1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Sin(x)/x via contour integration
Loading...