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Homework Help: Integral of (x-sinx)/x^3 (contour integration)

  1. Feb 22, 2010 #1
    The problem is [tex]\int _{-\infty}^\infty \frac{x-sinx}{x^3} dx [/tex]
    The answer is pi/2.
    Contour integration is to be used.

    Here are my questions about the method:

    Is the contour a semicircle with a bump?

    There is no z^-1 term in the expansion, so the residue at the origin is zero?

    Jordan's lemma=> large radius contribution ->0?

    So where does the actual contribution come from?

  2. jcsd
  3. Feb 22, 2010 #2
    If you are given an integral like this which you want to do using contour integration, then the first step could be to write down some meromorphic function of which the integral from minus to plus infinity will allow you to extract the value of the desired integral. Only then do you think about the details of a contour in order to compute that integral.

    Now, you do already know in advance that whatever contour you will need will extend to infinity. This then gives you a clue on how to choose the meromorphic function; sin(z) cannot appear in the integrand of any contour integral, as |sin(z)| will increase exponentially.

    Now, you probably know that in case of the integral of sin(x)/x you end up with replacing
    sin(x) by exp(ix) and then you consider the principal value of the integral and take the imaginary part.

    In this case, you can proceed as follows. You want to split up the integral in two parts, one containing 1/x^2 the other containing sin(x)/x^3, but that won't work because neither integral converges. You also cannot take the principal part, because they don't converge either. But what you can do is deform the integration contour to bypass the origin firstusing a half circle of radius epsilon containg the origin. After you do that you can split up the integral in the two parts.

    You can then replace sin(z) by 1/(2i) [exp(i z) - exp(-iz)] and split that part of the integral further in two. The contour of each part then has to be closed in a different way because |exp(iz)| decays exponentially in the upper half plane while |exp(-iz)| decays exponentially in the lower half plane.
    Last edited: Feb 22, 2010
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