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``.

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## Math on the Pololu website!

- 27 June 2014Ben 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``.

## Pocket-sized USB charger adapter

- 6 March 2014Thanks for the nice feedback, and please let us know when you post more projects!