Fourier Legendre expansion of Beta kernel $x^a (1-x)^b$ Preliminaries. I have difficulty computing FL expansion of Beta kernel $f_{a,b}(x)=x^a (1-x)^b$ where $4a, 4b \in \mathbb{Z}$. Here are two important examples:

*

*$a=s-1,b=0: x^{s-1}=\sum_{n=0}^\infty \frac{(-1)^n(3/2)_n(1-s)_n}{s (1/2)_n(1+s)_n} P_n(2x-1)$

*$a=b=s-1: (x(1-x))^{s-1}=B(s,s)\sum_{n=0}^\infty  \frac{(5/4)_n(1-s)_n(1/2)_n}{(1/4)_n(1/2+s)_n(1)_n} P_{2n}(2x-1)$
When $4s\in \mathbb Z$ these can be simplified further (see this article by M. Cantarini and J. D'Aurizio for examples and applications). By using these formulas as well as reflection and multiplication by $x$, all FL expansions of $f_{a,b}(x)$ where at least one of $a,b,a-b\in \mathbb Z$ are computable.

Problem. How can we calculate the FL expansion where none of $a,b,a-b$ is integral? For instance I have no idea how to calculate the expansion of $\sqrt[4]{\frac x{1-x}}$ or $\frac{1}{\sqrt{x}\sqrt[4]{1-x}}$.

Thoughts. By repeated IBP, whenever $f$ has no polymonial singularities at $0,1$:

$$I_n=\int_0^1 f(x) P_n(2x-1) dx=\frac 1{n!} \int_0^1 f^{(n)}(x) x^n (1-x)^n dx$$

Taking $f(x)=x^a (1-x)^b$ yields
$$I_n=\frac{1}{n!}\sum _{k=0}^n (-1)^{n-k} (a-k+1)_k \binom{n}{k} (b-(n-k)+1)_{n-k} B(a-k+n+1,b+k+1)$$
Here $(a)_k$ denotes Pochhammer symbol. In previous $2$ cases, this finite hypergeometric sum is evaluable via Dixon identity/residue calculus, but not for the general case.

Background. This problem arises from evaluation of $\int_0^1 x^a (1-x)^b \text{Li}_n(x) \, dx$. See this post for basic examples. More are given in this article.

Update. When $a+b\in\mathbb Z$, one may evaluate $\int_0^1 x^a (1-x)^b \text{Li}_n(x) \, dx$ by Beta derivatives directly, which circumvents calculation of FL expansion of rational terms like $x^k \sqrt[4]{\frac x{1-x}}$, etc.
 A: Expanding $(1-x)^b$ as $\sum_{k=0}^{\infty}\frac{\Gamma(-b+k)\,x^k}{k!\,\Gamma(-b)}$, we get
\begin{align*}
I_n(a,b)=\int_0^{1}x^a(1-x)^bP_n(2x-1)dx=&\sum_{k=0}^{\infty}\frac{\Gamma(-b+k)}{k!\,\Gamma(-b)}\int_0^{1}x^{a+k}P_n(2x-1)dx=\\=&\sum_{k=0}^{\infty}\frac{\Gamma(-b+k)\Gamma^2(1+a+k)}{k!\,\Gamma(-b)\Gamma(n+2+a+k)\Gamma(1+a+k-n)}=\\=&\frac{\Gamma^2(1+a)}{\Gamma(2+a+n)\Gamma(1+a-n)}{}_3F_2\left(\begin{array}{c}-b,1+a,1+a\\ 2+a+n,1+a-n\end{array};1\right)=\\=&(-1)^n B(1+a,1+b)\; {}_3F_2\left(\begin{array}{c}-n,1+n,1+a\\ 1,2+a+b\end{array};1\right)
\end{align*}
It is straightforward to deduce from this last representation that

*

*$\displaystyle I_n(0,b)=(-1)^n\frac{\Gamma^2(b+1)}{\Gamma(b+2+n)\Gamma(b+1-n)}$ (because $_3F_2$ reduces to $_2F_1$)

*$\displaystyle I_n(a,b)$ is a rational function of $a,b$ (because the hypergeometric series truncates) ; this should give the same answer as you get by using Rodriguez formula and integrating by parts

Also, from the identity 7.4.4.99 in Prudnikov-Brychkov-Marychev Vol. 3 it follows that

*

*$\displaystyle I_{2k+1}(a,a)=0$, $\displaystyle I_{2k}(a,a)=B(1+a,1+a)\frac{\left(\tfrac12\right)_k \left(-a\right)_k}{k!\left(a+\tfrac32\right)_k}$
The hypergeometric representation thereby reproduces all of the properties mentioned in the original post. It is not clear to me whether it counts as a closed-form answer but at least it may give a useful starting point for further analysis. In particular, the case $a=-b=\frac14$ boils down to finding a "closed" (better than the finite series) representation of ${}_3F_2\left(\begin{array}{c}-n,1+n,\tfrac54\\ 1,2\end{array};1\right)$. Unfortunately, the reference mentioned above does not seem to provide any further simplification.
