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Residue Theorem with real zero

  1. Jul 11, 2016 #1

    DCN

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    1. The problem statement, all variables and given/known data

    Find [tex]\int_{0}^{\infty} \frac{\cos(\pi x)}{1-4x^2} dx[/tex]

    2. Relevant equations

    The residue theorem

    3. The attempt at a solution

    The residue of this function at $$x=\pm\frac{1}{2}$$ is zero. Therefore shouldn't the integral be zero, if you take a closed path as a hemisphere in the upper half of the complex plane? Yet the integral evaluates to $$\pi/4$$

    I am completely lost.
     
    Last edited: Jul 11, 2016
  2. jcsd
  3. Jul 11, 2016 #2

    Ray Vickson

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    You cannot complete the integral along a semicircle in the upper half-plane because the function ##\cos(\pi x)## grows exponentially large when you let the imaginary part of ##x## go to infinity. However, you can instead write the integral as
    [tex] I = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\cos(\pi x)}{1-4x^2} \, dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{e^{i \pi x}}{1-4x^2} \, dx [/tex]
    because the imaginary part of the second integrand is an odd function, which integrates to zero.

    That last form can be completed in the upper half-plane, and when you do that the residues do not vanish.
     
    Last edited: Jul 12, 2016
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