evaluating an integral related to the volume of Hessenberg orthogonal matrices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:20:45Z http://mathoverflow.net/feeds/question/39673 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39673/evaluating-an-integral-related-to-the-volume-of-hessenberg-orthogonal-matrices evaluating an integral related to the volume of Hessenberg orthogonal matrices John Jiang 2010-09-23T01:24:25Z 2010-09-23T22:14:54Z <p>Consider the following integral, $$1/(4\pi^2) \int_0^{2\pi} \int_0^{2\pi} (9- \sin^2 \frac{\theta_1 }{2} \sin^2 \frac{\theta_2 }{2})^{1/2} d\theta_1 d\theta_2.$$ 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. </p> http://mathoverflow.net/questions/39673/evaluating-an-integral-related-to-the-volume-of-hessenberg-orthogonal-matrices/39687#39687 Answer by J. M. for evaluating an integral related to the volume of Hessenberg orthogonal matrices J. M. 2010-09-23T04:47:41Z 2010-09-23T04:47:41Z <p>Assuming that</p> <p>$$I=\int_0^{2\pi} \int_0^{2\pi}\sqrt{9-\sin^2 \frac{\theta_1 }{2} \sin^2 \frac{\theta_2 }{2}}\mathrm{d}\theta_1 \mathrm{d}\theta_2$$</p> <p>is correct,</p> <p>$$I=3\int_0^{2\pi} \int_0^{2\pi}\sqrt{1-\frac19 \sin^2 \frac{\theta_1 }{2} \sin^2 \frac{\theta_2 }{2}}\mathrm{d}\theta_1 \mathrm{d}\theta_2$$</p> <p>then,</p> <p>$$I=12\int_0^{2\pi} \int_0^{\frac{\pi}{2}}\sqrt{1-\frac19 \sin^2 \theta_1 \sin^2 \frac{\theta_2 }{2}}\mathrm{d}\theta_1 \mathrm{d}\theta_2$$</p> <p>$$I=12\int_0^{2\pi}E\left(\frac19 \sin^2 \frac{\theta_2 }{2}\right)\mathrm{d}\theta_2$$</p> <p>($E(m)$ is the complete elliptic integral of the second kind, with parameter $m$; for the Maple people, what you have is $E(k)$ where $k^2=m$)</p> <p>$$I=48\int_0^{\frac{\pi}{2}}E\left(\frac19 \sin^2 \theta_2\right)\mathrm{d}\theta_2$$</p> <p>and letting $m=\sin^2 \theta_2$,</p> <p>$$I=24\int_0^1 \frac1{\sqrt{m}\sqrt{1-m}} E\left(\frac{m}{9}\right)\mathrm{d}m$$</p> <p>which <em>Mathematica</em> evaluates to</p> <p>$$12\pi^2 {}_3 F_2\left(-\frac12,\frac12,\frac12 ; 1,1 ; \frac19\right)$$</p> <p>where ${}_3 F_2$ is a hypergeometric function; further "simplification" can be done using the formula <a href="http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/03/08/04/02/01/06/0001/" rel="nofollow">here</a>.</p>