10
$\begingroup$

The following integral $$\int\limits_0^\infty \frac{\cos{\left(\frac{1}{2}\sqrt{3}s\right)}}{\sqrt{\cosh{s}-\cos{\theta}}}\,ds$$ can be found in the paper

Tevian Dray and Gerard 't Hooft, The gravitational shock wave of a massless particle, Nuclear Physics B 253 (1985) 173--188, doi:10.1016/0550-3213(85)90525-5.

They write that they "have not attempted to perform the integration explicitly". Was this integral ever calculated explicitly?

$\endgroup$
4
  • $\begingroup$ Do you mean to find closed form or just numeric approximation? $\endgroup$ – joro Feb 12 '16 at 9:58
  • 2
    $\begingroup$ Wolfram alpha and sage don't seem to know how to compute this integral exactly. $\endgroup$ – Ben McKay Feb 12 '16 at 10:52
  • $\begingroup$ @joro I mean to find a closed form, if this is possible. $\endgroup$ – Zurab Silagadze Feb 12 '16 at 12:49
  • 1
    $\begingroup$ See Conical Function mathworld.wolfram.com/ConicalFunction.html $\endgroup$ – Alexey Ustinov Feb 13 '16 at 8:46
5
$\begingroup$

A closed form exists in Gradshteyn and Ryzhik, 8.842.1.

As a comment to this answer:

Referring to the Gradshteyn and Ryzhik, the following result is given in http://arxiv.org/abs/hep-th/9408169 (On Gravitational Shock Waves in Curved Spacetimes, by K. Sfetsos): $$\int \limits_0^\infty \frac{\cos{(\sqrt{c-1/4}\,s)}}{\sqrt{\cosh{s}-\cos{\theta}}}\,ds=\frac{\pi}{\sqrt{2}\cosh{(\sqrt{c-1/4}\,\pi)}}F\left(\frac{1}{2}-i\sqrt{c-1/4},\frac{1}{2}+i\sqrt{c-1/4},1,\cos^2{\frac{\theta}{2}}\right)=\frac{\pi}{\sqrt{2}\cosh{(\sqrt{c-1/4}\,\pi)}}P_{-1/2+i\sqrt{c-1/4}}(-\cos{\theta}). $$

$\endgroup$
3
  • 2
    $\begingroup$ Would you mind sharing it with those of us who don't have access to the book? In Google Book's preview the page with with the formula is unfortunately omitted. $\endgroup$ – eins6180 Feb 13 '16 at 6:22
  • $\begingroup$ See also mathworld.wolfram.com/ConicalFunction.html $\endgroup$ – Alexey Ustinov Feb 13 '16 at 8:45
  • $\begingroup$ If this is in terms of the Legendre function of the first kind (en.wikipedia.org/wiki/Legendre_function), then it's like saying "Erf is the solution to the integral...". Although one can (as usual) reduce to a certain hypergeometric function, which I guess as explicit as one can usually get in these cases. $\endgroup$ – theHigherGeometer Feb 13 '16 at 8:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.