1. Not finding help here? Sign up for a free 30min 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!

Find, using complex contour integrals, the function f(x)...

  1. May 15, 2016 #1
    1. The problem statement, all variables and given/known data
    whose Fourier transform is f~(p) = 1/(a2 + p2)

    2. Relevant equations
    f(x) = 1/√2π ∫-∞ eipx f~(p)


    3. The attempt at a solution
    First of all I let f(z) = eixz/(z2 + a2)
    and γ = γ1 + γ2
    with the ϒ's parametrised by:

    γ1 : {z=t, -R<t<R}
    γ2 : {z=Reit, 0<t<π}
    (So a semicircle of radius R)

    In this contour the only pole that lies inside is the z= +ia
    so using Cauchy's residue theorem:
    ϒf(z)dz = 2πi (Res(f, z=ia))
    I found the residue of z=ia to be Rez(f,z=ia)= -i e-ax/2a
    So ∫ϒf(z)dz= π e-ax/a

    And I have something in my notes about the fact you have to check that the integral of f(z) on γ2 goes to zero as R goes to infinity, which I vaguely understand because we actually want the integral from -∞ to +∞.
    So the checks I have in my notes are:
    Suppose γ has length L and on γ |f(z)|<M
    Then |∫γf(z)dz|≤M⋅L
    With γ ≡ {z=γ(t), a≤t≤b}
    L = ∫abdt|γ'(t)|
    On γ2: |z| =R
    Using ||a|-|b|| ≤ |a+b| ≤ |a|+|b|
    R2-a2 ≤ |z2+a2| ≤ R2+a2
    So 1/R2-a2 ≥ 1|z2+a2| ≥ 1R2+a2
    Therefore |∫γ2f(z)dz|≤πR/(R2-a2)
    So ∫γ2f(z)dz → 0 as R → ∞
    as you have |∫γ2 f(z)dz| ≤ e-Rxsin(t)/(R2 - a2)

    and if x ≥0 the exponent is, at most, equal to 1, but if x<0, it blows up and we need a different contour for x<0.

    Let γ = γ1 + γ3
    with ϒ3 : {z=Reit, π<t<2π}

    so a semicircle in the negative part, and this time it's a clockwise contour so Cauchy's residue must carry a minus sign on the right hand side.

    So doing everything as before I found ∫ϒ = πeax/a

    But I'm having trouble with the checks, I found that the inequality for the modulus of the integral over ϒ3 comes out the same? So effectively it still blows up?

    But I know this is the right answer I just can't follow through with the checks.

    I appreciate this is quite long, but a lot of it is just background to set the scene, and I think I just have a lack of understanding of the checks and just know the procedure.

    Thanks in advance
     
  2. jcsd
  3. May 15, 2016 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Looks like your notes have some mistakes. Assume ##R>a## since you're going to make the circle big in th end. You showed ##\left\lvert \frac{1}{z^2+a^2} \right\rvert \le \left\lvert \frac{1}{R^2-a^2} \right\rvert##, so you can say
    $$\left\lvert \frac{e^{izx}}{z^2+a^2} \right\rvert \le \left\lvert \frac{e^{izx}}{R^2-a^2} \right\rvert.$$ On the semicircle, you have ##z=R\cos\theta + iR\sin\theta##, so ##\lvert e^{izx} \rvert = \lvert e^{ixR\cos\theta}e^{-xR\sin\theta} \rvert = e^{-xR\sin\theta}##. Hence, an upper bound on the integrand is ##M = \frac{e^{-xR\sin\theta}}{R^2-a^2}##. With ##L=\pi R##, you get
    $$\left\lvert \int_{\gamma_2} \frac{e^{izx}}{z^2+a^2} \right\rvert \le \frac{\pi R e^{-xR\sin\theta}}{R^2-a^2}$$ where ##0<\theta<\pi##.

    Consider the exponential. What combinations of ##\theta## and ##x## cause it to go to 0 in the limit ##R \to \infty##?
     
  4. May 15, 2016 #3
    The only thing I can think of at the moment is if θ is -θ then the sin(-θ)= -sin(θ) so we'll have the extra minus sign and for x<0 this works? but I'm not sure how to word it and/or if that's right?
    Thank you for taking the time to help by the way, greatly appreciated :)!
     
  5. May 15, 2016 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I think you have it. If ##x>0##, you need ##\sin\theta>0## to make the sign of the exponent negative. That's why you close the contour in the upper half plane. If ##x<0##…
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Find, using complex contour integrals, the function f(x)...
  1. Finding y=f(x) (Replies: 4)

Loading...