# Thread: Proof regarding a complex line integral equality

1. ## 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.

2. 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 $\displaystyle \int_{\gamma^-}f(z)\mathrm dz=\int_b^af(\gamma(u))\gamma'(u)\mathrm du=-\int_a^bf(\gamma(u))\gamma'(u)\mathrm du=-\int_{\gamma}f(z)\mathrm dz$.