# Thread: Proof regarding a complex line integral equality

1. ## 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.

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