Integral evaluation

  • Thread starter nuuskur
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  • #1
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Homework Statement


[tex]\int_{-\pi}^{\pi}x^{2014}\sin x {\rm{d}}x[/tex]

Homework Equations




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?
 
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Answers and Replies

  • #2
STEMucator
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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.
 
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  • #3
Dick
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Homework Statement


[tex]\int_{-\pi}^{\pi}x^{2014}\sin x {\rm{d}}x[/tex]

Homework Equations




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?
x^2014 is an even function. sin(x) is an odd function. What kind of function is the product?
 
  • #4
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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)?
 
  • #5
Dick
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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)?
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?
 
  • #6
STEMucator
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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)?
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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