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Tricky integral

  1. Jan 23, 2012 #1
    I need a hint. Wolfram alpha says it equals pi/2, but I don't know how to get that.

    \int_{-\infty}^{\infty} \frac{\sin^4(x)}{x^2}\, dx
  2. jcsd
  3. Jan 23, 2012 #2
    Did you make a typo?
  4. Jan 23, 2012 #3
    I don't think so. Here's the Wolfram alpha solution.
  5. Jan 23, 2012 #4
    Are you familiar with contour integration?
  6. Jan 23, 2012 #5
    Yes. I thought about trying something like that, but the usual substitution z = e^{ix} to write the sin part as an exponentials doesn't work with the x^2 on the bottom.

    What were you thinking about with the contour integration?

    My other thought was to try to relate it back to \int sin(x)/x which is something I can do.
  7. Jan 23, 2012 #6
    I haven't worked it out but if you write out the sin^4 and then use contour integration it will probably work:
    [tex] \sin^4\theta = \frac{3 - 4 \cos 2\theta + \cos 4\theta}{8}\ [/tex]
    Last edited: Jan 23, 2012
  8. Jan 23, 2012 #7
    You're right, that would probably work. I'll give it a go. Thanks for your help!

    Alternatively, I figured out how to reduce it:

    \int_{-\infty}^{\infty} \frac{\sin^4{x}}{x^2}\, dx = \int_{-\infty}^{\infty} \frac{4 \sin^3{x} \cos{x}}{x}\, dx
    = \int_{-\infty}^{\infty} \frac{\sin{2x}(1 - \cos{2x})}{x}\, dx
    (trig identities)
    = \int_{-\infty}^{\infty} \frac{\sin{2x}}{x}\, dx - \int_{-\infty}^{\infty} \frac{\sin{4x}}{2x}\, dx
    = \int_{-\infty}^{\infty} \frac{\sin{x}}{x}\, dx - \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin{x}}{x}\, dx
    (change of variables)
    = \pi - \frac{\pi}{2}
    (sinc integral)
  9. Jan 23, 2012 #8
    Yes, well found!
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