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Real integral in complex plane

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data
    I want to find the value of the integral:
    ∫cos(x)/((x+a)2+1) dx from ]-∞;∞[


    2. Relevant equations
    Residue theorem


    3. The attempt at a solution
    My question is seeking more a conceptual understanding of why transforming to the complex plane works. According to Jordans Lemma the semi arc of your contour will go to zero if the maximum of lf(z)l -> 0 as lzl->∞. For the function f(z)= cos(z)/((z+a)2+1) - how can you check that this holds? If Jordans Lemma isn't satisfied is it still possible to calculate the real integral by transforming to the complex plane, calculating the integral, and subtracting the contribution from the semi arc?
     
  2. jcsd
  3. Jun 19, 2012 #2
    Jordan's lemma is necessary to ensure that the semiarc makes no contribution to the integral, but if I recall my complex analysis, you may have to pull off some sort of trick here, for it's clear the lemma is not satisfied--on the imaginary axis, for instance, [itex]\cos z \to \cosh y[/itex] which won't be bounded and grows faster than the denominator. I think there's a way around this, but I don't recall.
     
  4. Jun 19, 2012 #3

    Ray Vickson

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    We have [tex]\cos(x) = \text{Re}(e^{ix}),[/tex] so if you can compute the integral with exp(ix) instead of cos(x) you are done. Let x be extended to z into the upper half-plane, where [tex] e^{iz} = e^{i(x+iy)} = e^{ix} e^{-y},[/tex] and this goes to zero as y → +∞.

    RGV
     
  5. Jun 19, 2012 #4
    Okay that's quite neat. But one question (which maybe a bit stupid). You get a value for the integral which is complex. Are you then supposed to take the real value of this? How do you know that the complex part of exp(iz) will not contribute to the real value of the integral and the other way around:

    In other words:

    ∫exp(iz)dz = ∫cos(z)dz + i∫sin(z)dz

    The values of the total integral is, as far as I can see, complex. But if you take the real part of that: Are you then guarenteed that it only comes from the first integral in the sum above? And how do you know that? I do see the "i" in front of the second integral, but if the integral gives a complex value it would contribute to the real part.
     
  6. Jun 19, 2012 #5

    Ray Vickson

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    Of course you take the real part:
    [tex] I = \int_{-\infty}^{\infty} \frac{\cos(x)}{(x+a)^2+1}\, dx
    = \text{Re}\,J,\:\: J = \int_{-\infty}^{\infty} \frac{e^{ix}}{(x+a)^2+1}\, dx
    [/tex]
    We evaluate J by going out into the complex plane.

    RGV
     
    Last edited: Jun 19, 2012
  7. Jun 20, 2012 #6
    But how do you know that integrating over the imaginary part wont yield something complex and thus give a contribution to the real part? Is that impossible due to the definition of the integral of a complex function?
     
  8. Jun 20, 2012 #7

    vela

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    Because the imaginary part of J is
    $$Im \:J = i \int_{-\infty}^{\infty} \frac{\sin x}{(x+a)^2+1}\, dx. $$ The integral is real. There's no way it can yield a complex result.
     
  9. Jun 20, 2012 #8
    aha. Got it - thanks to both of you - you have been really helpful with this an other questions :)
     
  10. Jun 20, 2012 #9

    Ray Vickson

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    It doesn't in this case, but that does not matter! You want to extract the real part of some complex quantity, and if that quantity happens to have the form (a+ib)*(c+id) (with real a,b,c,d) then, of course, the answer you want is a*c-b*d.

    RGV
     
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