\section{Complex integration (contours)}
Many weird looking integrals in calculus make sense when you look over the
complex numbers, and interpret them as path integrals.

In particular, one can compute many integrals of the form
$$\int_{-\infty}^\infty f$$
because this is a closed path on the Riemann sphere, and thus is the sum of
residues in the upper half-plane.
(e.g., $f=\frac{1}{x^2+1}$)

In particular, note that if you have a function that is real on the real line,
then integrating in the opposite direction (along the real line) computes the
negative integral, which just says that the sum of residues on both sides
cancels (because you're counting the residues on the lower half plane when you
integrate in the other direction)
(FIXME: I think this is more general: seems to work for complex functions:
maybe it says that the sum of the residues is zero?)
(FIXME: so do functions real-valued on real line have nice symmetries of
poles?)

(FIXME: can I interprete ``$\int_{-1}^1 \frac{1}{x^2} = -2$'' in terms of
contour integrals? Yes: the integral over any path connecting those
points is $-2$ (b/c this has no residue about $0$):
 note that as you pass over the imaginary axis, you
pick up a huge negative amount, which cancels out the positive)
(what about $1/x$, which has a cute lil' pole...?
 Yes, this again makes sense -if- you are careful to let
 $\int 1/x = \log \abs{x} + i\arg x$)

-----------------------
visualizing various elementary functions as integrals,
  (over complex plane)
  e.g., $\arcsin = int 1/\sqrt{1-x^2}$ etc.
  key point: once you understand exp and log,
    and that log = int 1/x, then you can see that residue of 1/x = 2pi
i
    ...hence you can compute all residues.
    (incidentally, this shows another reason why arctan inf - arctan
-inf = pi;
     if you see it as int 1/(x^2+1), then it's residue of
(-i/2)/x-i=pi)

