# [Rd] minor flaw in integrate()

Duncan Murdoch murdoch at stats.uwo.ca
Tue Jul 3 19:21:19 CEST 2007

```On 7/3/2007 11:55 AM, Martin Maechler wrote:
>>>>>> "PetRd" == Peter Ruckdeschel <Peter.Ruckdeschel at uni-bayreuth.de>
>>>>>>     on Tue, 03 Jul 2007 17:26:43 +0200 writes:
>
>     PetRd> Thanks Martin and Duncan for your
>
>     PetRd> 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.
>
>     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

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
>
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