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Integration error

  • To: mathgroup at smc.vnet.net
  • Subject: [mg108885] Integration error
  • From: Jason Alexander <jalex at lse.ac.uk>
  • Date: Tue, 6 Apr 2010 07:23:03 -0400 (EDT)

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



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