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Homework Help: A sinusoid integrated from -infinity to infinity

  1. Apr 2, 2007 #1
    I had a sort of odd question on my homework,

    Sin(x)^3 dx, integrated over all reals (from negative infinity to infinity).

    The problem also gives this morsel of ambiguity:

    "Hint: think before integrating. this is easy"

    Now my initial guess because of the antisymmetry of the function is that it equals zero. Although the problem doesn't ask for a proof of any way shape or form however, I was baffled how I would argue that I reasoned it equaled zero if I was called upon in class.

    So I'm wondering whether my assumption is correct as well as maybe a brief explanation. No proof needed.
  2. jcsd
  3. Apr 3, 2007 #2

    Gib Z

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    Exactly correct.
  4. Apr 3, 2007 #3
    A pedant might ask for proof that you can use the antisymmetry of the integral in this way when the limits are +- infinity.

    But I guess that's why we have mathematicians.
  5. Apr 3, 2007 #4

    Gib Z

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    [tex]\lim_{a\to\infty}( \int^a_{-a} \sin^3 x dx )

    = \lim_{a\to\infty} (\int^a_0 \sin^3 x dx + \int^0_{-a} \sin^3 x dx)

    =\lim_{a\to\infty} (F(a) - F(0)) - (F(0) + F(-a))[/tex], Where dF(x)/dx=sin^3 x.

    Since the derivative of any odd function is an even function, F(-a)=F(a)

    [tex]\lim_{a\to\infty} (F(a) - F(0)) - (F(0) + F(-a)) = \lim_{a\to\infty} (F(a) - F(0)) - (F(0) + F(a))

    =\lim_{a\to\infty} (0)

    = 0[/tex].
  6. Apr 3, 2007 #5


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    What Gib Z gives is the "Cauchy Principal Value" of the integral. Of course, the limit is 0 because sin(x) is an odd function. Evaluating its integral at a and -a will give the same thing.

    Strictly speaking [itex]\int_{-\infty}^\infty f(x)dx[/itex] is
    [tex]\lim_{a\rightarrow -\infty}\int_a^0 f(x)dx+ \lim_{b\rightarrow \infty} \int_0^b f(x) dx[/tex]
    where the two limits are taken independently. Using that definition,
    [tex]\int_{-\infty}^\infty sin^3(x) dx[/tex]
    does not exist.
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