Complicated integration of complex number

[MissP.25_5]

Yes, but you already done the computation using CIF so need to bother doing it again.

The it means the this wasy is no simpler than CIF? Because I still haven't finished the derivation.

 

[CAF123]

The it means the this wasy is no simpler than CIF? Because I still haven't finished the derivation.

Well, it is a matter of taste but in some sense I find the residue theorem easier to actually get to that equation involving the derivative. You did not have to, for example, study f(z) to find a part that is analytic on the interior of the contour. Even though that is still quite a quick step...

Laurent expansions of functions also give you the residue - it is defined as the coefficient of the 1/z term (the first term in the expansion that 'knows' about the singularity). But again, you end up with the final equation and in fact this is how the general formula for the residue of a nth order pole was derived. If you have a textbook or notes, check it out.