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Solve a very complex line integral

  1. May 1, 2007 #1
    1. The problem statement, all variables and given/known data

    Solve [tex]I = \int_{\gamma} f(z) dz[/tex] where [tex]\gamma(t) = e^{i \cdot t}[/tex] and [tex]0 \leq t \leq \pi[/tex]

    2. Relevant equations

    Do I use integration by substitution??

    3. The attempt at a solution

    If I treat this as a line-integral I get:

    [tex]I = \int_{a}^{b} f(\gamma(t)) \cdot \gamma'(t) dt = \int_{0}^{\pi} cos(e^{i \cdot t}) \cdot (i \cdot e^{i \cdot t}) dt[/tex]

    then If I choose

    [tex]u = e^{it}[/tex] and [tex]dt = i \cdot e^{i \cdot u} du[/tex]

    where [tex]t=0, u=e^{i \cdot 0} = 1[/tex] and [tex]t=\pi, u=e^{i \cdot 0} = -1[/tex]

    [tex]\int_{0}^{\pi} cos(e^{i \cdot t}) \cdot (i \cdot e^{i \cdot t}) dt = \int_{1}^{-1} (i \cdot (cos(u) \cdot u) du = (i \cdot (cos(u) + u \cdot sin(u))) \cdot i \cdot e^{i \cdot u} ]_{1}^{-1} = 2 sin(1) \cdot ((cos(1) + sin(1)) \cdot i[/tex]

    Am I on the right path here??

    Best regards
    Last edited: May 2, 2007
  2. jcsd
  3. May 2, 2007 #2


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  4. May 2, 2007 #3
    Hi there,

    You are right :)

    Then I get


    and I let [tex]u=e^{it}[/tex]




    so that part is already in the integral so now it's just:


    right? I then get [tex]-2Sin[1][/tex]

    I got that result in Maple too. Had an idear that it was the right one :)

    By following the method above. I get:

    1)[tex]f(z) = sinh(z)[/tex] and 2) [tex]f(z) = (exp(z))^3[/tex] and 3)[tex]f(z) = tan(z)[/tex].

    1) Solution

    [tex]\int_{0}^{\pi} sinh(e^{it}) \cdot i \cdot e^{i \cdot t} dt = \int_{1}^{-1} sinh(u) du = 0[/tex]

    2) Solution

    [tex]\int_{0}^{\pi} (e^{e^{it}})^3 \cdot i \cdot e^{i \cdot t} dt = \int_{1}^{-1} (e^{u})^3 du = \frac{-2 sinh(3)}{3}[/tex]

    3) Solution

    [tex]\int_{0}^{\pi} tan(e^{it}) \cdot i \cdot e^{i \cdot t} dt = \int_{1}^{-1} tan(u) du = 0[/tex]

    How do they look? Am I following you guidelines correctly??

    Best regards Fred
  5. May 2, 2007 #4


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    Gold Member

    This should be

    [tex]du = i \cdot e^{i \cdot t} dt[/tex]

    I think....

    notice your integral becomes easier
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