27 June 2014
Ben wanted me to post the details of the Gaussian integral, so here we go.
It's simpler to think about the full integral from ``-oo`` to ``+oo``. Let's give that a name:
``lambda = int_-oo^oo e^(-x^2) dx``.
Then, instead of trying to compute ``lambda``, we work on ``lambda^2``:
``lambda^2 = ( int_-oo^oo e^(-x^2) dx )^2 = int_-oo^oo int_-oo^oo e^(-x^2) e^(-y^2) dx dy``.
We note that ``e^(-x^2) e^(-y^2) = e^(-(x^2+y^2))`` and rewrite in polar coordinates:
``lambda^2 = int_0^oo e^(-r^2) 2pi r \ dr``.
Luckily, that integral is easy, since ``d / (dr) pi e^(-r^2) = - 2 pi r \ e^(-r^2)``:
``lambda^2 = - [ pi e^(-r^2) ]_0^oo = pi``;
``lambda = sqrt(pi)``.
Finally, since the Gaussian is an even function, our desired integral is just half of that value:
``int_0^oo e^(-x^2) dx = sqrt pi / 2``.
6 March 2014
Thanks for the nice feedback, and please let us know when you post more projects!