A set of identities involving the Gaussian hypergeometric function and its generalizations to several variables.

Question

In my other question we derived a "closed form expression" for the spectral moments of Wishart random matrices (under the assumption that the underlying correlation matrix is an identity). Clearly the spectral moment of order zero has to be equal to the normalization of the eigenvalue distribution, i.e. it has to be a unity. Now, by taking $N=2,3,4$ in the formulae in question and using the normalization constraint we have obtained the following two puzzling identities below.

Let us take $a \in {\mathbb N}_+$. Then the following identities hold true:

\begin{eqnarray} &&\frac{\, _2F_1(2,-a;-2 a;2)}{2 a+1} = 1\\ &&\hline \\ && \frac{2^{2 a+2} 3^{-3 a-1} \Gamma (3 a+1) }{\Gamma (a+2) \Gamma (2 a+2)} \cdot \sum _{K_1=0}^a \sum _{K_2=K_1}^{2 a} \frac{2^{K_1-K_2} 3^{K_2} (-2 a)_{K_2} (-a)_{K_1} }{(-3 a)_{K_2} (-2 a)_{K_1}} \cdot \\ && \left(K_1+1\right) \left(-K_1+K_2+1\right) \left(\frac{1}{4} \left(K_1+2\right) +\frac{1}{8} \left(-K_1+K_2+2\right) \right) =1 \\ &&\hline \\ && \frac{2^{4 a+5-8 a-7} 3^{-4} (4 a)! }{\Gamma (2 a+2) \Gamma (2 a+4)} \cdot \sum\limits_{K_1,0}^{a} \sum\limits_{K_2,0}^{2 a} \sum\limits_{K_3,0}^{3 a} % \frac{2^{K_1-K_2} 3^{K_2-K_3} 4^{K_3} (-3 a)_{K_3} (-2 a)_{K_2} (-a)_{K_1}}{(-4 a)_{K_3} (-3 a)_{K_2} (-2 a)_{K_1}} \\ && % \left(K_1+1\right) \left(-K_1+K_2+1\right) \left(-K_2+K_3+1\right) \left(-9 K_1^3-9 \left(K_2+15\right) K_1^2+\left(K_2^2+\left(4 K_3-2\right) K_2+4 K_3 \left(K_3+20\right)-90\right) K_1+K_2^3+49 K_2^2+4 \left(K_2+6\right) K_3^2+618 K_2+4 \left(K_2+6\right) \left(K_2+20\right) K_3+1944\right) = 1 % \tag{1} \end{eqnarray} where $x_n := x (x+1) \cdot \cdots \cdot (x+n-1) $ is the upper Pochhammer symbol. The second identity above can be expressed in terms of Kampe de Ferriet functions.

As always, the code below verifies the identities above. We have:

In[622]:= (*NN=2*)
Table[1/(2 aa + 1) Hypergeometric2F1[2, -aa, -2 aa, 2], {aa, 0, 20}]
(*NN=3*)
Table[(2^(2 + 2 aa) 3^(-1 - 3 aa) Gamma[1 + 3 aa])/(Gamma[
      2 + aa] Gamma[2 + 2 aa]) (Sum[
    uPch[-aa, K1]/uPch[-2 aa, K1] uPch[-2 aa, K2]/uPch[-3 aa, K2] 2^
      K1 (3/2)^
      K2 (uPch[K1 + 1, 2] uPch[K2 - K1 + 1, 1]/4 + 
       uPch[K1 + 1, 1] uPch[K2 - K1 + 1, 2]/8), {K1, 0, aa}, {K2, K1, 
     2 aa}]), {aa, 0, 20}]
(*NN=4*)
NN = 4;
Table[(NN aa)!/(32 Gamma[2 + 2 aa] Gamma[4 + 2 aa]) 2^(10 + 4 aa) Sum[
   (Product[uPch[-(j + 0) aa, K[j]]/
        uPch[-(j + 1) aa, K[j]], {j, 1, NN - 1}] Product[1/xi^(
        K[xi] - K[xi - 1]), {xi, 1, NN}] /. {K[0] :> 0, 
       K[NN] :> NN aa}) 1/
    10368 (1 + K[1]) (1 - K[1] + K[2]) (1 - K[2] + K[3]) (1944 - 
      9 K[1]^3 + 618 K[2] + 49 K[2]^2 + K[2]^3 - 
      9 K[1]^2 (15 + K[2]) + 4 (6 + K[2]) (20 + K[2]) K[3] + 
      4 (6 + K[2]) K[3]^2 + 
      K[1] (-90 + K[2]^2 + 4 K[3] (20 + K[3]) + K[2] (-2 + 4 K[3])))
   , Evaluate[
    Sequence @@ 
     Table[{K[xi], If[xi == 1, 0, K[xi - 1]], xi aa}, {xi, 1, 
       NN - 1}]]], {aa, 0, 20}]
Out[622]= {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 

1, 1}
Out[623]= {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 

1, 1}
Out[625]= {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
  1}

My question is then how do we go about proving the identities in $(1)$?

