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Laplace math help

  1. Mar 13, 2014 #1
    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. jcsd
  3. Mar 13, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    What is your original integral? Is it
    [tex]
    I = \int_0^\infty \frac{1}{\sqrt x}e^{-px}\,dx
    [/tex]
    where the substitution [itex]px = s^2[/itex] is correct, yielding
    [tex]
    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.
    [/tex]

    You mean
    [tex]
    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
    [/tex]
    not
    [tex]
    \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
    [/tex]

    You need both [itex]\sin \theta[/itex] and [itex]\cos \theta[/itex] to be positive. What does that give you?
     
  4. Mar 19, 2014 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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 [itex]\theta[itex] from 0 to [itex]\pi/2[/itex].
     
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