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Gamma Functions

  1. Jan 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Questions are in picture.

    2. Relevant equations
    $$ \int _{0}^{\infty }x^{n}e^{-x}dx $$ = $$ Gamma (n+1) = n!
    $$ Gamma(P+1) $$ = $$Gamma(P)$$
    $$ Gamma(P) = (1/P) $$Gamma(P+1)$$


    3. The attempt at a solution
    2) I have found it from table.
    3) I have used recursion and table to find it.
    4) Again With recursion.
    5) $$ \Gamma(0.7) $$ = 1/p(p+1) with this formula.
    8) If $$ \ Gamma (p+1) $$ is equal to this integral,I think it can be written as $$ \ Gamma (2/3+1) $$ later we can found the value from table.Am I right?
    Same logic again for 9,10?
    But what next? How can I convert them to Gamma function?
     

    Attached Files:

  2. jcsd
  3. Jan 12, 2013 #2
    11) I found 1/2 but I'm not sure(?)

    Edit : No it's not true.I think true solution is on down.
     
    Last edited: Jan 12, 2013
  4. Jan 12, 2013 #3
    $$ \int _{0}^{\infty }x^{2}e^{-x^{2}}dx=2\int _{0}^{\infty }u^{2}e^{-u}du=2\cdot \Gamma 3 $$
     
  5. Jan 12, 2013 #4

    haruspex

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    No. How did you get that?
    What did you get for (5)?
    I don't follow your logic for (8). Where did 2/3 come from? I suspect what you posted is not what you meant.
     
  6. Jan 12, 2013 #5
    5. 1/0.7(Gamma1.7) = 1.30 ? Isn't it?
    I'm sorry questions mess up.It's solution of question 11.
    Here is for (8)

    For n= 2/3 Gamma(n+1) = Gamma(1+2/3)=Gamma(1.6) = 0.89
     
  7. Jan 12, 2013 #6

    haruspex

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    Yes.
    But you haven't done the substitution correctly. What will dx become?
    OK, but 1.6 is a bit inaccurate. The answer should be > 0.9.
     
  8. Jan 12, 2013 #7
    Yes.There's a problem.But I can't figure. if we say u=x^2 --> 2xdx=du --> dx = du/2x Where to go from here? I can't remember how can I escape from x variable?
     
  9. Jan 12, 2013 #8

    lurflurf

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    For the ones of the type

    [tex]\int_0^\infty t^a b \, e^{-t^b} \, \dfrac{\mathrm{dt}}{t}=\Gamma(a/b)[/tex]

    this can be derived by the change of variable u=t^b
     
  10. Jan 12, 2013 #9

    haruspex

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    Just replace the x using u=x2
     
  11. Jan 13, 2013 #10
    I understand but what about dx ? Would be? x^2=u 2xdx = du dx = du/2x (I'm asking about this x?)

    $$ \int _{0}^{\infty }ue^{-u}? $$
     
  12. Jan 13, 2013 #11

    lurflurf

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    dx = du/(2x)
    or
    dx = du/(2sqrt(u))
     
  13. Jan 13, 2013 #12
    Thanks!
     
  14. Jan 13, 2013 #13
    Can I use same logic for 12?

    x^3 = u
    3x^2dx= du
    dx = du / 3*√u

    ∫xe^-x^3 = ∫u^1/3*e^-u du/3*√u= 1/3 ∫ u^-1/6 e^-udu = 1/3 * Gamma -5/6 ...?


    + No idea for 15?
     
    Last edited: Jan 13, 2013
  15. Jan 13, 2013 #14

    haruspex

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    Not √u, a little more complicated than that. What is x as a function of u?
    In line with the other substitutions, u=8x looks obvious. Did you try that?
     
  16. Jan 14, 2013 #15
    Oh what I was did.it will be u^1/3 =?
    Not yet.But I will.
     
  17. Jan 14, 2013 #16

    lurflurf

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    ^Yes above I said

    [tex]\int_0^\infty t^a b \, e^{-t^b} \, \dfrac{\mathrm{dt}}{t}=\Gamma(a/b)[/tex]

    this can be derived by the change of variable u=t^b

    also

    [tex]\int_0^\infty (st)^a b \, e^{-(s t)^b} \, \dfrac{\mathrm{dt}}{t}=\Gamma(a/b)[/tex]

    s>0 this can be derived by the change of variable u=s t
     
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