Why the representation formula (poisson kernel) is $C^{\infty}$? (differentiation under integration sign)

Question

$$u(x) = \int_{\partial B(0,r)}\frac{r^2-|x|^2}{n\alpha(n)r|x-y|^n}g(y) dS(y)$$

This is the representation formula for the $u(x)$, that solves $\Delta u = 0$ in the interior of the ball and $u=g$ in the boundary of it. My teacher said that it is $C^{\infty}$ by differentiation under the integration sign. By Can a limit of an integral be moved inside the integral? I see that we can take the limit to the inside of the integral when the sequence of functions inside is bounded and converges pontwise (dominated convergence theorem).

I'm having trouble identifying in which variables are the sequence and in which it's a function, and why it is bounded.

Answer 1

Notice that $$u(x+te_i )-u(x)=\int_{B(0,r)}(\frac{r^2 -\vert x+te_i \vert^2 }{n\alpha (n)r\vert x+te_i -y\vert^n }-\frac{r^2 -\vert x\vert^2}{n\alpha(n)r\vert x-y\vert^n})g(y)\mathrm{d} S(y)$$thus when we do the differentiation, if we let $t$ sufficiently small, we can state that the term in the parenthesis is bounded and exchange the limit and the integral.