Numerically evaluating parameter derivatives of a hypergeometric function

Question

I am unable to obtain the numerical value of the derivative of the hypergeometric function. Please note that the (2,4,0,0) is the derivative with respect to the first and second argument

      N[Hypergeometric2F1(2,4,0,0)[0,1,2,1]]

Answer 1

If I use the integral representation of the Gaussian hypergeometric function, and then differentiate that before plugging in the arguments, I get the following expression:

$$\frac{2\pi^4}{5}+8 \int_0^1 \left(2(\operatorname{artanh}(1-2t))^2-\pi ^2\right)(\operatorname{artanh}(1-2t)\log(1-t))^2 \, \mathrm dt$$

which does not symbolically evaluate in Mathematica, but readily evaluates numerically:

(2 π^4)/5 + 8 NIntegrate[(2 ArcTanh[1 - 2 t]^2 - π^2) (ArcTanh[1 - 2 t] Log[1 - t])^2,
                         {t, 0, 1}, WorkingPrecision -> 30]
   291.989096054116602579595642919

---

In general, parameter derivatives of hypergeometric functions can get easily complicated, so I am not overly surprised that a symbolic route did not easily yield a known expression in terms of other special functions known to Mathematica.

Answer 2

Have a look at the definition of the hypergeometric function: you can see that $_2F_1(a,b;2;1)=\frac{\Gamma(2-a-b)}{\Gamma(2-a)\Gamma(2-b)}$:

h[a_,b_] = Sum[Pochhammer[a, k]*Pochhammer[b, k]/Pochhammer[2, k] * 1/k!, {k, 0, ∞}]
(*    Gamma[2 - a - b]/(Gamma[2 - a] Gamma[2 - b])    *)

The general form of this hypergeometric function at $z=1$ is described on the Wolfram Functions site. Thanks @DrWolfgangHintze!

The derivatives now become explicit:

D[h[a, b], {a, 2}, {b, 4}] /. {a -> 0, b -> 1} // FullSimplify
(*    68*π^6/315 + 24*(Zeta[3]^2+2*Zeta[5])    *)

% // N
(*    291.989    *)

Answer 3

Increasing MaxExtraPrecision to the value 1000 helps.

$MaxExtraPrecision = 1000;
N[Derivative[2, 4, 0, 0][Hypergeometric2F1][0, 1, 2, 1],20]
(*291.98909605411660258*)

Try this, you will see what is going on:

 $MaxExtraPrecision = 100;
 N[Derivative[2, 4, 0, 0][Hypergeometric2F1][0, 1, 2, 1],20]

 (*N::meprec: Internal precision limit $MaxExtraPrecision = 100.` reached while evaluating (Hypergeometric2F1^(2,4,0,0))[0,1,2,1].*)
 (*292*)