Show that if \(f\) has a primitive \(F\) on \(\Omega\) then \(\int_\gamma f = 0\) for every closed curve \(\gamma \subseteq \Omega\).
Solution
Goal: Show that if \(f\) has a primitive \(F\) on \(\Omega\) (i.e. \(F' = f\) on \(\Omega\)), then \(\int_\gamma f = 0\) for every closed curve \(\gamma \subseteq \Omega\).
1. If \(\gamma\) is a smooth curve from \(a\) to \(b\) parametrized by \(z(t)\), \(t \in [0,1]\), then \(\int_\gamma f = F(b) - F(a)\).
By the chain rule, \(\ddd{t}F(z(t)) = F'(z(t)) z'(t) = f(z(t)) z'(t)\), so \(\int_\gamma f = \int_0^1 f(z(t))z'(t)\,dt = \int_0^1 \ddd{t}F(z(t))\,dt = F(z(1)) - F(z(0)) = F(b) - F(a)\).
2. For a closed curve \(\gamma\), the endpoints coincide, so \(\int_\gamma f = F(a) - F(a) = 0\).
1 with \(b = a\): a closed curve has \(z(1) = z(0)\).
3. The claim extends to piecewise smooth closed curves by summing over the smooth pieces.
Subdivide \(\gamma\) into smooth arcs; the integral is the sum of the arc integrals, and the endpoint terms telescope to \(0\) around the closed loop.
4. Q.E.D.
<1>1–<1>3 prove the claim for (piecewise) smooth closed curves, which is the standard meaning of \(\int_\gamma f\) in this context.