# Path Integral over circle always $0$

Let's say we want to evaluate the integral $$\int_{\lvert x \rvert = R} f(x) dx$$ where $R \gt 0$ is the radius of a circle. Now we parameterize $\varphi(t) : [0, 2\pi] \rightarrow D$ :

$$\int_{\varphi(0)} ^{\varphi(2\pi)} f(\varphi(t)) \varphi'(t) \, dt \leq \, \, \sup(f(\varphi(t))) \cdot \int_{\varphi(0)} ^{\varphi(2\pi)} \varphi'(t) \,dt \, \,= \sup(f(\varphi(t))) (\varphi(2\pi)-\varphi(0)) = 0$$
since $\varphi(2\pi) = \varphi(0)$

Obviously there must be a mistake, but I can't figure out what is wrong. I suspect that I somehow misuse the supremum.

• math.stackexchange.com/questions/1489577/… – R.N Nov 1 '15 at 19:21
• @RaziehNoori I can't see how this is related to my question – Christian Nov 1 '15 at 19:33
• $\varphi$ have to be one to one. – R.N Nov 1 '15 at 20:16

I take it you are starting with an appropriately well behaved (not necessarily continuous) function $f : \mathbb R^2 \to \mathbb R$ where the $x$ in $f(x)$ is a vector in $\mathbb R^2$. In which case the integral over a path $P$ of

$$\int_P f(x) \ dx$$ is a vector quantity. Then for a path parameterization $\phi$ over $[0, 2\pi]$,

$$\left\| \int_{|x| = R} f(x) \ dx \right\| \leq \int_0^{2\pi} ||f(\phi(t)) \ \phi'(t)|| \ dt \leq \sup_{t\in [0,2\pi]} |f(\phi(t))| \int_0^{2\pi} \| \phi'(t)\| \ dt$$

That last integral is not equal to zero.

If instead you meant a scalar path integral

$$\int_P f(x) \ ds$$

then an almost identical analysis also results in the integral $\int_0^{2\pi} \| \phi'(t)\| \ dt$ which is again not zero.