Here is the idea: we’d like (relatively) easy ways to integrate functions around simple closed curves. For the purposes of this particular note, we assume the usual about the closed curve (piecewise smooth, once around in the standard direction) and that the function in question does not have singularities along the said closed curve and only a finite number of isolated singularities in the simply connected region enclosed by the curve. If we have no singularities in the said region, then our integral is zero and there is nothing more to do.

Now by the theory of cuts, we can assume that the integral around the given simple closed curve is equal to the sum of the integrals of the function around small circles surrounding each singular point.

So, integrals of this type just boil down to calculating the integrals of the function around these isolated singularities and then just adding.

So lets examine a typical singularity: say . Let done the circle of radius surrounding and assume that is small enough that we don’t enclose any other singularity.

Now on a punctured disk (excluding ) bounded by this circle, we have

. Now we have

Now the second integral is the integral of a power series (the regular part of the Laurent series) and therefore the integral is zero since we are integrating along a simple closed curve. So only the integral of the principal part matters.

Now we can interchange integration and summation by the work we did to develop the Laurent series to begin with.

Doing that we obtain:

and keep in mind that the are constants. So we have:

and so we need only evaluate

to evaluate our integral.

Now for we note that has a primitive in the punctured disk. OR..if you don’t like that, you can do the substitution and find that these integrals all evaluate to zero.

So only the integral matters and the aforementioned substitution shows that this integral evaluates to

So all of this work shows that ; only the coefficient of the term matters.

** is called the residue of at ** and sometimes denoted

So now we see that the integral around our original closed curve is just , that is, times the sum of the residues of which are enclosed by our closed curve.

**This means that it would be good to have “easy” ways to calculate residues of a function.**

Now of course, we could always just calculate the Laurent series of the function evaluated at each singularity in question. SOMETIMES, this is easy. And if the singularity in question is an essential one, it is what we have to do.

But sometimes the calculation of the series is tedious and it would be great to have another method.

**The idea**

Ok, suppose we have a simple pole at (we’ll build upon this idea). This means that Now note that so we obtain:

Example: calculate the residue of at .

.

Now sometimes this is straight forward, and sometimes it isn’t.

Now if where has an isolated zero at (as is the case when we have a pole), we can then write where both the numerator and denominator functions are going to zero.

This should remind you of L’Hopital’s rule from calculus. So we should derive a version for our use.

Now if both have isolated zeros of finite order at then

(because )

Now take a limit as to obtain which can be easier to calculate.

If we apply this technique to we get which is 1 at

Ok, my flight is boarding so I’ll write part II later.