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