Is there a simplified way to express the following function that is defined by two Gauss' hypergeometric functions?

Question

Consider the following function

\begin{align} f_k(x) = \left( 1 + 2k \left( 1-x^2 \right) \right) {}_2 F_1 \left( \tfrac{1}{2} ,k+\tfrac{1}{2}; k+\tfrac{3}{2}; x^2 \right) - 2(k+1) \, {}_2 F_1 \left( -\tfrac{1}{2} ,k+\tfrac{1}{2}; k+\tfrac{3}{2}; x^2 \right) \, , \end{align} where $k \in \mathbb{N}$ and $x \in [0,1]$. Here, ${}_2 F_1$ denotes Gauss' hypergeometric function.

For instance, it can be shown that \begin{align} f_0 (x) &= -(1-x^2)^{-1/2} \, , \\ f_1(x) &= \frac{3}{x^3} \left( (1-x^2)\arcsin(x) - x(1-x^2)^\frac{1}{2} \right) \, , \end{align}

i was wondering whether a closed form expression in terms of usual functions can be obtained for general value of $k$.

Any thoughts or hints are most welcome.

Thank you

Answer 1

I calculated that if $\,y:=\sqrt{1-x^2},\,$ then $\, f_0(x) = -y\,$ and for $\,k>0\,$ $$ f_k(x) = \frac{y}{2^{2k-1}(k-1)!x^{2k+1}} \left(y\sin^{-1}(x)\frac{(2k+1)!}{k!} + 2^k P_k(x) \right) $$ where $$ P_k(x) = (2k+1)!!\left(-x + \sum_{i=1}^{k-1}x^{2i+1} 2^{2i-1}\frac{i!(i-1)!}{(2i+1)!}\right) $$ is a polynomial in powers of $\,x\,$ of degree $\,2k-1\,$ with integer coefficients.

I used this Wolfram Mathematica code to check:

F[k_,a_] := Hypergeometric2F1[a/2, k+1/2, k+3/2, x^2];
f[k_] := (1+2k(1-x^2))F[k, 1] - 2(k+1)F[k, -1] // Simplify;
P[k_] := (2k+1)!!(-x + Sum[x^(2i+1)2^(2i-1)
    i!(i-1)!/(2i+1)!, {i, 1, k-1}];
y = Sqrt[1-x^2]; myf[0] = -y;
myf[k_] := y(y ArcSin[x](2k+1)!/k! +
    2^k P[k])/(2^(2k-1)(k-1)!x^(2k+1)) // Simplify;