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?

  • $\begingroup$ Do you mean to find closed form or just numeric approximation? $\endgroup$ – joro Feb 12 '16 at 9:58
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    $\begingroup$ 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
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    $\begingroup$ See Conical Function mathworld.wolfram.com/ConicalFunction.html $\endgroup$ – Alexey Ustinov Feb 13 '16 at 8:46

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): $$\int \limits_0^\infty \frac{\cos{(\sqrt{c-1/4}\,s)}}{\sqrt{\cosh{s}-\cos{\theta}}}\,ds=\frac{\pi}{\sqrt{2}\cosh{(\sqrt{c-1/4}\,\pi)}}F\left(\frac{1}{2}-i\sqrt{c-1/4},\frac{1}{2}+i\sqrt{c-1/4},1,\cos^2{\frac{\theta}{2}}\right)=\frac{\pi}{\sqrt{2}\cosh{(\sqrt{c-1/4}\,\pi)}}P_{-1/2+i\sqrt{c-1/4}}(-\cos{\theta}). $$

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    $\begingroup$ 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

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