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Re: Integration error

  • To: mathgroup at smc.vnet.net
  • Subject: [mg108908] Re: Integration error
  • From: dh <dh at metrohm.com>
  • Date: Tue, 6 Apr 2010 08:21:13 -0400 (EDT)
  • References: <hpf5ij$or8$1@smc.vnet.net>

On 06.04.2010 13:22, Jason Alexander wrote:
> Hello all,
>
> I'm getting a strange result when calculating what I thought was a relatively straightforward integral. If I evaluate the following expression, I receive an imaginary result when it should, in fact, equal 1. (This happens in Mathematica 7.0.1 on Mac OS X):
>
> In[1]:=
> Integrate[(
>   4 (8/(2320 + 3 x (-96 + 3 x)) + 8/(1460 + 3 x (-76 + 3 x)) + 6/(
>      800 + 3 x (-56 + 3 x)) + 4/(340 + 3 x (-36 + 3 x)) + 1/(
>      80 + 3 x (-16 + 3 x))))/(9 \[Pi]), {x, -Infinity, Infinity}]
>
> Out[1]:=
> 1/3 + (52 I)/9
>
> However, if I break the integral into two parts, as below, then I get an answer of 1.
>
> In[2]:=
> Integrate[(
>     4 (8/(2320 + 3 x (-96 + 3 x)) + 8/(1460 + 3 x (-76 + 3 x)) + 6/(
>        800 + 3 x (-56 + 3 x)) + 4/(340 + 3 x (-36 + 3 x)) + 1/(
>        80 + 3 x (-16 + 3 x))))/(9 \[Pi]), {x, -Infinity, 0}] +
>    Integrate[(
>     4 (8/(2320 + 3 x (-96 + 3 x)) + 8/(1460 + 3 x (-76 + 3 x)) + 6/(
>        800 + 3 x (-56 + 3 x)) + 4/(340 + 3 x (-36 + 3 x)) + 1/(
>        80 + 3 x (-16 + 3 x))))/(9 \[Pi]), {x, 0, Infinity}] // Simplify
>
> Out[2]:=
> 1
>
> Furthermore, if I compute the antiderivative, I get the following:
>
> (1/(9 \[Pi]))4 (-(1/24) ArcTan[1/4 (8 - 3 x)] -
>     1/4 ArcTan[1/4 (28 - 3 x)] - 1/3 ArcTan[1/4 (38 - 3 x)] +
>     2/3 ArcTan[3/4 (-16 + x)] + 1/3 ArcTan[3/4 (-6 + x)] +
>     1/3 ArcTan[1/4 (-38 + 3 x)] + 1/4 ArcTan[1/4 (-28 + 3 x)] +
>     1/24 ArcTan[1/4 (-8 + 3 x)])
>
> And if I define f[x] to be the above, and then ask Mathematica to calculate
>
> Limit[f[n] - f[-n], n ->  Infinity]
>
> I get 1, as expected.
>
> Is this a bug, or am I overlooking some subtle point in how Mathematica computes improper integrals?
>
>
> Cheers,
>
> Jason
>
> --
> Dr. J. McKenzie Alexander
> Department of Philosophy, Logic and Scientific Method
> London School of Economics
> Houghton Street, London WC2A 2AE
>
>
Hi Jason,
looks like  a bug to me. You have a rational real integrand without 
poles in R, therefore, the definite integral must be real.
Please report this bug to Wolfram.
Daniel


-- 

Daniel Huber
Metrohm Ltd.
Oberdorfstr. 68
CH-9100 Herisau
Tel. +41 71 353 8585, Fax +41 71 353 8907
E-Mail:<mailto:dh at metrohm.com>
Internet:<http://www.metrohm.com>



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