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!

Contour Integration

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data


    ∫[itex]_{\gamma}[/itex](x-y)dz where [itex]\gamma[/itex] has parametrization: z(t) = e[itex]^{it}[/itex] for [itex]\pi[/itex]/2 [itex]\leq[/itex] t [itex]\leq[/itex] 3[itex]\pi[/itex]/2
    2. Relevant equations
    the integral of the sum is the sum of the integral


    3. The attempt at a solution
    I tried to break it up and see if I could evaluate it as I normally would but it started to get really messy and I think I was going about it wrong.

    z=exp(it) and dz=iexp(it)
    ∫(x)dz - i∫(y)dz

    can someone please push me into the right direction? thank you
     
  2. jcsd
  3. Mar 20, 2012 #2

    lanedance

    User Avatar
    Homework Helper

    as you mean to evaluate directly you need to use a substitute for the parameterisation variable

    [tex]
    \cint_{\gamma} f(z) dz = \int_{t_a}^T_b f(z(t)) z'(t) dt[/tex]
     
  4. Mar 20, 2012 #3
    could you elaborate just a little more? I really appreciate your help
     
  5. Mar 20, 2012 #4

    lanedance

    User Avatar
    Homework Helper

    what don't you understand? I won't do the problem for you, but am happy to help
     
  6. Mar 20, 2012 #5
    I dont understand how to set up a substitution so I can get it into a form that I know how to work with
     
  7. Mar 20, 2012 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You intend that z= x+ iy, right? So if [itex]z= e^{it}= cos(t)+ i sin(t)[/itex] what are x and y in terms of t?
     
  8. Mar 20, 2012 #7
    Okay this is what I have so far:

    ∫[itex]_{\gamma}[/itex](x-y)dz where [itex]\gamma[/itex] has parametrization: z(t)=e[itex]^{it}[/itex] for [itex]\pi[/itex]/2[itex]\leq[/itex]t[itex]\leq[/itex]3[itex]\pi[/itex]/2


    ∫[itex]_{\gamma}[/itex](x-y)dz = ∫[itex]_{\gamma}[/itex](cos(t)-isin(t))dz

    which we can break up like,

    ∫[itex]_{\gamma}[/itex](cos(t))dz - i∫[itex]_{\gamma}[/itex](sin(t))dz

    since z(t)=e[itex]^{it}[/itex] then dz=ie[itex]^{it}[/itex]

    so we have

    ∫[itex]_{\gamma}[/itex](cos(t))(ie[itex]^{it}[/itex]) - i∫[itex]_{\gamma}[/itex](sin(t))(ie[itex]^{it}[/itex] )

    is this correct so far?

    or would my dz be -sin(t)+icos(t)?
     
    Last edited: Mar 20, 2012
  9. Mar 20, 2012 #8

    lanedance

    User Avatar
    Homework Helper

    you may want to check your "y" value, remember z= x+ iy,
     
  10. Mar 20, 2012 #9
    so instead of isin(t) just simply sin(t)

    ∫[itex]_{\gamma}[/itex](x-y)dz = ∫[itex]_{\gamma}[/itex](cos(t)-sin(t))dz

    ∫cos(t)dz-∫sin(t)dz

    every time I try evaluating the above expression I never can seem to get the correct answer. I'm using ie^(it) for my dz

    AHHH im getting so frustrated. I feel like I am missing an important concept, this problem should not take me this much time
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Contour Integration
  1. Contour integral (Replies: 3)

  2. Contour integrals (Replies: 22)

  3. Contour integral (Replies: 1)

  4. Contour integration (Replies: 4)

  5. Contour integrals (Replies: 2)

Loading...