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!

Homework Help: Complex Integration - Poles on the Imaginary axis

  1. Apr 27, 2012 #1
    1. The problem statement, all variables and given/known data

    evaluate the integral:
    [itex]I_1 =\int_0^\infty \frac{dx}{x^2 + 1} [/itex]

    by integrating around a semicircle in the upper half of the complex plane.

    2. Relevant equations

    3. The attempt at a solution

    first i exchange the real vaiable x with a complex variable z & factorize the denominator. Also, the contour of integration is a semicircle with radius= infinity

    [itex]I_2 = \int_{-\infty}^{\infty}\frac{dz}{(z+i)(z-i)}[/itex]

    the contour contains only the pole in the upper half, so from residue theorem we know:
    [itex]I_2 = 2\pi i R(i)[/itex]

    where R(i) is the residue at the point z=i

    R(i) = 1/(i+i) = 1/2i

    Hence [itex]I_2 = \pi[/itex]

    Now, i know the answer to the original integral is supposed to be pi/2.
    Can i say that: because the original limits range from 0 to infinity, and i have integrated twice this amount, my answer should be divided by 2? or is this reasoning flawed?
  2. jcsd
  3. Apr 27, 2012 #2
    I THINK you do divide by two, as noted, but how I'd approach this would be by using trig substitution, substituting x = tan(θ). :P But yep, I think your reasoning's right, or at least close!
  4. Apr 27, 2012 #3
  5. Apr 28, 2012 #4
    Thanks for the replies. Is this because the function is even in the upper half of the complex plane?

    I thought of doing this by integrating a contour in only the positive quadrent, ie:
    (0,0) to (R,0)
    (R,0) to (0,iR) along contour ω [a radial path of radius R from the real axis to the imaginary axis in the positive quadrent]
    (0,iR) to (0,i+δ) [where δ is a small value which we will take to 0 around the pole]
    (0,i+δ) to (0,i-δ) along a contour λ [radial path of radius δ in the clockwise direction such that the pole is not included in the enclosed path]
    (0,i-δ) to (0,0)

    we have:

    [itex]F(z) = \frac{1}{Z^2+1}=\frac{1}{(z+i)(z-i)}[/itex]

    [itex]\oint_C F(z)dz =\int_0^R F(z)dz + \int_\omega F(z)dz + \left[\int_{iR}^{i+\delta}F(z)dz+\int_{i-\delta}^0 F(z)dz\right] + \int_\lambda F(z)dz =0[/itex]

    as F(z) is analytic everywhere inside the contour we demand that the integral = 0 by Green's theorem.

    we know from residue theorem that integration on the contour λ will give a contribution of -πiR(i) as it is counterclockwise (& where R(i) is the residue at i, which is 1/2i)

    now let R→∞ and δ→0:

    First: consider the integral along the contour ω, if we have [itex]z=Re^{i\theta}[/itex]

    [itex]\int_\omega F(z)dz = \int_\omega \frac{iRe^{i\theta}d\theta}{R^2e^{i2\theta} + 1}[/itex]

    [itex]\lim_{R \to \infty}\int_\omega \frac{iRe^{i\theta}d\theta}{R^2e^{i2\theta}+1}≈ \frac{1}{R}\int_\omega \frac{id\theta}{e^{i\theta}}=0[/itex]

    Is this correct? I thought Jordan's Lemma worked only for semicircles but this seems to give the same result.

    we are left with:

    [itex]\int_0^\infty F(z)dz + \int_{i\infty}^0 F(z)dz - \frac{\pi}{2} = 0[/itex]

    so i know that the integral along the imaginary axis must be 0 but i'm not sure how to prove it
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook