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Antiderrivative of e^(-x)*sin(x)*x^n

  1. Mar 16, 2012 #1
    I have recently come up with a fascinating method of summing various divergent series, whether it has been done before I am unaware, but in some scenarios it involves the anti-derivative of e^(-x)*sin(x)*x^n with respect to x. I only need positive integer values of n, and of course a proof/explanation. Any help would be appreciated. Forgive me if this is in the wrong place or is phrased awkwardly... I also accidently hit the enter key before I was done... It does involve number theory at it's heart, but that was not presented in the problem...
     
    Last edited: Mar 16, 2012
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  3. Mar 16, 2012 #2

    micromass

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    Can you tell us what you've tried?? That way, we know how to help!!
     
  4. Mar 16, 2012 #3
    Well I have tried integration by parts on it, but it seemed to just explode to astronomical sizes... I have calculated it for certain values of n, and am rather sure that it can be expressed with common functions for any integer, and in particular I only need the Integral from 0 to infinity of this function, on occasion that helps to simplify it. Thank you whoever brought this out of number theory, that was a mistake, I was trying to move it.
     
  5. Mar 16, 2012 #4

    micromass

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    Wolfram alpha gives us following nice result

    [tex]\int_0^{+\infty} x^n e^{-x}\sin(x)dx=2^{-\frac{n}{2}-\frac{1}{2}}\sin( \pi(n+1)/4)n![/tex]

    I think you can reach it by induction. Basically, the induction hypothesis is that for each n, there are polynomial P(x) and Q(x) such that

    [tex]\int_0^{+\infty}=e^{-x}(P(x)\sin(x)+Q(x)\cos(x))[/tex]

    That this is the case is very easily seen. Indeed, partial integration will always give us such a form.

    Evaluating in 0 and infinity shows us that we are only interested in the constants of this polynomials. So try to calculate the integral while only caring for the constants.

    For example, look at

    [tex]\int x^3 e^{-x} \sin(x)dx[/tex]

    Partial integration will yield a term with [itex]x^3[/itex], which we can drop. Remembering that

    [tex]\int e^{-x}\sin(x)dx = -\frac{1}{2} e^{-x} (\sin(x)+\cos(x))[/tex]

    So we only care about the integrals

    [tex]-\frac{3}{2}(\int x^2e^{-x}\sin(x)dx + \int x^2e^{-x}\cos(x)dx)[/tex]

    Now we can perhaps apply some kind of induction hypothesis.

    Is this enough information for you to work it out?
     
  6. Mar 16, 2012 #5

    micromass

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    Perhaps an easier way is to calculate the integral

    [itex]\int x^n e^{(i-1)x}dx = \int x^n e^{-x}\cos(x)dx + i\int x^n e^{-x}\sin(x)dx[/itex]

    This integral seems a whole lot easier and is very approachable by integration by parts. The imaginary parts of the final integral will be the integral you're looking for.

    Another method is by contour integration. That should work as well.
     
    Last edited: Mar 16, 2012
  7. Mar 16, 2012 #6
    Ah, thank you very much, I should have consulted wolfram alpha. I'm sure I'll have enough for a proof of it now, thank you! My summing of divergent resulted in the common answer 1+2+3+4....=-1/12, though I can now compute 2*1-4*3+6*5-8*7..., or indeed any n!/1-(n+3)!/3!+(n+5)!/5!-(x+7)!/7!... with relative ease. I should have been able to do it by myself... thank you!
    edit:Ah, imaginary numbers! I remember them.. they got me interested in math in the first place, last year... I remember only 5 months ago when the simple chain rule proof for deriving was too tricky... I'll take a look at that integral sometime.
     
    Last edited: Mar 16, 2012
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