Given $f\in\mathcal{C}_{c}^2(\Bbb R^n)$ (i.e. twice real-differentiable with compact support) I'm in trouble with the following passage $$ \lim_{\delta\to0^+}\int_{|y|>\delta}|y|^{2-n}\frac{\partial}{\partial y_j}\left(\frac{\partial}{\partial y_k}f(x-y)\right)\,dy =\lim_{\delta\to0^+}\int_{S_{j\delta}^+-S_{j\delta}^-}|y|^{2-n}\frac{\partial}{\partial y_k}f(x-y)\,dy'\\ +(2-n)\int_{|y|>\delta}\frac{y_j}{|y|^{n}}\frac{\partial}{\partial y_k}f(x-y)\,dy $$ where $y=(y_j,y')$, i.e. $y'\in\Bbb R^{n-1}$ denotes the remaining coordinates of $y$, the ones different from $y_j$ and $$ S_{j\delta}^+:=\{y\in\Bbb R^n\;:\;|y|=\delta,\;y_j>0\}\\ S_{j\delta}^-:=\{y\in\Bbb R^n\;:\;|y|=\delta,\;y_j<0\}\;\;. $$
Now, this is a passage of the proof that $v(x)=-c_n|\cdot|^{2-n}\ast f(x)$ is a solution of the Poisson equation $\Delta v=f$ when $n\ge3$. Observe that $f$ is initially a function of the variable $x$; then the derivatives you see, were previously outside the integral and they were taken wrt $x$; then, once taken them inside, we must notice that $$ \frac{\partial}{\partial x_j}f(x-y)=-\frac{\partial}{\partial y_j}f(x-y) $$
I won't write down all details, otherwise the post would become too long. However, what I tried is to write
$$ \lim_{\delta\to0^+}\int_{|y|>\delta}|y|^{2-n}\frac{\partial}{\partial y_j}\left(\frac{\partial}{\partial y_k}f(x-y)\right)\,dy =\lim_{\delta\to0^+}\int_{\Bbb R^n}|y|^{2-n}\frac{\partial}{\partial y_j}\left(\frac{\partial}{\partial y_k}f(x-y)\right)\chi_{E_{\delta}}\,dy $$ where \begin{align*} E_{\delta} :&=\{y=(y_j,y')\in\Bbb R^n\;:\;y_j^2+|y'|^2>\delta^2\}\\ &=\{|y_j|>\sqrt{\delta^2-|y'|^2},\;|y'|\le\delta\}\sqcup\{|y'|>\delta\} \end{align*} from which the last integral can be expressed as $$ \lim_{\delta\to0^+}\int_{|y'|\le\delta}\left[\int_{|y_j|>\sqrt{\delta^2-|y'|^2}}|y|^{2-n}\frac{\partial}{\partial y_j}\left(\frac{\partial}{\partial y_k}f(x-y)\right)\,dy_j\right]\,dy'+\\ +\int_{|y'|>\delta}\left[\int_{\Bbb R}|y|^{2-n}\frac{\partial}{\partial y_j}\left(\frac{\partial}{\partial y_k}f(x-y)\right)\,dy_j\right]\,dy' $$
now integrating by parts the (unique) integral over $\Bbb R$, remembering $f$ is a compact support and the detail of the signs exposed above we should achieve the second part of RHS of the initial equality, since it seems reasonable to think that $$ \lim_{\delta\to0^+}\int_{|y'|>\delta}\int_{\Bbb R}= \lim_{\delta\to0^+}\int_{|y|>\delta}\;\;. $$ But for the first part I'm really in trouble; there must a mistake somewhere, since the integral is done wrt $y'\in\Bbb R^{n-1}$ but the domain seems to be a subset of $\Bbb R^n$. I tried to integrate by parts even this integral, but even adjusting the domain I can't obtain the same integrand function.
I worked on it for a couple of days. Can someone help me please?
Many thanks
The following is a "high level" proof of integration by parts, using differential forms. The version you want is in some sense "dual" to this one (as it uses vector fields).
Let $M$ be a smooth manifold with boundary $\partial M$, $\alpha\in\Omega^k(M)$, and $\beta\in\Omega^{n-k-1}(M)$, where $0\le k< n$. Then by Stokes' theorem we have $$\int_{\partial M}\alpha\wedge\beta = \int_Md(\alpha\wedge\beta) = \int_Md\alpha\wedge\beta + (-1)^k\int_M\alpha\wedge d\beta,$$ therefore, we have the formula $$\int_Md\alpha\wedge\beta = \int_{\partial M}\alpha\wedge\beta - (-1)^k\int_M\alpha\wedge d\beta.$$ Notice that the boundary term is given by integration on the boundary.
In your case, you have $M=\{y\in\mathbb{R}^n\mid\|y\|\ge\delta\}$, $\partial M$ is given by the sphere of radius $\delta$. You only need to identify the two differential forms and unwind the definitions to get to the formula you gave.