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Homework Help: Integrating e^-ax^2

  1. Nov 9, 2006 #1
    Hi, i have a problem which is confusing me :confused:


    Given that

    [tex]\int_{0}^{\infty}e^{-ax^2} dx = \frac{\sqrt{\pi}}{2\sqrt{a}}[/tex]

    What is

    (i) [tex]\int_{0}^{\infty}e^{-ax^2} x^2 dx[/tex]
    (ii) [tex]\int_{0}^{\infty}e^{-ax^2} x^3 dx[/tex]
    (iii) [tex]\int_{0}^{\infty}e^{-ax^2} x^4 dx[/tex]

    It tells me to use differentiation for (i) and (iii) with respect to the a paramter, and integration by parts for (ii)

    I tried (ii):


    [tex]\frac{du}{dx} = 3x^2[/tex]

    [tex]\frac{dv}{dx} = e^{-ax^2}[/tex]


    [tex]\int_{0}^{\infty}e^{-ax^2} x^3 dx = \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} - \int_{0}^{\infty}\frac{3x^2 \sqrt{\pi}}{2\sqrt{a}}[/tex]

    [tex]= \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} - \frac{x^3 \sqrt{\pi}}{2\sqrt{a}} = 0[/tex]

    I know limits should be in that last bit and the bit before but i thought it wouldn't matter as both parts are the same, but then i thought i can't say [itex]v=\frac{\sqrt{\pi}}{2\sqrt{a}}[/itex] anyway because the given expression wasn't general it was between limits, or does it not matter because i'm using the same limits?

    I don't know what the question means when it tells me to "use differentiation for (i) and (iii) with respect to the a paramter" how will that help me get to a solution?

    Thanks in advance
    Last edited: Nov 9, 2006
  2. jcsd
  3. Nov 9, 2006 #2


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    What happened to the exponential after you integrated by parts? And have you tried their suggestion for i and iii? You can move the d/da inside the integral (justifying this is a little tricky, but I'm sure you don't need to worry about that).
  4. Nov 9, 2006 #3


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    You can't say [itex]v=\frac{\sqrt{\pi}}{2\sqrt{a}}[/itex] for the part inside the integral
  5. Nov 9, 2006 #4


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    This is the definite integral from [itex]-\infty[/itex] to [itex]\infty[/itex], not the anti-derivative! [itex]e^{-ax^2}[/itex] has no simple anti-derivative. Try [itex] u= x^2[/itex], [itex]dv= xe^{-ax^2}dx[/itex] instead.

  6. Nov 10, 2006 #5


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    For (ii) compute first

    [tex] \int_{0}^{\infty} e^{-ax^2} \ x \ dx[/tex]

    And when you get the result, you can differentiate wrt "a" to get the needed integral.

    HINT:Make the sub x^{2}=t and of course assume a>0.

  7. Nov 10, 2006 #6
    Thanks for all these replies! :D

    I would of responded sooner but i thought i'd best come back with something to show i've been quite busy.

    I'll just show you what my answers are not sure if they're right but if they are credit to you all for helping me.

    [tex]\int_{0}^{\infty}e^{-ax^2} dx = \frac{\sqrt{\pi}}{2\sqrt{a}}[/tex]

    [tex]\int_{0}^{\infty}\frac{d}{da}\left(e^{-ax^2}\right) dx = \frac{d}{da}\left(\frac{\sqrt{\pi}}{2\sqrt{a}}\right)[/tex]

    [tex]\int_{0}^{\infty}x^2 e^{-ax^2} dx = \frac{\sqrt{\pi}}{4(\sqrt{a})^3}[/tex]

    [tex]\int_{0}^{\infty}e^{-ax^2} x dx = \frac{1}{2a}[/tex]

    [tex]\int_{0}^{\infty}\frac{d}{da}\left(e^{-ax^2} x \right) dx = \frac{d}{da}\left(\frac{1}{2a}\right)[/tex]

    [tex]\int_{0}^{\infty}e^{-ax^2} x^3 dx = \frac{1}{2a^2}[/tex]

    [tex]\int_{0}^{\infty}\frac{d}{da}\left(e^{-ax^2} x^2 \right) dx = \frac{d}{da}\left(\frac{\sqrt{\pi}}{4(\sqrt{a})^3}\right)[/tex]

    [tex]\int_{0}^{\infty}e^{-ax^2} x^4 dx = \frac{3\sqrt{\pi}}{8(\sqrt{a})^5}[/tex]
    Last edited: Nov 10, 2006
  8. Nov 10, 2006 #7
    You're missing a half in the last line, otherwise that looks fine.
  9. Nov 10, 2006 #8
    Ok edited to correct that thank you :)
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