Update I have de-contrived the example below. After some struggle, I've managed to accomplish this seemingly simple task: wrap a friendly open-source library inside python and make it a python module. Such wrapping is handy, because it retains the speed of c++ inside of python.
If you want to check out how to do more or less the same thing with cython instead of boost see this.
So here is the setup.py
// Comment
#! /usr/bin/python

from ez_setup import use_setuptools
from setuptools import setup, Extension

import sys, os
import glob

include_dirs = []

if __name__ == '__main__':

extensions = [Extension('_boostbabel',['src/boostbabel.cpp'],
include_dirs+['src/boostbabel/', '/usr/include/openbabel-2.0/'],
language="c++", libraries=['boost_python', 'openbabel']),
setup(name = 'boostbabel',
ext_package = 'boostbabel',
ext_modules = extensions,
zip_safe = False, # as a zipped egg the *.so files are not found (at least in Ubuntu/Linux)

Relative to setup.py we have the actual source: src/boostbabel.cpp Now I know nothing about building/linking/including, so getting to that stage was pretty hard. Being the lazy one, I wonder if there isn't a way to guess all of the includes and libs used to init the Extension object by being clever in parsing the actual boostbabel.cpp? Nooope, there isn't for the general case.