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Evaluate this (improper) integral

  1. Oct 18, 2008 #1
    1. The problem statement, all variables and given/known data

    ∫ x1/2 e-x dx
    (This is the gamma function evaluated at 3/2)

    2. Relevant equations

    3. The attempt at a solution
    I've tried integration by parts, but it doesn't seem to work.

    Thanks for any help!
  2. jcsd
  3. Oct 18, 2008 #2
    I tried integration by parts too, and it was giving me trouble. So I did some research. The solution I found hopefully is not the only one, since it's a little complicated (in my opinion). Here are the basic steps:

    1. Show that [tex]G(\frac{3}{2})=\frac{1}{2}G(\frac{1}{2})[/tex] where G is the gamma function.
    (This step is not difficult. Use integration by parts and it follows immediately.)

    2. Show that [tex]G(\frac{1}{2})=\sqrt{\pi}[/tex].
    To do this, you change variables: let [tex]u=\sqrt{x}[/tex]. After this, you'll see that you have exactly a Gaussian integral (which is where the [tex]\sqrt{\pi}[/tex] comes from).

    Have you seen a proof / calculation of the Gaussian integral before? If not, there's a really nice proof of it on Wikipedia:

    If you haven't seen any of this before, I'm hoping someone else will get a simpler solution for this integral.
  4. Oct 18, 2008 #3


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    The reason the "Gamma" function is given a special name is that it CANNOT be written in terms of elementary functions. If you let u= x1/2, then du= (1/2)x-1/2dx so dx= 2x1/2du. Of course [itex]e^{-x}= e^{-u^2}[/itex] so
    [tex]\int_0^\infty x^{1/2}e^x dx= 2\int_0^\infty e^{-u^2}du[/tex]
    The Gamma function and Gaussian function are pretty much interchangable.

    Here is the standard way of finding the Gaussian integral, which you will find in most (multi-variable) Calculus books.
    Let [tex]I= \int_0^\infty e^{-x^2} dx[/tex]

    (I see now, after all that work, that this is exactly what jjou's link gives!)
    Then it is certainly also true that [tex]I= \int_0^\infty e^{-y^2}dy[/tex]
    Multiplying those:
    [tex]I^2= \left(\int_0^\infty e^{-x^2}dx\right)\left(\int_0^\infty e^{-y^2}dy[/tex]
    which we can interpret as the double integral
    [tex] = \int_{x=0}^\infty \int_{y= 0}^\infty e^{-x^2-y^2}dydx[/itex]
    over the first quadrant of the plane. Switching to polar coordinates, that is
    [tex]\int_{\theta= 0}^{2\pi} \int_{r= 0}^\infty e^{-r^2} (r dr d\theta)[/tex]
    [tex]= 2\pi \int_{r=0}^\infty e^{-r^2} rdr[/tex]

    Now having that "r" in the integrand makes all the difference. Let u= r2 so that du= 2r dr or (1/2)du= r dr. The integral becomes
    [tex]2\pi (1/2)\int_{u= 0}^\infty e^{-u}du= -\pi e^{-u}\right|_0^\infty= \pi[/tex]
    and, since that was I2,
    [tex]I= \int_0^\infty e^{-x^2}dx= \sqrt{\pi}[/tex]

    [tex]\int_0^\infty x^{1/2}e^{-x}dx= 2\int_0^\infty e^{-u^2}du[/tex]

    [tex]\int_0^\infty x^{1/2}e^{-x}dx= 2\sqrt{\pi}[/tex]

    (After all that work I see that this is exactly what the jjou's link gives! Oh, well, consider it confirmation.)
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