I'm studying magnetic field and the mechanic actions that a vectorial field $\mathbf{B}$ impose to a circuit.
Now, when it comes to the resulting force on a closed circuit (closed line) with a current $I$ in a uniform magnetic field $\mathbf{B}$, it uses the formula: $$ \mathbf{F} = I\oint_{\gamma}\operatorname{d}\mathbf{l}\,\times\mathbf{B} $$ where the $\operatorname{d}\mathbf{l}$ stands for the infinitesimal vectorial length of the circuit, so that: $$ \oint_{\gamma}\operatorname{d}l = L(\gamma) $$ where $L(\gamma)$ is the total length of the closed curve. Because of $\mathbf{B}$ uniformity, one can write: $$ \mathbf{F} = I\left[\oint_{\gamma}\operatorname{d}\mathbf{l}\right]\,\times\mathbf{B} $$ as $\mathbf{B}$ doesn't depend on the position, so can be taken out from the integral. Here comes the problematic step. It's said that $$ \oint_{\gamma}\operatorname{d}\mathbf{l}=0 $$ for every closed curve. Even if I can understand heuristically why that integral should be zero, I cannot figure out what mathematical process bring to this result. I mean, how to prove it?
If you accept getting $\mathbf B$ get out of the integral, the remaining integral is simply adding vectors all of them lying along a loop. Consider a parametrization of $\gamma$, $(x(t),y(t))$ for which $(x(0),y(0))=(x(T),y(T))$
$$\oint_{\gamma}\operatorname{d}\mathbf{l}=\oint_{\gamma}(dx,dy)=\left(\int_0^T\dot x(t)dt,\int_0^T\dot y(t)dt\right)=\left(\left[x(t)\right]_0^T,\right[y(t)\left]_0^T\right)=(0,0)$$