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Gamma function calculation

  1. Oct 14, 2011 #1


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




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


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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


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    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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