Derivation of Euler-Lagrange equation from principle of least action

123hoedjevan gives you a wrong answer. The principle of least action states that the physical configuration of the system of fields realizes a minimum of the action with respect to compactly supported variations of the fields which, by the very definition of compactly supportedness, must then vanish on the boundary of the support itself. This in turn means that, in the formula

$$ \delta S=\int_{\Omega}d^{d}x\ \bigg[\bigg(\frac{\partial\mathcal{L}}{\partial x^{\mu}}-\partial_{\mu}\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\bigg)\,\delta\phi\bigg]+\int_{\Omega}d^{d}x\ \partial_{\mu}\bigg(\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\,\delta\phi\bigg) $$ ($d=\dim \Omega$), $\Omega$ must be compact. Now, as for the last term, we get $$ \int_{\Omega}d^{d}x\ \partial_{\mu}\bigg(\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\,\delta\phi\bigg)=\int_{\partial\Omega}d^{d-1}x_{\mu}\ \bigg(\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\,\delta\phi\bigg) $$ where I denoted by $d^{d-1}x_{\mu}$ the (oriented) volume element of the boundary $\partial\Omega$. The identity follows from Stoke's theorem, which (in one of its many forms) states that if you have a function $f$ defined on a compact set $\Omega$, then $$ \int_{\Omega} d^{d}x\ \big(\partial_{\mu}\,f\big)=\int_{\partial\Omega} d^{d-1}x\ \big(f\ n_{\mu}\big) $$ where $\partial\Omega$ is the boundary of $\Omega$ and $n_{\mu}$ are the components of the vector field normal to $\partial \Omega$ (notice that $n_{\mu}\,d^{d-1}x$ and my definition of $d^{d-1}x_{\mu}$ are the same thing). The proof of the theorem can be easily found on standard textbooks or on the internet. Going back to our integral, as $\delta\phi$ is by definition (i.e. as part of the hypotheses of the theorem) zero on the boundary of $\Omega$, $$ \int_{\Omega}d^{d}x\ \partial_{\mu}\bigg(\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\,\delta\phi\bigg)=\int_{\partial\Omega}d^{d-1}x_{\mu}\ \bigg(\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\,\delta\phi\bigg)=\int_{\partial\Omega}d^{d-1}x_{\mu}\ 0=0 $$ hence, as $\delta S=0$, $$ \delta S=\int_{\Omega}d^{d}x\ \bigg[\bigg(\frac{\partial\mathcal{L}}{\partial x^{\mu}}-\partial_{\mu}\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}\bigg)\,\delta\phi\bigg]=0 $$ and since this must hold for any compact $\Omega$ and any compactly supported $\delta\phi$, $$ \frac{\partial\mathcal{L}}{\partial x^{\mu}}-\partial_{\mu}\frac{\partial \mathcal{L}}{\partial\, \partial_{\mu}\phi}=0 $$ The compactness of $\Omega$ (and in turn the compactly supportedness of the variation of the fields) can indeed be removed from the hypotheses of the theorem, as long as the action integral is well-defined on $\Omega$. On the other hand, the fact that the variation must vanish on the boundary of the domain of integration cannot be removed from the hypotheses. Hence the former is still valid for non-compact $\Omega$'s.

When proving Noether's theorem (which is different from proving the equivalence between minimization and the Euler-Lagrange equations, and ultimately depends on this very proof), one allows for variations that do not vanish on the boundary of the domain of integration; moreover, in its usual formulation, Noether's theorem allows for the coordinates of the domain of integration to be varied. It is in this context that the Noether current arises as a divergenceless field, and the Noether current is defined as $$ j^{\mu}=-T^{\mu}_{\nu}\ \delta x^{\nu}+\frac{\partial\mathcal{L}}{\partial\,\partial_{\mu}\phi}\ \delta\phi $$ where $$ T^{\mu}_{\nu}=\frac{\partial\mathcal{L}}{\partial\,\partial_{\mu}\phi}\ \partial_{\nu}\phi-\mathcal{L}\ \delta^{\mu}_{\nu} $$ is the canonical energy-momentum tensor.