In general we have
Which can be proven from the power series of and the fact that while . (True by loop arguments, see figure below)``
The interesting bit is that defining the taylor coefficients like this also works, meaning a smooth complex function has a power series! You prove this with cursed analysis.
A residue takes a pole and writes
Where is regular, and this equality holds in some circle around . The residue is then defined as
The residue theorem is that
Prove this by contracting around each pole to get
Which, after writing around each pole and summing gives
We're going to show by computing a contour integral around the box, then letting .
First use the residue theorem on
There's a pole at every , first compute the residues. Consider and seperately.
Let be a pole. We can write
For regular . In other words . Computing the residue is which, using LHopital's rule gives
(This is where we get the in the sum from!)
Now consider the pole at . It's multiplicity ( from and from ) meaning we can write
Computing the residue means computing , taking derivatives gives
As the limit goes to (source: trust me bro.) meaning the integral (by the residue theorem) equals
If we can just show this integral goes to zero as that'll imply
We show the integral goes to zero by bounding the integrand via a constant, since all constants have zero contour integral. It's possible to show
Using Eulers formula and some algebra. First use the formula
Then set for the vertical ones, we have (separating real and imaginary parts)
At this point you either realize this equals and so is bounded in norm by , or you see that
Now for the horizontal ones set and simplify the exponents
The has norm , and as the terms dominate, meaning the ratio goes to , meaning for all there exists an large enough that
Setting and combining this with out previous result gives
Which multiplied by the perimeter gives an upper bound for the path integral around the boundary (which is the contour integral).
This goes to zero as