• ### Math on the Pololu website!

- 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.

• ### Pocket-sized USB charger adapter

- 6 March 2014

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