[Date Index]
[Thread Index]
[Author Index]
Re: Inconsistent behaviour of Integrate
*To*: mathgroup at smc.vnet.net
*Subject*: [mg112485] Re: Inconsistent behaviour of Integrate
*From*: Andreas Maier <andimai at web.de>
*Date*: Fri, 17 Sep 2010 06:40:44 -0400 (EDT)
*References*: <i6q0no$l8i$1@smc.vnet.net>
On Sep 15, 10:39 am, Bill Rowe <readn... at sbcglobal.net> wrote:
> On 9/14/10 at 5:12 AM, andi... at web.de (Andreas Maier) wrote:
>
> >I'm using Mathematica 7.0.1.0 on Linux x86 (64bit). I have a
> >notebook file, where I integrate the same integral twice:
> >In[1]:= Integrate[Sqrt[(x - 1/2)^2 + (y - 1/2)^2], {x, 0, 1}, {y, 0,
> >1}] Out[1]= 1/6 (Sqrt[2] + ArcSinh[1])
> >In[2]:= Integrate[Sqrt[(x - 1/2)^2 + (y - 1/2)^2], {x, 0, 1}, {y, 0,
> >1}] Out[2]= 1/24 (4 Sqrt[2] + Log[17 + 12 Sqrt[2]])
> >As you can see from the output, integrating the same integral a
> >second time gives a different result. If I integrate the same
> >integral a third and a fourth time I always get the second result
> >again. Only if I restart the mathematica kernel, I get the first
> >result again. The results are equivalent, since
> >Log[17 + 12 Sqrt[2]] = Log[(1 + Sqrt[2])^4] = 4* Log[(1 + Sqrt[2]) =
=
> >4* ArcSinh[1]
> >but somehow Mathematica seems to be able to do this simplification
> >only once. Is this inconsistent behaviour a bug?
>
> No. This issue has been discussed here previously. Basically,
> default time constraints on Integrate cause it to return an
> answer before some transformations have been done. The second
> time the same integral is done, Integrate makes use of the
> cached results from the first which allows further
> transformations to be done within the default time constraints.
>
> >Is there a possibility to give mathematica a hint, so that he always
> >find the first solution 1/6 (Sqrt[2] + ArcSinh[1]) to the integral?
>
> You can use ClearSystemCache to clear out the cached results
> from the previous evaluation. That is
>
> In[1]:= Integrate[
> Sqrt[(x - 1/2)^2 + (y - 1/2)^2], {x, 0, 1}, {y, 0, 1}]
>
> Out[1]= (1/6)*(Sqrt[2] + ArcSinh[1])
>
> In[2]:= ClearSystemCache[]
>
> In[3]:= Integrate[
> Sqrt[(x - 1/2)^2 + (y - 1/2)^2], {x, 0, 1}, {y, 0, 1}]
>
> Out[3]= (1/6)*(Sqrt[2] + ArcSinh[1])
>
> But given the time it requires to evaluate the integral, why do
> it twice if you get an acceptable answer the first time?
Call me conservative, but I expected to get the same result, when I'm
doing the same calculation again.
So I thought, this could be a bug in Mathematica and I gave the
simplest example, where I encountered this "bug".
Of course it doesn't make sense to compute this special integral
twice, especially since it's value is well
known in the literature (see http://mathworld.wolfram.com/SquarePointPickin=
g.html).
But I wanted to compute
a few more complex integrals of this kind (see
http://mathworld.wolfram.com/UnitSquareIntegral.html).
But since mathematica seems to generate not very consistent results
with such a simple box integrals, I'm
wondering how useful the results will be for more complex ones.
Andreas Maier
Prev by Date:
**Re: ANOVA question**
Next by Date:
**Re: Inconsistent behaviour of Integrate**
Previous by thread:
**Re: Inconsistent behaviour of Integrate**
Next by thread:
**Re: Inconsistent behaviour of Integrate**
| |