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Integral evaluation

  1. Dec 9, 2014 #1
    1. The problem statement, all variables and given/known data
    [tex]\int_{-\pi}^{\pi}x^{2014}\sin x {\rm{d}}x[/tex]

    2. Relevant equations


    3. The attempt at a solution
    For such problems there are probably some extremely clever solutions, but I can't see any easy way here.
    If I were to find the antiderivative of this: I would eventually come up with a sum: [tex]-\frac{2014!}{0!}x^{0} \sin x + \frac{2014!}{1!}x\cos x + \frac{2014!}{2!}x^2 \sin x -\frac{2014!}{3!}x^3 \cos x + ... +\\ \frac{2014!}{2013!}x^{2013} \sin x -\frac{2014!}{2014!}x^{2014} \cos x[/tex]
    [is there a way to write this sum in sigma notation?]
    Are there 2015 summands?
    All summands with sine in them sum up to 0, because [itex]\sin\pm\pi = 0[/itex]. and [itex]\cos\pm\pi = -1[/itex].
    I have some series left from evaluating at pi and I have exactly the same series evaluated at -pi. Is the integral equal to 0? EDIT: NO, I forgot the x terms.

    Is there any, more elegant solution to this problem, though?
     
    Last edited: Dec 9, 2014
  2. jcsd
  3. Dec 9, 2014 #2

    Zondrina

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    An integral of the form ##\int x^n sin(x) \space dx## would usually be handled with multiple applications of integration by parts, but when ##n## is large this is obviously out of the question without the aid of a computer.

    There is actually a standard form for such an integral, which relies on the notion of summing up the ##sin(x)## and ##cos(x)## terms (you have already noticed this). The standard form in terms of the gamma function is:

    $$\int x^n sin(x) \space dx = - \frac{i^n}{2} \left[ \Gamma(n+1, -ix) - (-1)^n \Gamma(n+1, -ix) \right]$$

    Substituting your limits should help you confirm.

    EDIT: Alternatively, express ##sin(x)## in terms of its power series, bring the ##x^{2014}## in and presto I believe.
     
    Last edited: Dec 9, 2014
  4. Dec 9, 2014 #3

    Dick

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    x^2014 is an even function. sin(x) is an odd function. What kind of function is the product?
     
  5. Dec 9, 2014 #4
    I'm not sure, going with my gut I would say it's an odd function, which means that F(-x) + F(x) = 0
    The evaluation would then look like F(pi) - F(-pi) = 2F(pi)?
     
  6. Dec 9, 2014 #5

    Dick

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    Yes, it's an odd function. The integral of any odd function on an interval like [-a,a] that's symmetric around the origin is zero. You haven't seen this before?
     
  7. Dec 9, 2014 #6

    Zondrina

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    Not quite. If ##f(x)## is odd on ##[-L, L]## or ##(-L, L)##, then ##\int_{-L}^{L} f(x) \space dx = 0##.
     
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