How to find a closed form formula for such integral using special functions I am trying to find a closed form formula for this integral $$\int_{}^{} \left(1-x^2\right)^n \frac{\mathrm{d^i} }{\mathrm{d} x^i} \left(1-x^2\right)^n dx $$ where
$i=0,1,2,...\in \mathbb{N}$ and $n\in\mathbb{R} :n\geq0 $.
I know that $\left(1-x^2\right)^n = F(a,-n;a;x^2)$ and this integral can be written as $$\int_{}^{} F(a,-n;a;x^2) \frac{\mathrm{d^i} }{\mathrm{d} x^i} F(a,-n;a;x^2) dx $$
which can be easily found for $i=0,1$ but I am looking for a closed form formula for any arbitrary $i$.
Thanks in advance
 A: If we first consider the polynomials separately, we can expand these quite simply
$$
\tag{1}
\left(1-x^2\right)^n=\sum_{k=0}^n\binom{n}{k}x^{2k}
$$
If we now find the $i$th derivative of this we get
$$
\left(1-x^2\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{\mathrm{d}^i}{\mathrm{d}x^i}x^{2k}
$$
This poses a slight problem because we can't just use the standard derivative of a monomial here. When $i>2k$, the derivative becomes 0, so we need to change our summation bounds to make sure this doesn't happen
$$
\sum_{k=\lceil i/2\rceil}^n\binom{n}{k}\frac{\mathrm{d}^i}{\mathrm{d}x^i}x^{2k}
$$
By changing the lower bound to start at $\displaystyle\left\lceil\frac{i}{2}\right\rceil$, we avoid the previously stated problem. This means we can expand the derivative of the monomial which gives
$$
\sum_{k=\lceil i/2\rceil}^n\binom{n}{k}\frac{n!}{(n-i)!}x^{2k-i}
$$
If we then multiply this with our original binomial expansion we arrive at
$$
\sum_{l=0}^n\sum_{k=\lceil i/2\rceil}^n\binom{n}{l}\binom{n}{k}\frac{n!}{(n-i)!}x^{2k+2l-i}
$$
If we now integrate this w.r.t. $x$ we get
$$
\sum_{l=0}^n\sum_{k=\lceil i/2\rceil}^n\binom{n}{l}\binom{n}{k}\frac{n!}{(n-i)!}\left[\frac{1}{2k+2l-i+1}x^{2k+2l-i+1}+c\right]
$$
This proof assumes that $n\leq i$. Is that's not the case, then you can simply integrate the sum in (1) to give
$$
\sum_{k=0}^n\binom{n}{k}\frac{x^{2k+1}}{2k+1}
$$
I hope that this helps in some way. There are probably more elegant solutions to your problem, this just happened to be the best one I could think of at the moment. Regardless, good question. Best of luck with it
