Re: Integration result depends on variable name / problem with BesselJ Integral representation

*To*: mathgroup at smc.vnet.net*Subject*: [mg127209] Re: Integration result depends on variable name / problem with BesselJ Integral representation*From*: JUN <noeckel at gmail.com>*Date*: Sun, 8 Jul 2012 19:05:08 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net

On Saturday, July 7, 2012 2:30:49 AM UTC-7, Kevin J. McCann wrote: > This is a real issue. I tried the integrals as you suggested below, and > indeed, I get the same results. > > If, however, I change psi to the Greek letter psi in the first version, > I get zero. Not sure what is going on, but it is a bug. > > Kevin > > On 7/2/2012 10:24 PM, richard wrote: > > Dear all, > > > > the following integral is the integral representation of the bessel function of first kind, second order. But mathematica (8.0.4.0) gives me wrong results, depending on the variable name, it seems: > > first with x: > > Integrate[Sin[2 x] Exp[I t Cos[x - psi]], {x, 0, 2 \[Pi]}] > > (8 I (-t Cos[t] + Sin[t]))/t^2 > > > > then x substituted with p: > > Integrate[Sin[2 p] Exp[I t Cos[p - psi]], {p, 0, 2 \[Pi]}] > > 0 > > > > How can the result depend on the variable name? There are no values assigned to x or p. The analytically obtained solution should be > > 2 \[Pi] I^2 BesselJ[2, t] Sin [2 psi] > > > > Regards, > > Richard > > Regarding Kevin's observations -- this is consistent with the lexicographical ordering of the variable names: Sort[{psi, x, \[Psi]}] Apparently the result is zero whenever the integration variable comes before the "shift" variable in the exponential -- e.g., Cos[x - psi] gets transformed to Cos[psi - x]. Anyway, this is a long-standing bug that inexplicably has been patched up only incorrectly so far. See the discussion under this post: http://mathematica.stackexchange.com/a/2743/245 where I also link to the earlier MathGroup post acknowledging the bug. One work-around for it is to replace the original integral by Integrate[Sin[2 (x + psi)] Exp[I t Cos[x]], {x, 0, 2 \[Pi]}] where I have made a substitution of variables to x - psi -> x. Jens