[Rd] minor flaw in integrate()
Peter.Ruckdeschel at uni-bayreuth.de
Tue Jul 3 17:26:43 CEST 2007
Thanks Martin and Duncan for your
Martin Maechler wrote:
>>>>>> "DM" == Duncan Murdoch <murdoch at stats.uwo.ca>
>>>>>> 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.
not quite convinced --- or more precisely:
[ Let's assume lower = upper = Inf here,
case lower = upper = -Inf is analogue ]
I'd say it depends on whether the (Lebesgue-) integral
integral(f, lower = <some finite value>, upper = Inf)
is well defined. Then, by dominated convergence, the integral should
default to 0.
But I admit that then a test
is.finite(integrate(f, lower = <some finite value>, upper = Inf)$value)
would be adequate, too, which makes evaluation a little more expensive :-(
integrate(f, lower = <some finite value>, upper = Inf)
throws an error, I agree, there should be a NaN ...
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