I am very new to this forum and I would like to ask you, if possible, one question I am stuck on for several days.
$\textbf{Problem statement}$
Let $S$ be an arbitrary open surface defined in $\mathbb{R}^3$, let ${\bf x}$ be the position vector and $m({\bf x})$ a well-behaved distribution of some quantity on $S$.
Let also
$$D({\bf x}) = \int_{\overline{S}}{m({\bf y})\,{\bf n}({\bf y})\cdot {\bf Grad}_{\bf y}\left(\frac{1}{|{\bf x}-{\bf y}|}\right)\,d\sigma({\bf y})} \qquad {\bf x}\neq{\bf y}$$
be the a double layer potential, solution of the $\textit{Laplace}$'s equation, and let us calculate its gradient: ${\bf grad}\,D({\bf x})$
Applying $\textit{Leibniz}$'s rule, and noting that $D({\bf x})$ can be rewritten as an integral of a divergence with respect to ${\bf x}$ coordinates and considering the vector identity ${\bf grad}(\textrm{div}\,{\bf \Phi}) = -{\bf curl}({\bf curl}\,{\bf \Phi})-\triangle\,{\bf\Phi}$, for any vector ${\bf \Phi}$ allow to obtain:
$${\bf grad}\,D({\bf x})=\int_{\overline{S}}{m({\bf y})\,{\bf curl}_{\bf x}{\bf curl}_{\bf x}\left(\frac{{\bf n}({\bf y})}{|{\bf x}-{\bf y}|}\right)\,d\sigma({\bf y})} \qquad {\bf x}\neq{\bf y}$$
Taking advantage of the property of the functions $f({\bf x}-{\bf y})$ that $${\bf grad}_{\bf x} f({\bf x}-{\bf y})= - {\bf grad}_{\bf y} f({\bf x}-{\bf y})$$ the following integral arises:
$${\bf grad}\,D({\bf x}) = -\int_{\overline{S}}{m({\bf y})\,{\bf curl}_{\bf x}\left[{\bf n}({\bf y})\times {\bf Grad}_{\bf y}\left(\frac{1}{|{\bf x}-{\bf y}|}\right)\right]\,d\sigma({\bf y})} \qquad {\bf x}\neq{\bf y} $$
where the operator ${\bf Grad}$ acts only in the surface.
But for reasons that will be clear afterwards, from the latter integral the following is obtained (with another one, unimportant for my question but of great interest for my proof)
$$I_1({\bf x}) = \int_{\overline{S}}{{\bf curl}_{\bf x}\left[{\bf n}({\bf y})\times {\bf Grad}_{\bf y}\left(\frac{m({\bf y})}{|{\bf x}-{\bf y}|}\right)\right]\,d\sigma({\bf y})} \qquad {\bf x}\neq{\bf y} \tag 1$$
My $\textbf{question}$ is the following: Is it completely legal to exchange the ${\bf curl}_{\bf x}$ and the integral in order to apply the $\textit{Stokes}$' theorem in the form presented
$$\int_{\overline{S}}{{\bf n}({\bf y})\times {\bf Grad}_{\bf y}\phi({\bf y})\,d\sigma({\bf y})}=\oint_{\partial \overline{S}}{\phi({\bf y})\,d{\bf s}({\bf y}})$$
Where $d{\bf s}$ is the arc of length vector, which is tangential to the boundary $\partial \overline{S}$, and has the usual direction with respect to the outer surface normal.
giving just $$I_1({\bf x})={\bf curl}_{\bf x}\int_{\partial \overline{S}}\frac{m({\bf y})}{|{\bf x}-{\bf y}|}\,d{\bf s}({\bf y})=\int_{\partial \overline{S}}m({\bf y})\,{\bf grad}_{\bf y}\left(\frac{1}{|{\bf x}-{\bf y}|}\right)\times d{\bf s}({\bf y}) \qquad {\bf x}\neq{\bf y} \tag2$$
$\textbf{Further comments}$
The version of the $\textit{Stokes}$' theorem is derived with ease from the original one supposing that the fector field to which the curl is applied has the form ${\bf c}\phi$, where ${\bf c}$ is an arbitrary constant vector.
I would like to know also, if the procedure I have followed is correct.
Thank you so much for your help, it would be greatly appreciated.