Re: Real integrand->complex result.

*To*: mathgroup at smc.vnet.net*Subject*: [mg20308] Re: [mg20293] Real integrand->complex result.*From*: David Withoff <withoff at wolfram.com>*Date*: Fri, 15 Oct 1999 20:20:40 -0400*Sender*: owner-wri-mathgroup at wolfram.com

> I get a complex result of the form f(t)=(a+bi)g(t) when evaluating a > real integrand F(x,t)*phi(x) over the interval {x:0<x<L). All of the > variables have been properly declared as real and the functions have > been declared as real for real arguments (using the package "ReIm"). > > To see the problem more clearly, the code can be copied into Mathematica > (3 or 4) and executed. (The Traditional or Standard Forms will not paste > into a text document such as this) > > When evaluating the integral > > \!\(\[Integral]\_0\%\[ScriptL]\( simpF\[CurlyPhi]\_n\) \[DifferentialD]x\) > > over the interval (0,ScriptL) where ScriptL=1. The integrand > \!\(simpF\[CurlyPhi]\_n\ is, for n=1, given by > > \!\(TraditionalForm > \`\(-\(1\/6\)\)\ \[ExponentialE]\^\(\(\(-3\)\ \[Tau]\)/2\)\ > \((\((3 + 2\ \[Pi]\^2)\)\ \(cos(\[Pi]\ x)\) - > 6\ \[ExponentialE]\^\(x/2\) + 3)\)\ \((cos(3.66558239083868908`\ x) + > sin(3.66558239083868908`\ x))\)\) > > The evaluation of the integral is also quite slow, even for n=2, even > when the expression is simplified as much as possible, and the final > program will need to evaluate a larger number of terms. > > Thanks in advance, Bill The imaginary 0.0 I parts are an artifact of numerical error. You can discard those parts using Chop. For example: In[1]:= Integrate[(-(1/6) (3 - 6 E^(x/2) + (3 + 2 Pi^2) Cos[Pi x])* (Cos[3.665582390838689 x] + Sin[3.665582390838689 x]))/ E^((3 tau)/2), {x, 0, L}] //Chop Out[1]= -(22.7991 + 1.38775 (-0.589747 Cos[3.66558 L] + 0.5 L > 1. E Cos[3.66558 L] - 16.839 Cos[3.14159 L] Cos[3.66558 L] - > 14.4319 Cos[3.66558 L] Sin[3.14159 L] + 0.589747 Sin[3.66558 L] - 0.5 L > 1.3159 E Sin[3.66558 L] + > 16.839 Cos[3.14159 L] Sin[3.66558 L] - (3 tau)/2 > 14.4319 Sin[3.14159 L] Sin[3.66558 L])) / (6 E ) That calculation finished in a fraction of a second when I tried it. The slowness in your calculation is almost certainly due to the ReIm package, which will consume a lot of time and will not be helpful, because there isn't anything that this or any other package can do to avoid numerical error. Dave Withoff Wolfram Research