# Re: [Rd] minor flaw in integrate()

From: Peter Ruckdeschel <Peter.Ruckdeschel_at_uni-bayreuth.de>
Date: Tue, 03 Jul 2007 17:26:43 +0200

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 :-(

If, otoh

integrate(f, lower = <some finite value>, upper = Inf)

throws an error, I agree, there should be a NaN ... Best, Peter

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