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Consider the following integral, $$ {1 \over 4\pi^{2}}\int_{0}^{2\pi}\int_{0}^{2\pi} \sqrt{\, 9 -\sin^{2}\left(\theta_{1} \over 2\right) \sin^{2}\left(\theta_{2} \over 2\right)\,} \,{\rm d}\theta_{1}\,d\theta_{2} $$ This integral comes up in computing the volume of $3$-dimensional special orthogonal matrices of Hessenberg form, i.e., the bottom left entry is $0$. Mathematica isn't able to produce close form solution. Numerically it's about $2.95$.

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I'll point out the obvious (which you may have already tried). You can do one of the $\theta_i$ integrals to get an elliptic integral, probably of the second kind. Not sure what happens next, are there any results on integrals of elliptic integrals? – j.c. Sep 23 '10 at 1:31
That's what I found out in mathematica also. Unfortunately mathematica doesn't know what to do next. I was hoping there is some multivariate change of variable that simplies the integral. – John Jiang Sep 23 '10 at 1:52
Is there a particular reason you think this integral might have a closed form in special functions or that you need one? In any case you might want to hit the tables of integrals, e.g. Gradshteyn and Ryzhik. – j.c. Sep 23 '10 at 2:21
Even faster: 12 NIntegrate[EllipticE[(Sin[th/2]/3)^2], {th, 0, 2Pi}] – J. M. Sep 23 '10 at 3:43
You are right. I forgot to divide by $4 \pi^2$, because I was viewing it as an integral over the torus. – John Jiang Sep 23 '10 at 4:43
up vote 4 down vote accepted

Assuming that

$$I=\int_0^{2\pi} \int_0^{2\pi}\sqrt{9-\sin^2 \frac{\theta_1 }{2} \sin^2 \frac{\theta_2 }{2}}\mathrm{d}\theta_1 \mathrm{d}\theta_2$$

is correct,

$$I=3\int_0^{2\pi} \int_0^{2\pi}\sqrt{1-\frac19 \sin^2 \frac{\theta_1 }{2} \sin^2 \frac{\theta_2 }{2}}\mathrm{d}\theta_1 \mathrm{d}\theta_2$$


$$I=12\int_0^{2\pi} \int_0^{\frac{\pi}{2}}\sqrt{1-\frac19 \sin^2 \theta_1 \sin^2 \frac{\theta_2 }{2}}\mathrm{d}\theta_1 \mathrm{d}\theta_2$$

$$I=12\int_0^{2\pi}E\left(\frac19 \sin^2 \frac{\theta_2 }{2}\right)\mathrm{d}\theta_2$$

($E(m)$ is the complete elliptic integral of the second kind, with parameter $m$; for the Maple people, what you have is $E(k)$ where $k^2=m$)

$$I=48\int_0^{\frac{\pi}{2}}E\left(\frac19 \sin^2 \theta_2\right)\mathrm{d}\theta_2$$

and letting $m=\sin^2 \theta_2$,

$$I=24\int_0^1 \frac1{\sqrt{m}\sqrt{1-m}} E\left(\frac{m}{9}\right)\mathrm{d}m$$

which Mathematica evaluates to

$$12\pi^2 {}_3 F_2\left(-\frac12,\frac12,\frac12 ; 1,1 ; \frac19\right)$$

where ${}_3 F_2$ is a hypergeometric function; further "simplification" can be done using the formula here.

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Thanks so much! That suggested to me that volumes of higher dimensional Hessenberg circular ensemble might be computable in terms of hypergeometric functions. – John Jiang Sep 23 '10 at 22:16
John, technically it's still an elliptic integral (I haven't been able to find a further simplification of the identity in the link I gave for this special case). – J. M. Sep 23 '10 at 23:54

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