# Drawing Blanks

Premature Optimization is a Prerequisite for Success

## Obvious identity

Marquis de Laplace said that those to whom this identity is not obvious are clueless in math:

$\int_{-\infty}^{\infty} e^{-x^2}dx=\sqrt{\pi}$

Seriously, come on:

$\int\int e^{-y^2}e^{-x^2}dx dy = \int_0^{\infty} e^{-r^2} 2\pi r dr =\pi\int_0^{\infty}e^{-r^2}d(r^2)=\pi$

(just testing latex notation)

Another try

$\int \limits_{-\infty}^{+\infty}\int \limits_{-\infty}^{+\infty}e^{-(x^2+y^2)}dxdy=\int \limits_0^{2{\pi}}\int \limits_0^{+\infty} e^{-r^2}rdr d \varphi =\pi \int \limits_0^{+\infty} e^{-r^2}d(r^2)=\pi$