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Image of 1/2 under gamma function

  1. Sep 25, 2005 #1
    I am trying to follow a proof of [tex]\Gamma(\frac{1}{2}) = \sqrt{\pi}[/tex] but in the way i have found this equality:

    [tex]\int_{0}^\frac{\pi}2sin^n xdx = \int_{0}^\frac{\pi}2cos^n xdx[/tex].

    I have tried unsucessfully integration for parts and I don't see how can I make some substitution. Maybe you can help me to justify this equation?

    (Note.- The fraction in the upper limit has a pi)

  2. jcsd
  3. Sep 25, 2005 #2


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    Do you understand it geometrically? It's the area under the sine (and cosine) function from 0 -> pi/2. Try to see it for n = 1, and then perhaps it's easy to see that it applies to higher powers too.
  4. Sep 25, 2005 #3
    It should be obvious graphically, but you can also let cos(x) = sin(x + (pi/2)).
  5. Sep 25, 2005 #4
    TD, I see the equality for n = 1, but how I pass to any n? By induction?

    Hypermorphism, with the equality you give me, what should i try? a substitution? integration by parts?

    Hm, if these are moronic questions, excuse them...

  6. Sep 25, 2005 #5
    Just try a substitution [tex]y = \pi/2 - x[/tex] in either one of the integrals.
  7. Sep 25, 2005 #6
    Hi Castilla,
    Timbuktu gives a better substitution. Letting u=x-pi/2, we get:
    [tex]\int_{-\frac{\pi}{2}}^0 \pm \sin^n u du[/tex]
    for the RHS depending on whether n is even or odd. Either way, it is trivially equivalent to the LHS integral. Algebraically:
    [tex]\int_{-\frac{\pi}{2}}^0 -\sin^n u du = \int_{-\frac{\pi}{2}}^0 \sin^n (-u) du = -\int_\frac{\pi}{2}^0 \sin^n u du = \int_0^\frac{\pi}{2} \sin^n u du[/tex]
    for odd n and for even n:
    [tex]\int_{-\frac{\pi}{2}}^0 \sin^n u du = -\int_\frac{\pi}{2}^0 \sin^n (-v) dv = \int_0^\frac{\pi}{2} \sin^n (-v) dv[/tex]
    But since n is even, [itex]\sin^n (-v) = sin^n v[/itex].
    Last edited: Sep 25, 2005
  8. Sep 25, 2005 #7
    Thanks for your kindly answers, Hypermorphism, Timbuktu, and TD.

  9. Sep 30, 2005 #8

    out of curiousity, how do you prove that [tex]\Gamma(\frac{1}{2}) = \sqrt{\pi}[/tex] ?
  10. Sep 30, 2005 #9
    I followed the proof in a book of analysis of one variable functions. So the methods are very elementary and clumsy. There are shorter ways which use double integrals.

    First you proof that [tex] \int_0^{+\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2}[/tex] (inside this cumbersome proof appeared the equality of my first post) Then, puting x = t^2 you get

    [tex] \Gamma (\frac{1}{2}) = \int_0^{+\infty}e^{-x}x^\frac{-1}{2}dx = 2\int_0^{+\infty}e^{-t^2}dt = \sqrt{\pi}}[/tex].
  11. Oct 1, 2005 #10

    George Jones

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    I = \int_0^{+\infty}e^{-x^2}dx.


    2I = \int_{-\infty}^{+\infty}e^{-x^2}dx,


    (2I )^2 = \int_{-\infty}^{+\infty}e^{-x^2}dx \int_{-\infty}^{+\infty}e^{-y^2}dy.

    Now write the iterated integral as a double integral and change to polar coordinates:

    (2I )^2 = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-x^2} e^{-y^2}dxdy = \int_{0}^{+\infty} \int_{0}^{2\pi} r e^{-r^2} d\theta dr = 2\pi \int_{0}^{+\infty} r e^{-r^2} dr = 2\pi \left[ -\frac{1}{2} e^{-r^2} \right]_{0}^{+\infty} = (2\pi) \left( \frac{1}{2} \right).

  12. Oct 1, 2005 #11
    How do you change to polar coordinates??

    I only learned the theorem of substitution for simple integrals. Could you be able to justify logically the change to polar coordinates? Thanks.
  13. Oct 1, 2005 #12

    George Jones

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    Have you studied double integrals and iterated integrals?

  14. Oct 1, 2005 #13
    Yes George, I have studied this version of Fubini's theorem:

    If [tex]R = [a, b] x [c, d] [/tex] and certain conditions, then

    [tex] \int\int_R f(x,y)dxdy = \int_a^{b}(\int_c^{d}f(x,y)dx)dy = \int_c^{d}(\int_a^{b}f(x,y)dx)dy [/tex].
  15. Oct 1, 2005 #14

    George Jones

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    I am a bit puzzled - you have studied Fubini's theorem, but you have not studied how to change from Cartesian to polar coordinates in a double integral?

    This change of coordinates invloves: 1) appropriately changing the limits of integration; 2) the substitutions [itex]x = r cos \theta[/itex] and [itex] y = r sin \theta[/itex]; the Jacobian [itex]r[/itex].

  16. Oct 2, 2005 #15
    What happens, George, is that I study analysis on my own account, learning without teacher, directly from some books. I have these ones:

    - Calculus (Apostol)
    - Mathematical Analysis (idem).
    - Elementary Classical Analysis (Marsden).

    If you review Fubini's theorem on Apostol's Analysis (chapter about multiple integrals), you'll see that it is only requested to know some concepts defined in the same chapter.

    Nevertheless, I have found in Apostol's Calculus a proof of the theorem of change of variables for double integrals (from which I would deduce the changing from cartesian to polar coordinates), but Apostol request his reader to previously have learned:

    1) Contour integrals (I know absolutely nothing about this)
    2) Green Theorem (nothing also)

    So, as the change from cartesian to polar coordinates is a particular case of changing variables in double integrals, my curiosity is if you know some logical proof of change of variables for double integrals which don' request contour integrals or Green Theorem.


  17. Oct 2, 2005 #16
    Spivak does this in his "Calculus on Manifolds" text, after an exercise asking the reader to prove a simpler statement, that the (signed) volume of the image of a paralleletope under a linear transformation is simply the volume of the paralleletope scaled by the determinant of the linear transformation (In other texts, the determinant is introduced as a signed area operator, and the properties and form are derived from that, instead of the other way around). The generalization to nonlinear functions is not hard, but requires care. Contour integrals and Green's theorem come much later.
  18. Oct 2, 2005 #17
    Well, maybe I can stole some money to my wife and buy a used copy of Spivak's book. Thanks for the info.
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