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

    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 $\displaystyle \phi$ be a differentiable 1-1 mapping of $\displaystyle [c,d]$ onto $\displaystyle [a,b]$ such that $\displaystyle \phi(c)=b$ and $\displaystyle \phi(d)=a$ and $\displaystyle \gamma^-(t)=(\gamma\circ\phi)(t)$.

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

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