Let $P_{m}^{\left(a,b\right)}\left(s\right)$ the $m$-th Jacobi polinomial. An old result of Bateman shows that $$ \left(1-x\right)^{\alpha+\mu}\frac{P_{m}^{\left(\alpha+\mu,\beta-\mu\right)}\left(x\right)}{P_{m}^{\left(\alpha+\mu,\beta-\mu\right)}\left(1\right)}=\frac{\Gamma\left(\alpha+\mu+1\right)}{\Gamma\left(\alpha+1\right)\Gamma\left(\mu\right)}\int_{x}^{1}\left(1-y\right)^{\alpha}\frac{P_{m}^{\left(\alpha,\beta\right)}\left(y\right)}{P_{m}^{\left(\alpha,\beta\right)}\left(1\right)}\left(y-x\right)^{\mu-1}dy \tag1 $$ which is a special case of the relation $$ \,_{2}F_{1}\left(a,b;c+\mu;x\right)=\frac{\Gamma\left(c+\mu\right)}{\Gamma\left(c\right)\Gamma\left(\mu\right)}\int_{0}^{1}y^{c-1}\left(1-y\right)^{\mu-1}\,_{2}F_{1}\left(a,b;c;xy\right)dy,\quad 0<c<c+\mu,\,x<1 \tag2 $$ and $(2)$ follows integrating term by term the $2F1$. For a work, I need to evaluate the following integral: $$ \int_{x}^{1}\left(1-s\right)^{-1/2}P_{m}^{\left(1/2,-1/2\right)}\left(s\right)ds \tag3 $$ and clearly $(3)$ resembles $(1)$ very closely. I tried to use the recurrence relations of $P_{m}^{\left(a,b\right)}\left(s\right)$ to exploit $(1)$ in some way but I falied. So the question is:
Is it possible to find a closed form in the sense of $(1)$ of $(3)$?
My request seems reasonable given that there are, in the literature, articles that generalize results like $(1)$. For example, see:
- Chen, S., Shen, J., & Wang, L. L. (2016). Generalized Jacobi functions and their applications to fractional differential equations. Mathematics of Computation, 85(300), 1603-1638.
- John T. Conway (2021) Indefinite integrals involving Jacobi polynomials from integrating factors, Integral Transforms and Special Functions, 32:10, 801-811, DOI: 10.1080/10652469.2020.1844197
- Askey, R., & Fitch, J. (1969). Integral representations for Jacobi polynomials and some applications. Journal of Mathematical Analysis and Applications, 26(2), 411-437.
Any help is appreciated, thanks.
Update: from the formula $$\frac{\partial}{\partial\alpha}P_{m}^{\left(\alpha,\beta\right)}\left(y\right)=P_{m}^{\left(\alpha,\beta\right)}\left(y\right)\sum_{k=0}^{m-1}\frac{1}{\alpha+\beta+k+m+1}$$ $$+\sum_{k=0}^{m-1}\frac{\left(\alpha+\beta+2k+1\right)\left(\beta+k+1\right)_{m-k}}{\left(\alpha+\beta+k+m+1\right)\left(\alpha+\beta+k+1\right)_{m-k}}P_{k}^{\left(\alpha,\beta\right)}\left(y\right)$$ we can in some sense use $(1)$ to evaluate $(3)$, because differentiating $(1)$ with respect $\alpha$ we get $$\int_{x}^{1}\left(1-y\right)^{\alpha-1}P_{m}^{\left(\alpha,\beta\right)}\left(y\right)dy=\frac{1}{\alpha}\int_{x}^{1}\left(1-y\right)^{\alpha}\frac{\partial}{\partial\alpha}P_{m}^{\left(\alpha,\beta\right)}\left(y\right)dy$$ $$+\frac{1}{\alpha}\frac{\partial}{\partial\alpha}\frac{\left(1-x\right)^{\alpha+1}}{\alpha+m+1}P_{m}^{\left(\alpha+1,\beta-1\right)}\left(x\right) \tag4$$ and the integrals in the RHS of $(4)$ can be evaluated using $(1)$. It is not clear if such expression is more interesting than the sum involving the Chebyshev polynomials obtained by Paul Enta.
Update2: Since we have the connection formula $$P_{n}^{\left(\gamma,\beta\right)}\left(x\right)=\frac{\left(\beta+1\right)_{n}}{\left(\alpha+\beta+2\right)_{n}}\sum_{k=0}^{n}\frac{\left(\gamma-\alpha\right)_{n-k}\Gamma\left(\alpha+\beta+k+1\right)\left(\alpha+\beta+2k+1\right)\left(\beta+\gamma+n+1\right)_{k}}{\left(n-k\right)!\left(\beta+1\right)_{k}\Gamma\left(\alpha+\beta+2\right)\left(\beta+\alpha+n+2\right)_{k}}P_{k}^{\left(\alpha,\beta\right)}\left(x\right)$$ we get, taking $\gamma=1/2,\,\alpha=-1/2,\beta=-1/2$, that $$P_{m}^{\left(1/2,-1/2\right)}\left(s\right)=\left[\frac{1}{4^{m}}\binom{2m}{m}\right]\left(1+2\sum_{k=1}^{m}\frac{P_{k}^{\left(-1/2,-1/2\right)}\left(s\right)}{\left[\frac{1}{4^{k}}\binom{2k}{k}\right]}\right)$$ so now we can apply $(1)$, hence now the problem is to understand if $$\sum_{k=1}^{m}\frac{P_{k}^{\left(1/2,-3/2\right)}\left(x\right)}{\left[\frac{1}{4^{k}}\binom{2k}{k}\right](2k+1)}$$ has a closed form in terms of the Jacobi polynomials.
