Helmholtz's Theorem Derivation Clarification In the derivation for Helmholtz's decomposition on Wikipedia , the Laplacian is taken out from the integral because "it does not affect the variables of integration" as follows:
$\int_{V}F(r')(-\frac{1}{4 \pi})\nabla^2 \frac{1}{|r-r'|} dV' = \nabla^2\int_{V}F(r')(-\frac{1}{4 \pi}) \frac{1}{|r-r'|} dV' $
How does this apply? Doesn't $\nabla^2$ affect $\frac{1}{|r-r'|}$ within the integral?
 A: Well, This is actually confusing, But yes I think quiet a bit and I realized it's TRUE. Even This the case of 3d.
Basically, I wanna start up with the the following Theorem called Newton-Lebniz
$$\boxed{\text{Theorem:}\frac{\partial}{\partial x}\int_{b(x)}^{a(x)} f(x,t) dt=\int_{a(x)}^{b(x)}\frac{\partial}{\partial
x}f(x,t)+\frac{\partial b(x)}{\partial x}f(x,b(x))-\frac{\partial a(x)}{\partial x}f(x,a(x))}$$Now Whenever $a(x), b(x)$ are constant no matter hom many time you operate , $\frac{\partial}{\partial x}$ The extra term will be zero. So even we can Say that There exist an awesome generalization what I can think of is Whenever $a(x),b(x)$ are not constant you get an expression for .
$$\frac{\partial^n}{\partial x^n}\int_{b(x)}^{a(x)} f(x,t) dt$$ by just repeated applying the Theorem Depending on the How many times $f(x,t),a(x),b(x)$ are differentiable.
Hope it helps.
For better note:Go to Wikipedia to look over the proof by googling Newton-Lebniz
A: The Laplacian is taken with respect to $r$, which is a parameter in the integral. The region of integration is fixed, so you can take the derivatives with respect to the parameters out of the integral,
$$
\int_V\frac{\partial f}{\partial t}(x,t)\, dx=\frac{\partial}{\partial t}\int_Vf(x,t)\, dx
$$
in particular the Laplacian w.r.t. $r$.
