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    Principal Value and Principal Branch of an Integrand

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    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.