Answer 1

This is an answer for $N=2$ only. Note that the hypergeometric function is evaluated outside its circle of convergence so it needs to be mapped onto that circle before doing anything. Fortunately there exists an appropriate identity here which we are going to use. Therefore we have:

\begin{eqnarray} &&m_p = \\ &&\left. \frac{T^{-p} (2 a+p)! (\, _2F_1(2,-a;-2 a-p;2)+\, _2F_1(2,-a-p;-2 a-p;2))}{2 (2 a+1)!} \right. = \\ &&\frac{T^{-p} (2 a+p+1)! }{3 (2 a+1)!} \cdot \left( \right. \\ && \left. \frac{2^{a-1} (a+1)! (a+p)! \, _2F_1\left(-a,a+p+1;-a-1;\frac{1}{2}\right)}{(2 a+p+1)!}+\right. \\ &&\left.\frac{a! 2^{a+p-1} (a+p+1)! \, _2F_1\left(a+1,-a-p;-a-p-1;\frac{1}{2}\right)}{(2 a+p+1)!}+\right.\\ &&\left. \frac{(2 a+p+2) \, _2F_1\left(2,2 a+p+3;a+3;\frac{1}{2}\right)}{4 (a+1) (a+2)}+\right. \\ &&\left.\frac{(2 a+p+2) \, _2F_1\left(2,2 a+p+3;a+p+3;\frac{1}{2}\right)}{4 (a+p+1) (a+p+2)}\right. \\ &&\left.\right) \tag{1} \end{eqnarray} The first line above is just the definition of the moment and in the second line we used the aforementioned identity along with the reflection formula for the Gamma function to eliminate indeterminate expressions.

Now we set $p=0$ and we get:

\begin{eqnarray} &&m_0=\frac{2^a a! (a+1)! \, _2F_1\left(-a,a+1;-a-1;\frac{1}{2}\right)}{3 (2 a+1)!}+\frac{2 a+2}{3 (a+1)} \\ &&= \frac{1}{3} + \frac{2}{3} \\ &&=1 \end{eqnarray}

In the first line the first & the last term in the right hand side come from the first two terms and the last two terms in $(1)$ and from the usage of the Gauss' second summation theorem . In the second line we used the identity $_2F_1\left(-a,a+1;-a-1;\frac{1}{2}\right) = (2 a+1)!/(2^a a! (a+1)!)$.

Update:

As a matter of fact it is possible to give a closed form expression for all the moments of negative order ($p\le 0$ and $p\in {\mathbb Z}$ and $a+p \ge 0$). This is what we are going to do now. We start from the second equality in $(1)$ and we have:

\begin{eqnarray} m_p &=& \frac{T^{-p} \, _2F_1\left(2,2 a+p+3;a+3;\frac{1}{2}\right) \Gamma (2 a+p+3)}{4 (a+2) \Gamma (2 a+3)}+\frac{T^{-p} \, _2F_1\left(2,2 a+p+3;a+p+3;\frac{1}{2}\right) \Gamma (2 a+p+3)}{8 (a+p+1) (a+p+2) \Gamma (2 a+2)} \\ &=& 2^{p-1} T^{-p} (a+1)_p \cdot \sum\limits_{l=0}^{-p} \left((-1)^{l+p}+1\right) \binom{-p}{l} \frac{ \left(a+\frac{3}{2}\right)_{\frac{l+p}{2}}}{\left(\frac{3}{2}\right) _{\frac{l+p}{2}}} \end{eqnarray}

In the top line above we used the following identities: $_2F_1\left(a+1,-a-p;-a-p-1;\frac{1}{2}\right)=\frac{2^{-a-p-2} \, _2F_1\left(2,2 a+p+3;a+p+3;\frac{1}{2}\right) \Gamma (2 a+p+3)}{(a+p+1) \Gamma (a+1) \Gamma (a+p+3)}+\frac{2^{a+1} p}{a+p+1}$ and $ _2F_1\left(-a,a+p+1;-a-1;\frac{1}{2}\right),\frac{2^{-a-2} \, _2F_1\left(2,2 a+p+3;a+3;\frac{1}{2}\right) \Gamma (2 a+p+3)}{(a+1) \Gamma (a+3) \Gamma (a+p+1)}-\frac{p 2^{a+p+1}}{a+1}$. Those identities are being derived just from the definition of the hypergeometric series. Note that since $a \in {\mathbb N}_+$ and $a+p \ge 0 $ the series are truncated at a certain finite value of summation index rather than extending to infinity.

In the bottom line above we used the following identities $_2F_1\left(2,2 a+p+3;a+3;\frac{1}{2}\right)= \sum\limits_{l=0}^{-p} \frac{2^{2 a+p+2} \Gamma (a+3) (-1)^{l+p} \Gamma (a+p+1) \binom{-p}{l} \Gamma \left(a+\frac{1}{2} (l+p+3)\right)}{\Gamma (a+1) \Gamma (2 a+p+3) \Gamma \left(\frac{1}{2} (l+p+3)\right)}$ and $_2F_1\left(2,2 a+p+3;a+p+3;\frac{1}{2}\right)=\sum\limits_{l=0}^{-p} \frac{2^{2 a+p+2} \Gamma (a+p+3) \binom{-p}{l} \Gamma \left(a+\frac{1}{2} (l+p+3)\right)}{\Gamma (2 a+p+3) \Gamma \left(\frac{1}{2} (l+p+3)\right)}$. Again, those identities are being derived by using equation $(8)$ in page 510 in

Choi, Junesang; Rathie, Arjun K.; Purnima, A note on Gauss’s second summation theorem for the series (_2 F_1 (1/2)), Commun. Korean Math. Soc. 22, No. 4, 509-512 (2007). ZBL1168.33304.