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Need help, soon please Integrating

  1. May 18, 2007 #1
    Need help, soon please!! Integrating

    I am struggling to integrate this function since last night but,it seems like it is a dead end doing it.

    how would you integrate this function

    integral x^n e^(-x^(n-1)) dx
    i tried the parcial method, and went like this
    integ x^(n-2) x^2 e^(-x^(n-1))= then i did this u=x^2,du=2xdx

    v=integ x^(n-2) e^(-x^(n-1)), then i took the substitution
    -x^(n-1)=t
    -(n-1)x^(n-2)dx=dt
    x^(n-2)dx=-1/(n-1)dt, than i get

    -1/(n-1)integ e^t dt=-1/(n-1) e^(-x^(n-1))

    then using the formula uv-integ vdu, i get
    -1/(n-1)x^2 e^(-x^(n-1))+ 2/(n-1) integ e^(-x^(n-1)) xdx,

    so the problem is how to integrate now this
    integ e^(-x^(n-1)) xdx, i could expand it using taylor formula, but i just have a feeling that it should be done using some other methods.!!
    I do not know if my approach is correct at first place, i might be missing something here.
    Any help please?
     
  2. jcsd
  3. May 21, 2007 #2
    So no one is going to help me>>>????
     
  4. May 21, 2007 #3

    cristo

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    Is this homework, or just a general problem that's troubling you? Maple gives the answer in terms of gamma functions, so I'd imagine that the integral is not that nice to compute!
     
    Last edited: May 21, 2007
  5. May 21, 2007 #4

    Gib Z

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    Maple will give the answer in terms of gamma functions, that integral is pretty similar to the definition of the gamma function itself.

    The integral can be computed by repeated integration by parts which is the same why the integral definition of the gamma function yields a generalized factorial function.
     
  6. May 21, 2007 #5
    NO it is not homework!!
    And sorry but i do not understand what mapple means, and what is gamma function??
    Maybe i have heard or used gamma function before, but just i am not familiar with this term, so any more explanations on this??
     
  7. May 21, 2007 #6
    ANd i do not think the repeated integration by parts works here, because look here

    integ e^(-x^(n-1)) xdx

    how would one go about integrating it by parts???

    if we take here u=x, du=dx, and v=integ e^(-x^(n-1))dx , so how to find v??
    And if we get u=e^(-x^(n-1)). and v=integ x dx, after we use the formula of integraton by parts we come to a form of the integral which is the integral wich we beagan with at the very begining, wich is positive, and we get nothing!!!!!!!!
     
  8. May 21, 2007 #7

    cristo

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    Ok.. It wasn't maple I used before, it was the online version of mathematica, called "the integrator." Maple returns a result including a Whittaker M function, which I have never seen before.

    Is there any particular reason why you have to do this integral by hand?
     
    Last edited: May 21, 2007
  9. May 21, 2007 #8
    First of all i just like to know all the steps needed to come up to the result. Second, sometimes the proffesor puts it as one of five problems in the final exam, if it is not identical with this one, one of the five problems in most cases has been something like this. And the exam is paper based, so no calculators are allowed.
     
  10. May 21, 2007 #9

    cristo

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    Hmm, well I'm afraid I can't help you. Are you sure the integral isn't something more like [tex]\int x^{n-1}e^{x^n}dx [/tex] ? I can't see the use of such an integral as one you give being put on an exam, but then I could be wrong.
     
  11. May 21, 2007 #10
    No it is exactly like i wrote it.
    And our proffesors, always put a really hard problem in every exam, thats why no one could ever get more than 85 points out of 100, in the exam!!!
     
  12. May 22, 2007 #11
    Thankyou cristo for your attempt to help me.

    Any other ideas or suggestions on how to go about integrating this function by hand????
     
  13. May 22, 2007 #12
    Are there no limits to this integration? One could consider the case of n=0,1,2 and solve it with standard integrals if the limits are 0 to infinite. I don't know about the general case though.
     
  14. May 23, 2007 #13
    I guess it is an indefinite integral.
     
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