# Laplace math help

1. Mar 13, 2014

### vanceEE

px = t
t = s^2

$$I = \int_0^∞ e^{-s^2}ds$$
$$I*I = \int_0^∞ e^{-s^2}ds * \int_0^∞ e^{-u^2}du = \int_0^∞\int_0^∞ e^{-(s^2+u^2)}du ds$$
$$s = rsin\theta$$
$$u = rcos\theta$$
$$r = s^2 + u^2$$
$$I*I = \int_0^∞\int_\alpha^\beta e^-{r^2}rdrd\theta$$
How can I find my limits of integration in polar coordinates?

Last edited: Mar 13, 2014
2. Mar 13, 2014

### pasmith

What is your original integral? Is it
$$I = \int_0^\infty \frac{1}{\sqrt x}e^{-px}\,dx$$
where the substitution $px = s^2$ is correct, yielding
$$I = \int_0^\infty \frac{1}{\sqrt x}e^{-px}\,dx = \int_0^\infty \frac{\sqrt {p}}{s} e^{-s^2} \frac{2s}{p}\,ds = \frac 2{\sqrt{p}} \int_0^\infty e^{-s^2}\,ds = \frac 1{\sqrt p} \int_{-\infty}^\infty e^{-s^2}\,ds.$$

You mean
$$I^2 = \int_\alpha^\beta \int_0^\infty e^{-r^2} r\,dr\,d\theta = \int_\alpha^\beta \left(\int_0^\infty e^{-r^2} r\,dr\right)\,d\theta$$
not
$$\int_0^\infty \int_\alpha^\beta e^{-r^2} r\,dr\,d\theta = \int_0^\infty \left(\int_\alpha^\beta e^{-r^2} r\,dr\right)\,d\theta$$

You need both $\sin \theta$ and $\cos \theta$ to be positive. What does that give you?

3. Mar 19, 2014

### HallsofIvy

Staff Emeritus
Your two integrals, in x and y, were from 0 to infinity so the two integrals cover the first quadrant. In order to do that in polar coordinates you have to have r from 0 to infinity and $\theta[itex] from 0 to [itex]\pi/2$.