Integral to Infinity of Exponential of -t^2

Theorem

 * $\displaystyle \int_0^\infty \map \exp {-t^2} \rd t = \dfrac {\sqrt \pi} 2$

Proof
Let $\displaystyle I = \int_0^\infty \map \exp {-t^2} \rd t$.

Let $\displaystyle I_P = \int_0^P \map \exp {-x^2} \rd x = \int_0^P \map \exp {-y^2} \rd y$.

Then we have:
 * $I = \displaystyle \lim_{P \mathop \to \infty} I_P$

Hence:

where $\mathscr R_P$ is the square $\Box OACE$ in the figure below:


 * Integral to Infinity of Exponential of -t^2.png

Because the integrand is positive:

where $\mathscr R_1$ and $\mathscr R_2$ are the regions in the first quadrant bounded by the circles with centers $O$ and radii $P$ and $P \sqrt 2$ respectively.

Using polar coordinates, we can express $(1)$ as: