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Sin(x)/x via contour integration

  1. Jul 16, 2010 #1
    1. The problem statement, all variables and given/known data
    Evaluate [itex]\int_{0}^{\infty} \frac{\sin x}{x}\,dx[/itex]

    2. Relevant equations

    1. Cauchy's integral theorem
    2. [tex]\int_{|z| = r} \frac{dz}{z} = 2\pi i[/tex]

    3. The attempt at a solution
    I want to see if the following works. The hint suggests the use of the contour here: http://upload.wikimedia.org/wikibooks/en/9/9e/ContourSinzz.gif" [Broken]. We denote by [itex]\gamma _R ^+[/itex] and [itex]\gamma _{\epsilon} ^+[/itex] the semicircles of radius R and [itex]\epsilon[/itex] with positive and negative orientations, respectively.

    We consider the function
    [tex]f(z) = \frac{e^{iz}-1}{z}.[/tex]
    By Cauchy's theorem, we have
    [tex]\int_{-R}^{-\epsilon} + \int_{\gamma _{\epsilon} ^+} + \int_{\epsilon}^{R} + \int_{\gamma _{R} ^+}.[/tex]

    First we note that
    [tex]\int_{\gamma _{\epsilon} ^+} f(z)\, dz = \int_{\gamma _{\epsilon} ^+} \frac{e^{iz}}{z}\, dz \,-\, \int_{\gamma _{\epsilon} ^+} \frac{dz}{z} = \int_{\pi}^{0} \frac{\exp \left(i \epsilon e^{it} \right)}{\epsilon e^{it}}\cdot i \epsilon e^{it} dt \,+\, \pi i = \int_{\pi}^{0} i\exp \left(i\epsilon e^{it}\right)dt + \pi i.[/tex]

    Since [itex]|e^{iz}| = e^{-y},[/itex] applying Euler's formula gives
    [tex] |i\exp{(i\epsilon e^{it})}| \leq e^{-\epsilon \sin t}.[/tex]
    Letting [itex]\epsilon \rightarrow 0[/itex] and applying Lebesgue's dominated convergence theorem, we have
    [tex]\lim_{\epsilon \rightarrow 0}\int_{\gamma _{\epsilon} ^+} f(z)\, dz = \lim_{\epsilon \rightarrow 0} \int_{\pi}^{0} i\exp \left(i\epsilon e^{it}\right) dt + \pi i = 0.[/tex]

    Applying the same arguments to the larger semicircle, we have
    [tex]\left|\int_{\gamma _{\R} ^+} \frac{e^{iz}}{z}\,dz\right| = \left|\int_{0}^{\pi} i\exp \left(iRe^{it} \right) dt \right| \leq \int_{0}^{\pi} \exp(-R \sin t)\,dt.[/tex]
    Clearly this last integrand is dominated over the interval [itex][0,\pi][/itex], so applying DCT again shows that this last integral vanishes as [itex]R \rightarrow \infty.[/itex]

    Thus the contribution from the integrals over each semicircle is [itex]-\pi i[/itex] as we let R approach infinity and epsilon approach 0. Hence
    [tex]\int_{-\infty}^{\infty} \frac{e^{ix}-1}{x} dx = \pi i.[/tex]
    Taking imaginary parts and noting parity symmetry gives the desired result.

    *EDIT* Hopefully I'm not missing anymore differentials.
    Last edited by a moderator: May 4, 2017
  2. jcsd
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