10 for the oriented circle C0 there to show that....where a is any real number other than zero and where the principal branch of the integrand and where the principal value of R^G are taken.
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Ok the change of varibale is this:
z=z0+Rexp(i*theta) ---> dz=i*R*exp(i*theta)*dtheta and naturally:
-Pi < theta < Pi, so:
int((z-z0)^(a-1)dz)= ...
Solution Summary
An integral is solved using the principal value and principal branch of an integrand.
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