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Log(z) gives an anti-derivative of 1/z

  1. May 27, 2010 #1
    How do I calculate
    [tex]\int_C[/tex] dz/1-z when C is a line segment from 1/2 to [tex]\sqrt{3i}[/tex].
    I know that in the complement of the negative real axis, log(z) gives an anti-derivative of 1/z.
     
  2. jcsd
  3. May 27, 2010 #2
    Re: Integral

    you need to use calculus of residues
     
  4. May 28, 2010 #3
    Re: Integral

    I don't see a convinient way to use residues since this is not an integral of a close curve.

    What you need to do, is to do a parametrization z(t) of the line segement.
    Then you substitute z with a corresponding expression of t, and dz with a corresponding expression of dt, and then the integral comes down to integration over a real variable.
     
  5. May 28, 2010 #4

    HallsofIvy

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    Re: Integral

    I hope that last was a typo. The derivative of log(z) is 1/z. The anti-derivative is zlog(z)- z. Also, was that supposed to be [tex]i\sqrt{3}[/itex] rather than [tex]\sqrt{3i}[/tex]? That is, is the i supposed to be inside the square root or not?

    Looks like tedious but straightforward computation from me. Assuming you mean [tex]i\sqrt{3}[/itex], C can be written [itex]z= (1/2)(1+ t)+i\sqrt{3} t[/itex] so that [itex]dx= (-1/2+ i\sqrt{3})dt[/itex] and [itex]1- z= (1/2)(1+ t)- i\sqrt{3}t[/itex] with t going from 0 to 1.

    Put those into the integral and you should be able to reduce it to two integrals, one of the form [itex]1/z^2[/itex] and the other [itex]1/(z^2+ 1)[/itex].

    If you really meant [tex]\sqrt{3i}[/itex], then
    [tex]\sqrt{3i}= \sqrt{\frac{3}{2}}+ i\sqrt{\frac{3}{2}}[/itex]
     
  6. May 28, 2010 #5
    Re: Integral

    It was a typo. Oopsy daisys.
    dz not dx
     
  7. May 28, 2010 #6
    Re: Integral

    Can we not use the Cauchy integral formula because it is not a closed curve.
     
  8. May 28, 2010 #7
    Re: Integral

    I didn't do it that way but i got -0.944 + 0.117i
     
  9. May 28, 2010 #8
    Re: Integral

    You can only use Cauchy's Integral Formula if it's a closed contour. Of course you could add two more legs and make it closed but that's not necessary. It helps if you write it clearly as an integral:

    [itex]\mathop\int\limits_{1/2}^{\sqrt{\frac{3}{2}}+i\sqrt{\frac{3}{2}}} \frac{1}{z}dz[/itex]

    Keep in mind it's logarithmic antiderivative can be made analytic along a straight-line contour between the end points so you can just evaluate the antiderivative at those endpoints as long as you use an analytic extension of log in a region containing the contour, say the branch that has a cut along the negative real axis. Can you then use that branch and compute:

    [itex]\log(z)\biggr|_{1/2}^{\sqrt{\frac{3}{2}}+i\sqrt{\frac{3}{2}}[/itex]
     
    Last edited: May 28, 2010
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