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Gamma Function

  1. Oct 30, 2005 #1
    I am wondering how the following statement holds true:

    [tex]\Gamma\left(\frac{1}{2}\right)=\int_0^{\infty}e^{-x}x^{-\frac{1}{2}}\,dx=\sqrt{\pi}[/tex]

    I know how to show that:

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

    But I can't seem to apply that method (converting to a double-integral) to the gamma function. Any ideas?

    Thanks.
     
    Last edited: Oct 30, 2005
  2. jcsd
  3. Oct 30, 2005 #2

    Hurkyl

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    [tex]\int_0^{+\infty} e^{-x^2} \, dx[/tex]

    and

    [tex]\int_0^{+\infty} e^{-x} x^{1/2} \, dx[/tex]

    look a lot different... after all, one has [itex]-x^2[/itex] in the exponent, and the other has [itex]-x[/itex]...
     
  4. Oct 30, 2005 #3
    Yeah, I just figured that since the results look similar, there may be a way to rearrange that gamma function to make it look like the other. --I guess not.

    Would you suggest integration by parts to solve this? That is the only way I can think of, but it doesn't seem too promising because I can't get rid of that fractional exponent. Is there some sort of trick to this?

    Thanks again.
     
  5. Oct 30, 2005 #4

    Hurkyl

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    I never said you couldn't. :tongue: My post was a hint!
     
  6. Oct 30, 2005 #5

    saltydog

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    One method is to use Laplace Transforms:
    First calculate:
    [tex]\Gamma(x+1)=\int_0^{\infty}e^{-\beta}\beta^{x}d\beta[/tex]
    let:
    [tex]\beta=st[/tex]
    [tex]\Gamma(x+1)=s^{x+1}\int_0^{\infty}e^{-st}t^x dt[/tex]
    rearrange, equate to the Laplace transform of [itex]t^{x}[/itex] and then let x=-1/2. Can you finish it?
    Also, shouldn't your integral have [itex]x^{-1/2}[/itex]?
     
  7. Oct 30, 2005 #6

    benorin

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    yep

    Yes, it should, for [itex]\Gamma(z)=\int_{t=0}^{\infty}e^{-t}t^{z-1}dt[/itex] for [itex]z\in\mathbb{C}[/itex] such that [itex]\Re(z)>0[/itex].
    Hence [itex]\Gamma\left(\frac{1}{2}\right)=\int_{t=0}^{\infty}e^{-t}t^{-\frac{1}{2}}dt[/itex], apply the substitution [itex]x=\sqrt t \Rightarrow dx=\frac{1}{2}t^{-\frac{1}{2}}dt[/itex] to get
    [itex]\Gamma\left(\frac{1}{2}\right)=2\int_{t=0}^{\infty}e^{-x^{2}}dx = \sqrt\pi[/itex]
     
    Last edited: Oct 30, 2005
  8. Oct 30, 2005 #7
    Thanks a lot everyone for the help. I like that last substitution to change it into the form I am familiar with. Thank you :smile:
     
  9. Oct 31, 2005 #8

    Hurkyl

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    Do you understand why he thought to use that substitution in the first place? (Hint: my hint is why, though he probably knew it without my hint) Will you think to look at that next time you get stuck on an integral?
     
  10. Oct 31, 2005 #9
    Yes, thank you both very much.
     
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