A semi-algebraic generalization of the Steiner surface has appeared,

$$S = \left\{(x,y,z,t) \space \vert \space t^2(1-x^2-y^2-z^2-t^2) - (x^2 y^2 + x^2 z^2 + y^2 z^2 - 2 x y z) \geq 0 \right\}$$

which has a hypervolume

$$V_S = \frac{\pi^4+12\Gamma^8(3/4)}{24 \Gamma^4(3/4)}$$

where $\Gamma(x)$ is the gamma function. A potentially physically relevant subset of S has only non-negative elements,

$$S^+ =\\ \left\{(x,y,z,t) \space \vert \space t^2(1-x^2-y^2-z^2-t^2) - (x^2 y^2 + x^2 z^2 + y^2 z^2 - 2 x y z) \geq 0 , x\geq0,y\geq0,z\geq0,t\geq0\right\}$$

which has hypervolume

$$V_{S^+} = \frac{1}{16}\left(\frac{\pi^4+12\Gamma^8(3/4)}{24 \Gamma^4(3/4)} \right) + \frac{1}{27} \,_4 F_3(1,1,1,3; 3/2,5/2,5/2;1).$$

I wonder that since S has a closed form for its volume, the non-negative subset might also have a closed form for its volume. By closed form, I mean expressible in terms of known numerical constants.

I have tried to apply some of the magic that appeared in the related question: What is $\, _4F_3\left(1,1,1,\frac{3}{2};\frac{5}{2},\frac{5}{2},\frac{5}{2};1\right)$?

but for the hypergeometric in my question, the additional whole number in the top row of arguments seems to drain some luck out of an analogous approach.

Is there a closed form for $\,_4 F_3(1,1,1,3; 3/2,5/2,5/2;1)$?


By the way, I arrived at the above expression for $V_{S^+}$ through the following. The region can be parametrized,

$$x = \frac{\cos\theta_1\sin\theta_2\sin\theta_3}{r-\cos\theta_1\cos\theta_2\cos\theta_3}$$

$$y = \frac{\sin\theta_1\cos\theta_2\sin\theta_3}{r-\cos\theta_1\cos\theta_2\cos\theta_3}$$

$$z = \frac{\sin\theta_1\sin\theta_2\cos\theta_3}{r-\cos\theta_1\cos\theta_2\cos\theta_3}$$

$$t = \frac{\sin\theta_1\sin\theta_2\sin\theta_3}{r-\cos\theta_1\cos\theta_2\cos\theta_3}$$

with $r\in(1,\infty)$ and $\theta_i\in(0,\pi/2)$. Hence by computing the Jacobian,

$$V_{S^+} = \int_{S^+} dx \,dy \,dz \,dt = \int_1^\infty dr\int_0^{\pi/2}d\theta_1\int_0^{\pi/2}d\theta_2\int_0^{\pi/2}d\theta_3 \frac{\sin^2\theta_1\sin^2\theta_2\sin^2\theta_3}{(r-\cos\theta_1\cos\theta_2\cos\theta_3)^5}$$

where the $r$ integral is elementary, and one could expand the denominator as a binomial series and integrate term by term.

In any case, after applying the methods of the linked question, with a handful of additional manipulations, I reduced the hypergeometric to the following,

$$V_{S^+} = \frac{1}{12}+\frac{1}{8}\left(\frac{\pi^4+12\Gamma^8(3/4)}{24 \Gamma^4(3/4)} \right) +\frac{3}{24} \mathcal{I}_1 - \frac{2}{24}\mathcal{I}_2$$


$$\mathcal{I}_1 = \int_0^1 dk \sqrt{1-k^2} \arcsin (k) K(k)$$

$$\mathcal{I}_2 = \int_0^1 \frac{dk}{\sqrt{1-k^2}} \arcsin( k) K(k)$$

where $K(k)$ is the complete elliptic integral of the first kind as a function of the elliptic modulus. The above integrals are quite similar to one given in Jack D'Aurizio's preprint (linked below, bottom of p. 16):

$$\int_0^1 \frac{dk}{\sqrt{k}} \arcsin(\sqrt{k})K(\sqrt{k}) = 2 \pi G - \frac{\pi^2}{2} \ln(2)$$

  • $\begingroup$ In other terms, you want a closed form for $$\frac{9}{2}\sum_{n\geq 0}\frac{(n+2)(n+1)64^n}{\binom{2n}{n}^3 (2n+1)^3 (2n+3)^2}. $$ I'll see what I can do. It would be useful to have the triple or double integral leading to such series written down in a explicit way. $\endgroup$ – Jack D'Aurizio Dec 23 '17 at 10:13
  • $\begingroup$ Anyway, have a look at this preprint of mine, since you might spot something useful before me. $\endgroup$ – Jack D'Aurizio Dec 23 '17 at 10:33
  • $\begingroup$ A path which seems promising to me is to compute integrals of the $$ \iint_{(0,1)^2}\frac{\arcsin(xy)}{\sqrt{1-x^2}\sqrt{1-y^2}}\,dx\,dy $$ kind by regarding $\arcsin(xy)$ as $\arcsin(axy)|_{a=1}$ and by applying differentiation under the integral sign. This reduces the problem to the evaluation of an integral involving $\text{Li}_2$, which should be doable. $\endgroup$ – Jack D'Aurizio Dec 23 '17 at 10:49

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