I have an integral as follows: $$ -2\pi\int_0^{\infty}\int_{\lvert r_e-r \rvert}^{r_e+r} \rho_p(r_p) r_p dr_p dr $$ where $$ \rho_p(r_p)=\frac{2\sqrt{2}\gamma^{3/2}}{\pi^{3/2}}*\exp^{-2r_p^2 \gamma} $$ where $\gamma$ is a constant and finally I should obtain a function in term of $r_e$. I tried it by Mathematica but it couldn't solve it and just gives the initial following answer: $$ -2\pi\int_0^{\infty} \frac{-\exp^{-2(r+r_e)^2\gamma}+\exp^{-2\gamma \lvert r-r_e \rvert ^2}}{\sqrt{2}\pi^{3/2}} dr $$ I'm not an expert in mathematics so have no idea. However I'm sure there are some masters in this forum which can help me. Any suggestion?
Addendum
Answer of @CVolk gives the correct result, I have compared this answer with numerical integration by Mathematica in the following photo and as you can see they are match
$\space\space\space$I think I see the problem that Mathematica is having....
Let $m={2\sqrt{2}\gamma^{3/2}\over\pi^{3/2}}$ Then we have,
$$-2m\pi\int_{0}^{\infty}\int_{|r_e-r|}^{r_e+r}e^{-2r_p^2\gamma}r_pdr_pdr$$
Letting u=$2r_p^2\gamma$ yields the following,
$${-2m\pi\over4\gamma}\int_{0}^{\infty}\int_{2(r-r_e)^2\gamma}^{2(r_e+r)^2\gamma}e^{-u}dudr$$
Which gives us,
$${-2m\pi\over4\gamma}\int_{0}^{\infty}\big(-e^{-2\gamma(r_e+r)^2}+e^{-2\gamma(r-r_e)^2}\big)dr$$
Splitting the integral over the sum and another round of substitutions...
$${2m\pi\over4\gamma}\int_{0}^{\infty}e^{-2\gamma(r_e+r)^2}dr-{2m\pi\over4\gamma}\int_{0}^{\infty}e^{-2\gamma(r-r_e)^2}dr$$
Lettting k=$\sqrt{2\gamma}(r_e+r)$ and j=$\sqrt{2\gamma}(r-r_e)$...
$${2m\pi\over4\gamma\sqrt{2\gamma}}\int_{r_e\sqrt{2\gamma}}^{\infty}e^{-k^2}dk-{2m\pi\over4\gamma\sqrt{2\gamma}}\int_{-r_e\sqrt{2\gamma}}^{\infty}e^{-j^2}dj$$
$\space\space\space$These integrals are both special cases of the "Complementary Error Function" which has no elementary antiderivative. You can, however, get a somewhate nice answer...
The Complementary Error Function $"erfc(x)"$ is defined as:
$${2\over\sqrt{\pi}}\int_{x}^{\infty}e^{-t^2}dt$$
There is a relation between this function and the Error Function "$erf(x)$" which states that $$erfc(x)=1-erf(x)$$ Where the Error Function is defined as:
$${2\over\sqrt{\pi}}\int_{0}^{x}e^{-t^2}dt$$
Applying the previous definitions yields;
$${2m\pi\over4\gamma\sqrt{2\gamma}}\bigg(-\int_{0}^{r_e\sqrt{2\gamma}}e^{-k^2}dk+\int_{0}^{-r_e\sqrt{2\gamma}}e^{-j^2}dj\bigg)$$
Which we can write thusly;
$${-2m\pi\over4\gamma\sqrt{2\gamma}}\bigg(\int_{-r_e\sqrt{2\gamma}}^{r_e\sqrt{2\gamma}}e^{-x^2}dx\bigg)$$
Expanding the exponential as an infinite sum gives us,
$${-2m\pi\over4\gamma\sqrt{2\gamma}}\sum_{n=0}^{\infty}{(-1)^n\over{n!}}\int_{-r_e\sqrt{2\gamma}}^{r_e\sqrt{2\gamma}}x^{2n}dx$$
Which finally gives us an answer to the original integral;
$${-m\pi\over\gamma\sqrt{2\gamma}}\sum_{n=0}^{\infty}{(-1)^n\over{n!}(2n+1)}\big(r_e\sqrt{2\gamma}\big)^{2n+1}$$
Unpacking m gives us a little simplification as well...
$${-2\over\sqrt{\pi}}\sum_{n=0}^{\infty}{(-1)^n\over{n!}(2n+1)}\big(r_e\sqrt{2\gamma}\big)^{2n+1}$$