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Math Help - Proof regarding a complex line integral equality

  1. #1
    MHF Contributor arbolis's Avatar
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    Proof regarding a complex line integral equality

    Let f be a continuous function over a curve C parametrized by [a,b]\to C.
    If \gamma ^-(t):[c,d]\to C is a reparametrization of \gamma that inverts the orientation, show that \int _{\gamma ^-} f(z) dz = - \int _{\gamma} f(z)dz.

    My attempt : I still didn't find the main idea of the proof. Stuck at starting. I think I've seen a similar proof in calculus 3, but I don't remember it.
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  2. #2
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    Let \phi be a differentiable 1-1 mapping of [c,d] onto [a,b] such that \phi(c)=b and \phi(d)=a and \gamma^-(t)=(\gamma\circ\phi)(t).

    Then \int_{\gamma^-}f(z)\mathrm dz=\int_c^d f(\gamma^-(t)){\gamma^-}'(t)\mathrm dt=\int_c^d f((\gamma\circ\phi)(t))(\gamma'\circ\phi)(t)\phi'(  t)\mathrm dt by the chain rule.

    Now substitute u=\phi(t), giving \int_{\gamma^-}f(z)\mathrm dz=\int_b^af(\gamma(u))\gamma'(u)\mathrm du=-\int_a^bf(\gamma(u))\gamma'(u)\mathrm du=-\int_{\gamma}f(z)\mathrm dz.
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