General expression for polynomials related to ${_2 F_1} (n,n,2n,1-x)$

Question

I've been searching for elementary expression for this class of hypergeometric functions, and found experimentally the following result:

$${_2 F_1} (n+1,n+1,2n+2,1-x)= \frac{-(2n+1) \log x}{(1-x)^{2n+1}} \binom{2n}{n} \sum_{k=0}^n \binom{n}{k}^2 x^k-\frac{p_{n-1}(x)}{(1-x)^{2n}} \tag{1}$$

Where $n \in \mathbb{Z}$ and $n \geq 1$.

This result is pretty useful, because it allows us to separate the part with the logarithm and we are left with a rational function.

However, I haven't been able to find a general expression for $p_{n-1}(x)$. From experiments, it's always a symmetric polynomial with positive coefficients. For even $n$ it is divisible by $(1+x)$.

$$p_0=12 \\ p_1=90(1+x) \\ p_2 = \frac{140}{3} (11+38 x+11 x^2) \\ p_3 = 525 (1+x)(5+32 x+5x^2) \\ p_4 = \frac{462}{5} (137+1762 x+3762 x^2+1762 x^3+132 x^4) \\ p_5 = \frac{42042}{5} (1+x) \left(7 +132 x+382 x^2+132 x^3+7x^4\right) \\ p_6 = \frac{5148}{7} \left(363 +10310 x+58673 x^2+101548 x^3+58673 x^4+10310 x^5+363x^6 \right) \\ p_7 = \frac{21879}{14} (1+x) \left(761+28544 x+209305 x^2+423680 x^3+209305 x^4+28544 x^5+761 x^6\right)$$

So far I see no pattern here, which is strange to me, since there's such a nice pattern in the logarithmic part.

Is it possible to find an explicit general expression for $p_n(x)$?

How does one prove that (1) is correct?

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We can also rewrite (1) as:

$${_2 F_1} (n+1,n+1,2n+2,1-x)= \frac{-(2n+1) \log x}{(1-x)^{n+1}} \binom{2n}{n} P_n \left(\frac{1+x}{1-x} \right)-\frac{p_{n-1}(x)}{(1-x)^{2n}} \tag{2}$$

Where $P_n$ are Legendre polynomials.

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Another identity I derived from an integral, which may be useful here:

$${_2 F_1} \left(n+1,n+1,2n+2, \frac{2 \sqrt{x^2-y^2}}{x+\sqrt{x^2-y^2}} \right)= \\ = \frac{(-1)^n (2n+1) \binom{2n}{n} (x+\sqrt{x^2-y^2})^{n+1}}{2^{n+1} n!} \frac{\partial^n }{\partial x^n} \left(\frac{1}{\sqrt{x^2-y^2}} \log \frac{x+\sqrt{x^2-y^2}}{x-\sqrt{x^2-y^2}} \right) \tag{3}$$

Answer 1

Using the integral representation for $_2 F_1$, we see that \begin{align}f_n(x)&:={_2 F_1}(n+1,n+1;2n+2;1-x)\\&=\frac{(2n+1)!}{n!^2}\int_0^1\big(y(1-y)\big)^n\big(1-(1-x)y\big)^{-n-1}\,dy\\&=\frac{2n+1}{(1-x)^{2n+1}}\binom{2n}{n}\int_x^1 t^{-1}(1-t)^n(1-x/t)^n\,dt\end{align} after substituting $y=(1-t)/(1-x)$. Multiplying binomial expansions, $$(1-t)^n(1-x/t)^n=\sum_{k=0}^{2n}(-t)^{k-n}\sum_j\binom{n}{j}\binom{n}{k-j}x^{n-j}$$ (more precisely, the inner sum is over $\max\{k-n,0\}\leqslant j\leqslant\min\{k,n\}$).

In particular, the logarithmic part corresponds to $k=n$, and coincides with $(1)$.

