On 7/3/2007 11:55 AM, Martin Maechler wrote: >>>>>> "PetRd" == Peter Ruckdeschel <[EMAIL PROTECTED]> >>>>>> on Tue, 03 Jul 2007 17:26:43 +0200 writes: > > PetRd> Thanks Martin and Duncan for your > PetRd> comments, > > PetRd> Martin Maechler wrote: > >>>>>>> "DM" == Duncan Murdoch <[EMAIL PROTECTED]> > >>>>>>> on Mon, 02 Jul 2007 21:56:23 -0400 writes: > >> > DM> On 28/06/2007 5:05 PM, Peter Ruckdeschel wrote: > >> >> Hi, > >> >> > >> >> I noticed a minor flaw in integrate() from package stats: > >> >> > >> >> Taking up arguments lower and upper from integrate(), > >> >> > >> >> if (lower == Inf) && (upper == Inf) > >> >> > >> >> or > >> >> > >> >> if (lower == -Inf) && (upper == -Inf) > >> >> > >> >> integrate() calculates the value for (lower==-Inf) && (upper==Inf). > >> >> > >> >> Rather, it should return 0. > >> > DM> Wouldn't it be better to return NA or NaN, for the same reason > Inf/Inf > DM> doesn't return 1? > >> > DM> Duncan Murdoch > >> > >> Yes indeed, I think it should return NaN. > > PetRd> not quite convinced --- or more precisely: > > PetRd> [ Let's assume lower = upper = Inf here, > PetRd> case lower = upper = -Inf is analogue ] > > PetRd> I'd say it depends on whether the (Lebesgue-) integral > > PetRd> integral(f, lower = <some finite value>, upper = Inf) > > PetRd> is well defined. Then, by dominated convergence, the integral > should > PetRd> default to 0. > > PetRd> But I admit that then a test > > PetRd> is.finite(integrate(f, lower = <some finite value>, upper = > Inf)$value) > > PetRd> would be adequate, too, which makes evaluation a little more > expensive :-( > > No, that's not the Duncan's point I agreed on. > The argument is different: > > consider Int(f, x, x^2) > Int(f, x, 2*x) > Int(f, x, exp(x)) > etc, > These could conceivably give very different values, > with different limits for x --> Inf > > Hence, Int(f, Inf, Inf) > > is mathematically undefined, hence NaN
In the case Peter was talking about, those limits would all be zero. But I don't think we could hope for the integrate() function in R to recognize integrability. For example, > integrate(function(x) 1/x, 1e8, Inf) 1.396208e-05 with absolute error < 2.6e-05 where the correct answer is Inf, since the integral is divergent. So I'd be fairly strongly opposed to returning 0. Whether we return NaN or NA is harder: I suspect the reason Inf-Inf or Inf/Inf is NaN is because this is handled on most platforms by the floating point hardware or the C run-time, rather than because we've made a deliberate decision for that. Do we have other cases where NaN is used to mean "unable to determine the answer"? Duncan Murdoch > Martin > > PetRd> If, otoh > > PetRd> integrate(f, lower = <some finite value>, upper = Inf) > > PetRd> throws an error, I agree, there should be a NaN ... > PetRd> Best, Peter > > ______________________________________________ > R-devel@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-devel ______________________________________________ R-devel@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-devel
