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Homework Help: Contour integral

  1. Jun 15, 2012 #1
    You can do integrals of real functions like:

    [itex]\oint[/itex]1/(3-sinθ) by transforming to a complex contour, which enloses the origin, and then using the residue theorem. Normally you would transform to the unit circle, but in principal you could use any contour (right?). Now, sometimes you find that some rediues are inside the unit circle and some are not. If you picked a difference contour this could be changed. However, the integral above must give the same for every contour, so what is it that still makes the complex integrals give the same (even though different contours are involved.)
  2. jcsd
  3. Jun 15, 2012 #2


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    No, that's not true, for exactly the reason you said. The integral depends on which poles of the function are inside the contour. What is true is that you can find a specific definite integral by finding a contour that includes the interval of integration. The integral over that interval will be constant while the integral over the entire contour and the integral over the rest of the contour vary- there difference being the constant.
  4. Jun 15, 2012 #3
    hmm I might need you to repeat that for me. Say we have some kind of sinusoidal function that is to be integrated from [0;2[itex]\pi[/itex]]. There are infinitely many contours in the complex plane that encloses the origin and they may well have different poles inside them. How do you then know which of those will give the correct value for the real valued integral of the sinusoidal function? I always did the unit circle (easiest!) but guessing the others will be just as good, it is weird that they would yield different results for the integral.
    Or is that exactly what compensates for the fact that when you substitute [itex]\theta[/itex] = [itex]\theta[/itex](z) you will get different expressions for sin([itex]\theta[/itex]) as a function of z? For instance sin[itex]\theta[/itex] = 1/(2i)(z/r + r/z) for the general substitution z=rexp(i[itex]\theta[/itex])
  5. Jun 15, 2012 #4
    You havin' trouble understanding that. The basis of the principle is the Residue Theorem: the integral of a meromorphic function (an analytic function with poles) over a closed contour is equal to 2pi i times the sum of the enclosed residues. Now, that's what's not going to change no matter what the contour is, big, small, circle, square, triangle. So you're referring to a real integral like:

    [tex]\int_0^{2\pi} \frac{1}{3-\sin(t)}dt[/tex]

    Now, the easiest thing to do is to let [itex]z=e^{it}[/itex] but you could let [itex]z=5e^{it}[/itex] or any size circle but those others are just more difficult to work with. Nevertheless, they would still give you the correct answer as long as your complex algebra is correct and you correctly identify the enclosed poles for the residue calculation. For example, if I let [itex]z=5e^{it}[/itex], I get:

    [tex]\int_0^{2\pi} \frac{dt}{3-\sin(t)}=\mathop\oint\limits_{|z|=5}\left(-i\frac{dz}{z}\right) \frac{10iz}{30 iz-z^2+1}[/tex]

    but I did that fast so you'd have to verify it. It's more messy though than just letting [itex]z=e^{it}[/itex].
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