Generating function of the square of Jacobi polynomial The generating function of the Jacobi polynomials is given by
$$
\sum_{n=0}^{\infty} P_{n}^{(\alpha, \beta)}(z) t^{n}=2^{\alpha+\beta} R^{-1}(1-t+R)^{-\alpha}(1+t+R)^{-\beta}
$$
where
$$
R=R(z, t)=\left(1-2 z t+t^{2}\right)^{\frac{1}{2}}.
$$
However, we need a similar result or alternative representations for the square of the Jacobi polynomial given by
$$
\sum_{n=0}^{\infty}( P_{n}^{(\alpha, \beta)}(z))^2 t^{n}
$$
or more specifically,
$$
\sum_{n=0}^{\infty}( P_{n}^{(1, 0)}(z))^2 t^{n}.
$$
I have researched this in detail, but have not found an answer. If someone could give me some hints, that would be really helpful.
 A: I don't know if this is useful to you, but Bailey gave the identity
\begin{multline*}\sum_{n=0}^\infty\frac{n!(\alpha+\beta+1)_n}{(\alpha+1)_n(\beta+1)_n}\,P_n^{(\alpha,\beta)}(x)P_n^{(\alpha,\beta)}(y)\,t^n\\
=\frac 1{(1+t)^{\alpha+\beta+1}}\sum_{m,n=0}^\infty\frac{(\alpha+\beta+1)_{2m+2n}}{m!n!(\beta+1)_m(\alpha+1)_n}\left(\frac t{4(1+t)^2}\right)^{m+n}((1+x)(1+y))^m((1-x)(1-y))^n.\end{multline*}
The original reference is  J. London Math. Soc. 13 (1938), 8-12, but I copied it from Stanton,  Proc. Amer. Math. Soc. 80 (1980), 398–400. When $\beta=0$ and $x=y$, this gives an expression for
$$\sum_{n=0}^\infty P_n^{(\alpha,0)}(x)^2\,t^n$$
as a double series.
A: Here is an expression that involves elliptic functions and a portion represented as an integral that (I think) cannot be reduced to elliptic functions.
$$ \sum_{n=0}^\infty P^{(1,0)}_n(x)^2 t^n = \frac{1}{(1-x)^2}\big( F(x,t)+\frac{1}{t} (F(x,t)-1) \big) -\frac{2x}{t}\big(F(x,t)-1\big)  + 
 \frac{1+x}{1-x}\frac{G(x,t)}{t} $$
where
$$F(x,t) = \sum_{n=0}^\infty P_n(x)^2t^n$$
and has a closed-form in elliptic functions from Bailey's formula
$$ \tag{B} \sum_{n=0}^\infty P_n(x) \, P_n(y) t^n = r \cdot {}_2F_{1}\big(1/2,1/2;1;-4\sqrt{(1-x^2)(1-y^2)} \,t \, r^2\big)$$
$$ r= \big(1+t(t-2\sqrt{(1-x^2)(1-y^2)} - 2xy)\big)^{-1/2} $$
I've found two integral representations for $G(x,t):$
$$G(x,t)=\frac{2x}{\pi (1-x^2)}\int_0^t \frac{du}{u(1-u)}\Big( \frac{(u-1)^2}{(1+u)^2-4ux^2} \,E\big(\frac{4u(x^2-1)}{(u-1)^2}\big) - K\big(\frac{4u(x^2-1)}{(u-1)^2}\big) \Big) $$
and
$$G(x,t)=\frac{-4 t}{\pi \, x}\int_0^1 du\sqrt{\frac{u}{1-u}}\frac{1}{\rho(u;x,t)}
\Big((1+t+ \rho(u;x,t))^{-1} + (1-t+ \rho(u;x,t))^{-1} \Big)$$
$$ \text{with } \rho(u,x,t)=((1-t)^2+4tu(1-x^2))^{1/2} $$
In the first integral, $E$ and $K$ are elliptic functions in Mathematica notation.
The following is a proof sketch:
$$ P^{(1,0)}_n(x) = \frac{P_n(x)-P_{n+1}(x)}{1-x} $$
where the $P_n(x)$ are the ordinary Legendre polynomials.  Square this expression within the original sum, and there are three sums involving summands with $(P_n)^2,$ $(P_{n+1})^2$ and $P_n P_{n+1}.$  The 'perfect squares' can be dealt with eq. (B), although an index shift is needed for the $(P_{n+1})^2.$ This gives two of the $F(x,t)$ terms in the answer. The cross-term series can be dealt with a relationship among Legendre polynomials
$$\sum_{n=1}^\infty P_n(x)P_{n-1}(x)t^n=
\sum_{n=1}^\infty P_n(x)\Big(xP_n(x) + \frac{1-x^2}{n} \frac{d}{dx}P_n(x) \Big)t^n $$
Again, we have a 'perfect square' case that can be dealt with eq. (B).  The last sum to deal with is
$$  \sum_{n=1}^\infty P_n(x) \frac{d}{dx} P_n(x) \frac{t^n}{n}
=\frac{1}{2} \frac{d}{dx} \sum_{n=1}^\infty P_n(x)^2 \frac{t^n}{n}  $$
The first integral representation I gave for $G(x,t)$ involves integrating eq. B (specialized to y=x) with respect to $t.$  Then do the derivative with respect to $x$ afterwards.  The second integral relationship started from an integral expression for the square of a Legendre polynomial, Gradshteyn and Rhyzik 7.137.5.  Again, after getting the integral, differentiate with repsect to $x.$
I doubt the integrals simplify to elliptic functions, but would love to be proved wrong.
