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!

Proving that an integral is a pure imaginary

  1. Feb 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that [itex]\int_{\gamma}\ f^*(z)\ f'(z)\ dz[/itex] is a pure imaginary for any piecewise smooth closed curve [itex]\gamma[/itex] and any [itex]C^1[/itex] function [itex]f[/itex] whose domain contains an open set containing the image of [itex]\gamma[/itex]


    2. The attempt at a solution

    I have tried to approach it from some different angles. I've tried to Taylor expand everything, to see if some interesting things would happen, but I found none. I've also tried to expand it as such:

    [itex]\int_{\gamma}\ f^*(z)\ f'(z)\ dz = \int_{\gamma}\ [(u - iv)(\frac{\partial u}{\partial x} + i\frac{\partial v}{\partial x})](dx + idy) = [/itex]

    [itex]\int_{\gamma}\ (u\frac{\partial u}{\partial x}+v\frac{\partial v}{\partial x})dx+(u\frac{\partial u}{\partial y} + v\frac{\partial v}{\partial y})dy + i\int_{\gamma}\ (u\frac{\partial v}{\partial x}-v\frac{\partial u}{\partial x})dx+(u\frac{\partial v}{\partial y}-v\frac{\partial u}{\partial y})dy[/itex]

    What can I do with this? Is this the way to go?

    Cheers
     
  2. jcsd
  3. Feb 2, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It's a lot simpler than that. A number is pure imaginary if its complex conjugate is equal to its negative. Hint: use integration by parts. You are starting from the integral of f*df.
     
    Last edited: Feb 2, 2013
  4. Feb 2, 2013 #3
    I thought of that, but integration by parts produces [itex]-\int_\gamma\ f\cdot(f^*)'dz[/itex], but I have no guarantee that [itex]f^*[/itex] is differentiable... And even if it were, I'd still have to show that [itex]-\int_\gamma\ f\cdot(f^*)'dz = -\int_\gamma\ f\cdot(f')^*dz[/itex], which doesn't seem trivial...
     
  5. Feb 2, 2013 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Your are making this seem a LOT harder than it is. If f is differentiable then f* is differentiable. That's almost trivial. As is (f*)'=(f')*. We are just talking C^1 differentiabilty here, the existence of continuous partial derviatives, not complex differentiability. If your curve is z=γ(t) then you can express all of the derivatives as d/dt.
     
    Last edited: Feb 2, 2013
  6. Feb 3, 2013 #5
    Oh, that makes a lot of sense! So one gets:

    [itex]\int_{\gamma}\ f^*(z)\ f'(z)\ dz = \int_a^b\ f(\gamma(t))^*\cdot f'(\gamma(t))\cdot \gamma'(t)\ dt = \int_a^b\ ((f\circ\gamma)(t))^*\cdot \frac{d}{dt}(f\circ\gamma)(t)\ dt[/itex]
    [itex] = [\ |\ f(\gamma(t))\ |^2\ ]_a^b - \int_a^b\ \frac{d}{dt}((f\circ\gamma)^*)(t) \cdot (f\circ\gamma)(t)\ dt = -\int_a^b\ (f'(\gamma(t)))^* \cdot \gamma'(t) \cdot f(\gamma(t))\ dt[/itex]
    [itex] = -\int_{\gamma}\ f(z)\ (f'(z))^*\ dz[/itex]
     
    Last edited: Feb 3, 2013
  7. Feb 3, 2013 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Right! It's actually (ff*)=|f|^2 in the difference term for the integration by parts, but that doesn't make any difference.
     
  8. Feb 3, 2013 #7
    You're totally right, my mistake. I've already corrected it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proving that an integral is a pure imaginary
  1. Prove the integral (Replies: 9)

  2. Prove the integral (Replies: 0)

Loading...