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Gamma function calculation

  1. Oct 14, 2011 #1
    [tex]\Gamma(x)=\int^{\infty}_0t^{x-1}e^{-t}dt[/tex]

    [tex]\Gamma(\frac{1}{2})=\int^{\infty}_0\frac{e^{-t}}{\sqrt{t}}dt=[/tex]

    take [tex]t=x^2[/tex]

    [tex]dt=2xdx[/tex]

    [tex]x=\sqrt{t}[/tex]

    [tex]=\int^{\infty}_0\frac{e^{-x^2}}{x}2xdx[/tex]

    Why here we can here reducing integrand by [tex]x[/tex]?
     
  2. jcsd
  3. Oct 14, 2011 #2

    phyzguy

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    What is your question? You've done the substitution correctly, so you see how the x's cancel. What is the problem?
     
  4. Oct 14, 2011 #3
    Lower limit is [tex]0[/tex]. Why I may cancel x's?
     
  5. Oct 14, 2011 #4

    dextercioby

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    Homework Helper

    Because the functions are defined on intervals not containing the point 0, that's why you can have the x in the denominator and simplify it through.
     
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