- Let \(\gamma\) be a smooth curve in \(\mathbb { C }\) parametrized by \(z ( t ) : [ a , b ] \to \mathbb { C }\) . Let \(\gamma ^ { - }\) denote the curve with the same image as \(\gamma\) but with the reverse orientation. Prove that for any continuous function \(f\) on γ
\[ \int_ {\gamma} f (z) d z = - \int_ {\gamma^ {-}} f (z) d z. \]
Solution
If \(\gamma\) is parametrized by \(z:[a,b]\to\mathbb C\), then the reverse curve is parametrized by \[ z^-(t)=z(a+b-t),\qquad a\le t\le b. \] Therefore \((z^-)'(t)=-z'(a+b-t)\), and \[ \begin{aligned} \int_{\gamma^-}f(z)\,dz &=\int_a^b f(z(a+b-t))\,[-z'(a+b-t)]\,dt\\ &=-\int_a^b f(z(a+b-t))z'(a+b-t)\,dt. \end{aligned} \] With the substitution \(u=a+b-t\), the last integral becomes \[ -\int_a^b f(z(u))z'(u)\,du=-\int_\gamma f(z)\,dz. \] Hence \[ \int_\gamma f(z)\,dz=-\int_{\gamma^-}f(z)\,dz. \]