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Integral of sinc(x)

  1. Jun 1, 2009 #1
    1. The problem statement, all variables and given/known data
    I'm trying to prove the following definite integral of [tex]sinc(x)[/tex]

    [tex]\int_{-\infty}^{\infty}\frac{\sin(x)}{x}dx=\pi[/tex]


    2. Relevant equations



    3. The attempt at a solution
    I've tried power series expansions as well as trigonometric identities like

    [tex]\frac{\cos 2x}{x}=\frac{\cos^2 x}{x}-\frac{\sin^2 x}{x}[/tex]

    I also looked at techniques used to integrate the definite integral
    [tex] \int_{-\infty}^{\infty}e^{-x^2}dx [/tex]

    which I know is solved by double integration and changing to polar coordinates. However, this does not help me integrate [tex]sinc(x)[/tex].
     
  2. jcsd
  3. Jun 1, 2009 #2
    Well, I suppose you could do it by making a closed curve in the complex plane and using Caychy's theorem (and Jordan's lemma). There might be an easier way, but I can't think of any.
     
  4. Jun 1, 2009 #3
    Think about euler's formula and leibniz. A 'simple' proof can be made this way.
     
  5. Jun 1, 2009 #4
    [tex]\int_{0}^{\infty}\sin(x)\exp(-sx)dx=\frac{1}{1+s^2}[/tex]

    Integrate both sides from s = 0 to infinity to obtain the result.
     
  6. Jun 2, 2009 #5
  7. Jan 19, 2011 #6
    sinc(x) = sin(x)
    x
    has no anti-derivative
     
  8. Jan 19, 2011 #7
    no elementary anti-derivative :wink:
     
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