I got this problem in page 321 from Advaced Calculus, written by Friedman:
"Let $\sigma(x,y,z)$ be a continuous function, and let $S$ be a continuously differentiable surface. The Newtonian potential at a point $(x,y,z)$ outside $S$ due to a mass distribution $\sigma$ on $S$ is the integral
$$U(x,y,z) = \int \int_S \frac{\sigma(\xi,\eta,\zeta)}{[(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2]^{\frac{1}{2}}}dS$$
Prove the U is harmonic, that is, $U_{xx}+U_{yy}+U_{zz}=0$"
I want to know if I can use Leibniz Integral Rule in this case... I know that writing $\dfrac{\sigma(\xi,\eta,\zeta)}{[(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2]^{\frac{1}{2}}} = f(x,y,z)$, we have that $\frac{d^2f}{dx^2}+\frac{d^2f}{dy^2}+\frac{d^2f}{dz^2}=0$. So if
$$U_{xx}=\int \int_S \frac{d^2}{dx^2}\frac{\sigma(\xi,\eta,\zeta)}{[(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2]^{\frac{1}{2}}}dS$$ $$U_{yy}=\int \int_S \frac{d^2}{dy^2}\frac{\sigma(\xi,\eta,\zeta)}{[(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2]^{\frac{1}{2}}}dS$$ $$U_{zz}=\int \int_S \frac{d^2}{dz^2}\frac{\sigma(\xi,\eta,\zeta)}{[(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2]^{\frac{1}{2}}}dS$$ then, $$U_{xx}+U_{yy}+U_{zz} =$$ $$\int \int_S \left(\frac{d^2}{dx^2}\frac{\sigma(\xi,\eta,\zeta)}{[(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2]^{\frac{1}{2}}} + \frac{d^2} {dy^2}\frac{\sigma(\xi,\eta,\zeta)}{[(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2]^{\frac{1}{2}}} + \frac{d^2}{dz^2}\frac{\sigma(\xi,\eta,\zeta)}{[(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2]^{\frac{1}{2}}}\right)dS$$ $$U_{xx}+U_{yy}+U_{zz}=\int \int_S 0 dS =0$$
But I'm not sure if I can use Leibniz Rule here, because I don't have the integration limits, and beyond that if $S=r(u,v)=(x(u,v), y(u,v), z(u,v))$ does the second derivatives of $f(r(u,v))$ still be 0 when added to each other? So I was hoping someone could clarify this for me, and if the rule can't be used here, does someone can give a hint or a solution, please?