Answer 2

Using @metamorphy's idea, but different substitution for the integral, I was able to derive another double sum expression:

$$f_n(x)=\frac{2 n+1 }{(1-x)^{n+1}} \binom{2 n}{n} \sum _{k=0}^n \sum _{l=0}^{n+k} \frac{(-1)^{k+l} }{(1-x)^k} \binom{n}{k} \binom{n+k}{l} \begin{cases}-\log (x), & l=n \\ \frac{1-x^{l-n}}{l-n}, & l \neq n \end{cases} \tag{4}$$

For the logarithmic part we have:

$$g_n(x)=-\log (x)\frac{2 n+1 }{(1-x)^{n+1}} \binom{2 n}{n} \sum _{k=0}^n \frac{(-1)^{k+n} }{(1-x)^k} \binom{n}{k} \binom{n+k}{n}$$

Wikipedia gives the so called shifted Legendre polynomials:

$$P(2y-1)=(-1)^{n}\sum _{k=0}^{n}{\binom {n}{k}}{\binom {n+k}{k}}(-y)^{k}$$

Which is the same as our sum with $y=1/(1-x)$, or $2y-1=(1+x)/(1-x)$, so finally:

$$g_n(x)=-\log (x)\frac{2 n+1 }{(1-x)^{n+1}} \binom{2 n}{n} P_n \left(\frac{1+x}{1-x} \right)$$

Which proves $(2)$.

For $p_n$ we can write:

$$p_{n-1}(x)=-(2n+1) \binom{2 n}{n} \sum _{k=0}^n \sum _{l=0 \\ l\neq n}^{n+k} \frac{(-1)^{k+l}}{l-n} \binom{n}{k} \binom{n+k}{l}(1-x)^{n-k-1} (1-x^{l-n}) \tag{5}$$

A closed form might exist for (5), or at least reduction to a single sum. I'll look into this more.

Experimentally I found the following simplification:

$$p_{n-1}(x)=-2 (2n+1) \binom{2 n}{n} \sum _{k=0}^n \sum _{l=0 \\ l\neq n}^{n+k} \frac{(-1)^{k+l}}{l-n} \binom{n}{k} \binom{n+k}{l}(1-x)^{n-k-1} \tag{6}$$

Now we can search for:

$$Q_{n,k}=\sum _{l=0 \\ l\neq n}^{n+k} \frac{(-1)^l}{l-n} \binom{n+k}{l} $$

Separating the sum in two parts we have:

$$\sum _{l=0}^{n-1} \frac{(-1)^l}{l-n} \binom{n+k}{l}= (-1)^n \binom{n+k}{n} (H_{n+k}-H_k)$$

$$\sum _{l=n+1}^{n+k} \frac{(-1)^l}{l-n} \binom{n+k}{l}= -(-1)^n \binom{n+k}{n} (H_{n+k}-H_n)$$

Which means that:

$$Q_{n,k}=(-1)^n \binom{n+k}{n} (H_n-H_k)$$

So now we have:

$$p_{n-1}(x)=-2 (-1)^n (2n+1) \binom{2 n}{n} \sum _{k=0}^n (-1)^k \binom{n}{k} \binom{n+k}{n} (H_n-H_k) (1-x)^{n-k-1} \tag{7}$$

I don't know if this can be simplified further, but it's already pretty good (even if most of my proof is done by Mathematica).

Curiously, we have the following identities:

$$\sum _{k=0}^n (-1)^k \binom{n}{k} \binom{n+k}{n} (H_n-H_k)=(-1)^{n+1} H_n$$

$$\sum _{k=0}^n (-1)^k \binom{n}{k} \binom{n+k}{n} (H_n-H_k) \frac{1}{2^k}=\frac{1-(-1)^n}{2} \frac{(-1)^{\frac{n-1}{2}} 2^{n-1}}{n} \binom{n-1}{\frac{n-1}{2}}^{-1}$$

Now we only need to generalize it somehow.