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Integration over a discontinuity

  1. Feb 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Evaluate ∫(1/x)sin^2(x)dx from -a to a


    3. The attempt at a solution

    Mathematica doesn't want to evaluate this because of the lack of convergence.

    I think it is zero. When we consider non-zero values of x in the associated riemann sum, the integrand is odd and so contributions from x and -x cancel out. I can make the Δx in the riemann sum arbitrarily small, so when we consider the step near zero, sin^2 x is almost exactly equal to x^2 and that term in the sum gets replaced to xΔx where x is very nearly zero as is delta x. In other words, the integral is zero.

    Is this kind of reasoning valid?
     
  2. jcsd
  3. Feb 16, 2013 #2
    Yea, it is zero by symmetry. The integrand is odd (it obeys [itex]f(-x)=-f(x)[/itex]) so the "negative area" accumulated from -a to 0 cancels out the "positive area" accumulated from 0 to a.
     
  4. Feb 16, 2013 #3

    jbunniii

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    Yes, your reasoning is OK. More formally, we may write the integrand as
    $$\frac{\sin^2(x)}{x} = \left(\frac{\sin(x)}{x}\right)\sin(x)$$
    Both factors on the right hand side are continuous on [-a,a] (we continuously extend ##\sin(x)/x## to equal 1 at ##x = 0##), so their product is also continuous, hence integrable. Therefore, since the product is an odd function and we are integrating over [-a,a], the result is zero.
     
  5. Feb 16, 2013 #4

    lurflurf

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    since f(0-)=f(0+)=0

    this is no more of a problem than x^2/x would be

    in fact x~sin(x) near zero
     
  6. Feb 16, 2013 #5

    haruspex

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    That's not enough by itself. You also have to show that the integral is bounded on all subintervals, otherwise the symmetry argument might equate to saying ∞-∞ = 0. Other posts in this thread cover that.
     
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