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