Re: False result with Integrate ?

*To*: mathgroup at smc.vnet.net*Subject*: [mg8745] Re: [mg8711] False result with Integrate ?*From*: David Withoff <withoff>*Date*: Sat, 20 Sep 1997 22:28:11 -0400*Sender*: owner-wri-mathgroup at wolfram.com

> Why : Integrate[Integrate[ Sqrt[(x-y)^2],{x,0,1}],{y,0,1}] > gives 0 with math2.2 or math3.0. > > The exact result is 1/3 !! > > Gilles > EDF/DER You can get a correct result for this integral by breaking it up into pieces, to allow for the singularity in the range of integration. For example: In[1]:= Integrate[Integrate[ Sqrt[(x-y)^2],{x,0,y}],{y,0,1}] + Integrate[Integrate[ Sqrt[(x-y)^2],{x,y,1}],{y,0,1}] 1 Out[1]= - 3 The easiest way to understand what goes wrong if you don't break the integral up into pieces is to look at the integrals one at a time. Here is the inner integral: In[2]:= Integrate[ Sqrt[(x-y)^2],{x,0,1}] 2 Sqrt[(-1 + y) ] (-1 + 2 y) Out[2]= -------------------------- 2 (-1 + y) In this integral, the symbol y is an unknown parameter. The Integrate function does not know if this parameter is positive or negative or complex, and must make assumptions about this parameter in order to proceed. A close look at this result shows that the Integrate function effectively assumes that y is negative. This assumption causes a problem when that result is used in the integral over y, since the integral over y requires a result that is valid for y between 0 and 1. There have been dramatic improvements in the Integrate function in Version 3.0, both in reporting assumptions about parameters, and in the handling of singularities in the range of integration. This is a very difficult problem, however, and there are a few examples, such as this one, where singularities are missed, or where assumptions are not reported. There are some additional notes about definite integration in the technical support section of the Wolfram Research web site (http://www.wolfram.com/support/Kernel/Symbols/System/Integrate.html). Dave Withoff Wolfram Research