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Contour integration on sin^2x/x

  1. Feb 24, 2010 #1
    SOLVED. (Sorry I worked out the solution when I was typing it and I can't work out how to remove the post; for some reason I assumed sin(z)<1 for complex z which is very false.)

    1. The problem statement, all variables and given/known data
    [tex]I = \int_{-\infty}^{\infty} \frac{\sin^2(x)}{x^2} \,dx=\pi.[/tex]

    2. Relevant equations

    3. The attempt at a solution
    For a first attempt I tried to integrate
    [tex]\int_{0}^{\infty} \frac{\sin^2(\beta x)}{x^2} e^{-\alpha} \,dx[/tex] by differentiating twice under the integral, then integrating twice with respect to [tex]\alpha[/tex]. I wasn't very careful with the branches of my logarithms and got an answer for the integral [tex]I = i \ln(-1)[/tex]; so if I was more careful I think I could prove it that way.

    However I though this would be a good integral to evaluate by contour integration; integrating along the semi-circle lying on the upper half plane, centre the origin, radius R anti-clockwise (then take the limit as R goes to infinity).

    Now [tex]\frac{\sin^2(z)}{z^2}[/tex] is holomorphic, so the integral around this contour must be zero. Thus we get
    [tex]\int_{-R}^{R} \frac{\sin^2(x)}{x^2} \dx=- \int_{0}^{\pi} \frac{sin^2(R e^{i\theta})}{(Re^{i\theta})^2} (iRe^{i\theta} \,d \theta)[/tex]
    So it follows
    [tex]\left| \int_{-R}^{R} \frac{\sin^2(x)}{x^2} \dx \right | <= \frac{\pi}{R} [/tex].
    Thus the integral I is zero.
    What have I done wrong here? I am certain the function is holomorphic, and it makes sense the integral along the boundary should decay at infinity - why am I getting the wrong value for the integral?

    Last edited: Feb 24, 2010
  2. jcsd
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