Vector field is path-independent if and only if the integral of the vector field over every closed loop in its domain is zero

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While trying to understand the proof of the converse of the gradient theorem I have a hard time to understand why a vector field is path-independent if and only if the integral of the vector field over every closed loop in its domain is zero. Wikipedia says that should be "straight forward to show" but I don't get it.


Let's say $F$ is a path-independent vector field. How can we do anything with $\int_\gamma \langle F,dx \rangle$ since we don't know if it has a potential at all? It would be easy if there would be a potential $\phi$ with property $\int_\gamma \langle F,dx \rangle = \phi(\gamma(a)) - \phi(\gamma(b))$. In this case we could easily conclude that every closed loop must be zero. But since $F$ is only a path-indpendent (the integral of $F$ over some piecewise-differentiable curve is dependent only on end points) vector field and not a gradient field this argument is invalid.

I was able to prove the converse statement though.

Where is the "straight forward"-part I'm missing?