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Re: Int(exp*BesselI0(sqrt)): Gradshteyn & Ryshik, 6.616.5 ?
Posted:
Oct 5, 2013 9:07 AM


In article <524FF9CF.6F7CD843@freenet.de>, <clicliclic@freenet.de> wrote:
> Axel Vogt schrieb: > > > > The book says: Int( exp(x*a) * BesselI(0, b*sqrt(x^2+1)), x = 1 .. 1) = > > = 2*sinh(a^2+b^2)/sqrt(a^2+b^2) > > > > For a=0 I obtain that it is 2*sinh(b)/b, using Maple (sketch: write Bessel > > as series and integrate termwise; evaluating the infinite series gives it). > > Numerical tests do confirm that. > > > > Is somebody aware of the root of the proof in G&R (can not find an erratum, > > but some usage of the formula) to eliminate the possible typo? > > There is no entry 6.616.5 for > > INT(exp(x*a)*BesselI(0, b*sqrt(1x^2)), x, 1, 1) > > in my 1981 edition of Gradshteyn & Ryzhik; it must have been added > later. I suggest to try the modified evaluation > > 2*sinh(sqrt(a^2+b^2))/sqrt(a^2+b^2) > > which holds at b=0 as well as a=0, since BesselI(0, 0) = 1. It is also > in full agreement with a numerical evaluation of the integral on Derive > giving 2.199018052 for a = 0.3 and b = 0.7. Doing the general integral > symbolically may be difficult; related integrals under 6.616 in my > edition are all taken from Magnus and Oberhettinger, 1948. > > You could file an error report with the G&R editors (after checking the > latest edition). > > Martin.
In my 1980 copy of G&R (English translation of the fourth Russian edition), there is such an entry, with note 3 meaning "added in the third edition". However (as Axel suggests) the value is 2*sinh(sqrt(a^2+b^2))/sqrt(a^2+b^2)
 G. A. Edgar http://www.math.ohiostate.edu/~edgar/



