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Complex analysis integral

  1. Aug 3, 2016 #1
    1. The problem statement, all variables and given/known data
    calculate ## \int_{-\infty}^{\infty}{\frac{2}{1+x^2} e^{-ikx}dx}##, where k is any positive number

    2. Relevant equations


    3. The attempt at a solution
    So first consider the closed contour ##I= \int{\frac{2}{1+z^2} e^{-ikz}dx}##

    We can choose the contour to be along the real axis ## [-R,R]## then along the semi circle for which ##z=Re^{i \theta}, [0, \pi] ##

    then ##I= \int_{-\infty}^{\infty}{\frac{2}{1+x^2} e^{-ikx}dx} +\int_{0}^{\pi} {\frac{2}{1+(Re^{i \theta})^2} e^{-ikRe^{i\theta}}dx}##

    I calculated the residue to be ## \pi e^{k} ##, with the poles being at ## z=i## and ##z=-i##

    I think the integral around the semi-circle is supposed to go to zero in the limit of R going to infinity, but I am struggling to see why. I can't use Jordan's lemma (?) since it is not in the usual formal, and if i expand

    ##e^{-ikRe^{i\theta}} = e^{-ikR(cos(\theta)+isin(\theta))}##
    ##= e^{-ikRcos(\theta)}e^{kRsin(\theta)}##

    but since ##\theta ## is between ##[0, \pi], sin(\theta)## is greater than, or equal to zero, meaning that the integrand does not drop off as R goes to infinity..

    Any help would be greatly appreciated- thanks in advance :)
     
    Last edited: Aug 3, 2016
  2. jcsd
  3. Aug 3, 2016 #2

    Math_QED

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    You should add dx to your integrals.
     
  4. Aug 3, 2016 #3
    Noted and corrected. Thank you :)
     
  5. Aug 3, 2016 #4

    Ray Vickson

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    For ##k > 0## completing in the upper x-plane won't work, because for ##x = t + i y## we have ##e^{-ikx}= e^{-ikt}\, e^{ky}##. which behaves badly for large ##y > 0##. However, we can complete in the lower x-plane instead.

    Alternatively, we can see that for real ##x## the integrand has a real part that is an even function of ##x## and an imaginary part that is odd (so will integrate to 0); that is, we could obtain the same result if we replace ##e^{-ikx}## by ##\cos(kx)##. But then, we might as well extend this to ##e^{ikx}## without changing its value.

    So, let's look at the integral with integrand ##f(x) = e^{ikx}/(1+x^2)##. Now we can complete in the upper x-plane, because##e^{ik(t+iy)} = e^{ikt} e^{-ky}##. However, completing along a semi-circle is not very convenient; using a rectangle is much easier. So, let's take the original integral from ##x = -R## to ##x = +R##, taking ##R \to \infty## eventually. Complete in a rectangle as follows: append the integral from ##R+i0## to ##R + iR## along the vertical, then from ##R + iR## to ##-R + iR## along an upper horizontal, then from ##-R + iR## down to ##-R + i0## along the other vertical. It is pretty easy to get bounds on each of the vertical and upper-horizontal integrals, and to show that they ##\to 0## as ##R \to \infty##.
     
    Last edited: Aug 3, 2016
  6. Aug 3, 2016 #5
    Closing in the lower half plane produces the same problem. Though the exponential power will be negative, over the limit ##[ 0, -\pi], sin(\theta)## is negative, meaning the resultant power is positive, and the same problem occurs. (This may also be seen by making the substitution ##x \to (-x)## in my initial working).

    Edit:
    But x (more accurately z) itself is complex, not real, over either the semi-circle or rectangle so taking the real part of the function will not simply be a matter of changing the exponential into its ##cosx+isinx## form. For example the integrand is actually ##\frac{1}{1+R^2e^{2i\theta}} e^{-ikRe^{i\theta}}Rie^{i\theta} ## taking the real part of this is not as straightforward, surely ?

    Sorry if I'm missing something obvious- thanks for the help
     
    Last edited: Aug 3, 2016
  7. Aug 3, 2016 #6

    Ray Vickson

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    No: we start with the integral over the real line, and show that we can replace ##e^{-ikx}## by ##e^{ikx}## in that integral. Then we extend ##x## into the upper complex plane. Alternatively, complete in the lower x-plane, and use arguments similar to the one below.

    Taking ##e^{ikx} = e^{ik(t + iy)} = e^{ikt} e^{-ky}## for ##t = R## and ##0 \leq y \leq R##, the right-hand vertical integral becomes
    $$ I_1 = \int_0^R \frac{e^{ikR} e^{-ky}}{(R + iy)^2 + 1} i \, dy,$$.
    Use the facts that ##|\int f| \leq \int |f|##, ##|a/b| = |a|/|b|## and ##|e^{ikt} e^{-ky}| = |e^{ikt}| e^{-ky} = e^{-ky}##. The magnitude of the denominator in ##|f|## is
    $$|(R+iy)^2 + 1| = \sqrt{R^4+2R^2 y^2+y^4+2 R^2-2y^2+1} \geq \sqrt{R^4 + 2 R^2},$$
    since ##2R^2 y^2 \geq 0## and ##y^4 - 2 y^2 \geq -1## for ##y \geq 0##. We thus have
    $$|I_1| \leq \int_0^R \frac{e^{-ky}}{ \sqrt{R^4+2R^2}} dy \to 0 $$
    as ##R \to \infty##.

    Similar types of arguments apply to the other vertical integral and to the upper horizontal integral.

    Trying something like the above on a semi-circular arc of radius ##R## would be much harder, so that is why I suggested not doing it that way.
     
    Last edited: Aug 3, 2016
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