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Re: Integration error
*To*: mathgroup at smc.vnet.net
*Subject*: [mg108922] Re: Integration error
*From*: DrMajorBob <btreat1 at austin.rr.com>
*Date*: Wed, 7 Apr 2010 03:20:54 -0400 (EDT)
You (and all of us) are overlooking MANY subtle points buried in
Integrate. (We have no choice.)
Suffice it to say, always check your results another way. Break things
down into smaller intervals, as you have done.
And don't trust ANYBODY.
Bobby
On Tue, 06 Apr 2010 06:23:03 -0500, Jason Alexander <jalex at lse.ac.uk>
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
>
>
--
DrMajorBob at yahoo.com
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