An equivalent representation of the particular Jacobi polynomial is given here \begin{equation} P_n^{(a,-a)}(z)=\frac{\Gamma(a+n+1)}{\Gamma(n+1)} \frac{(1+z)^{a / 2}}{(1-z)^{a / 2}} P_n^{-a}(z) \end{equation} then \begin{equation} P_{m}^{\left(1/2,-1/2\right)}\left(s\right)=\frac{\Gamma(m+3/2)}{m!} \frac{(1+s)^{1/4}}{(1-s)^{1/4}} P_n^{-1/2}(s) \end{equation} An explicit form of the associated Legendre polynomial is \begin{equation} P_n^\mu(z)=\frac{(1+z)^{\mu / 2}}{(1-z)^{\mu / 2}} \sum_{k=0}^n \frac{(-n)_k(n+1)_k}{\Gamma(k-\mu+1) k !}\left(\frac{1-z}{2}\right)^k \end{equation} Then, \begin{equation} P_{m}^{\left(1/2,-1/2\right)}\left(s\right)=\frac{\Gamma(m+3/2)}{m!}\sum_{k=0}^m \frac{(-m)_k(m+1)_k}{\Gamma(k+3/2) k !}\left(\frac{1-s}{2}\right)^k \end{equation} We have thus \begin{align} I&= \int_{x}^{1}\left(1-s\right)^{-1/2}P_{m}^{\left(1/2,-1/2\right)}\left(s\right)ds\\ &=\frac{\Gamma(m+3/2)}{m!}\sum_{k=0}^m \frac{2^{-k}(-m)_k(m+1)_k}{\Gamma(k+3/2) k !}\int_{x}^{1}(1-s)^{k-1/2}\\ &=\frac{\Gamma(m+3/2)}{m!}\sum_{k=0}^m \frac{(-m)_k(m+1)_k}{\Gamma(k+3/2) k !}\frac{2^{1-k}}{2k+1}(1-x)^{k+1/2} \end{align} which can be written as \begin{equation} I=\frac{4}{\sqrt\pi}\frac{\Gamma(m+3/2)}{m!}\sqrt{1-x}\,{}_3F_2\left( -m,m+1,1/2;3/2,3/2;\frac{1-x}{2} \right) \end{equation}
Edit
A somewhat nicer expression can be obtained by remarking that \begin{equation} Z\,{}_2F_1\left( -m,m+1,1/2;3/2,3/2;Z^2 \right)=\int_0^Z {}_2F_1\left( -m,m+1;3/2;Z^2 \right)\,dZ \end{equation} and that \begin{equation} {}_2F_1\left( -m,m+1;3/2;Z^2 \right)=\frac{(-1)^m}{2m+1}\frac{T_{2m+1}(Z)}{Z} \end{equation} where $T_n$ are the Chebyshev plynomials. They verify the recurrence relations \begin{align} \frac{T_{n+1}}x&=-\frac{T_{n-1}}x+2T_n\\ \int T_n(x)\,dx&=\frac12\left[\frac{T_{n+1}(x)}{n+1}-\frac{T_{n-1}(x)}{n-1}\right] \end{align} from which one deduces \begin{equation} \int_0^Z\frac{T_{2m+1}(z)}{z}\,dz=\frac{T_{2m+1}(Z)}{2m+1}+2(-1)^m\sum_{k=0}^{m-1} (-1)^k\frac{T_{2k+1}(Z)}{2k+1} \end{equation} Then \begin{equation} Z\,{}_2F_1\left( -m,m+1,1/2;3/2,3/2;Z^2 \right)=\frac{(-1)^m}{2m+1}\left[\frac{T_{2m+1}(Z)}{2m+1}+2(-1)^m\sum_{k=0}^{m-1} (-1)^k\frac{T_{2k+1}(Z)}{2k+1}\right] \end{equation} and thus \begin{equation} I=\frac{(-1)^m4\sqrt{2}}{\sqrt\pi}\frac{\Gamma(m+3/2)}{(2m+1)m!}\left[\frac{T_{2m+1}\left(\sqrt{\frac{1-x}2}\right)}{2m+1}+2(-1)^m\sum_{k=0}^{m-1} (-1)^k\frac{T_{2k+1}\left(\sqrt{\frac{1-x}2}\right)}{2k+1}\right] \end{equation} By denoting $x=\cos\theta$ with $0\le\theta\le\pi$, we have $\sqrt{(1-x)/2}=\sin\theta/2$ and $T_{2k+1}=(-1)^k\sin((k+1/2)\theta)$ which gives \begin{equation} I=\frac{4\sqrt{2}}{\sqrt\pi}\frac{\Gamma(m+3/2)}{(2m+1)m!}\left[\frac{\sin\left( (m+1/2)\theta \right)}{2m+1}+2\sum_{k=0}^{m-1} \frac{\sin\left( (k+1/2)\theta \right)}{2k+1}\right] \end{equation} which, in turn, can be expressed in terms of a Lerch-Zeta function.