I am learning from Jackson (3r edition), where I found one concept very confusing, that is Taylor expansion of charge density. (This is given in section "1.7 Poisson and Laplace equations" p.n:35)
I will write some equations first.
$$ {\Phi}_a(x) = \frac{1}{4{\pi}{\epsilon_0}}\int \frac{{\rho}(x')}{\sqrt{(x - x')^2 + a^2}}d^3x' $$
Now we want to find out potential such that a tends to zero. $$ \nabla^2{\Phi}_a(x) = \frac{1}{4{\pi}{\epsilon_0}}\int{\rho}(x') \nabla^2\frac{1}{(r^2 + a^2)^\frac{1}{2}}d^3x' $$
$$ \nabla^2{\Phi}_a(x) = -\frac{1}{4{\pi}{\epsilon_0}}\int{\rho}(x') \frac{3a^2}{(r^2 + a^2)^\frac{5}{2}}d^3x' $$
I understood this until above step, but now I didn't got the next step
$$ \nabla^2{\Phi}_a(x) = -\frac{1}{{\epsilon_0}}\int_0^R \frac{3a^2}{(r^2 + a^2)^\frac{5}{2}} \left[{\rho}(x) + \frac{r^2}{6}\nabla^2{\rho} + ....... \right]r^2 dr + O(a^2) $$
Jackson says that we gonna expand ${\rho}(x')$ around x' = x., but the expansion of ${\rho}(x')$ should also contain the first order derivative of ${\rho}(x')$ like $\nabla{\rho}$, also the taylor expansion of the second term should contain 2 at the denominator but it's 6, and how the last term of $O(a^2)$
So what I am thinking is that Taylor expansion should be $\left[{\rho}(x) + r\nabla{\rho} + \frac{r^2}{2}\nabla^2{\rho}\right]$
I know I am wrong but I don't know what is the answer. Any help is appreciated.
