Re: Boolean Integral

*To*: mathgroup at smc.vnet.net*Subject*: [mg62396] Re: Boolean Integral*From*: "antononcube" <antononcube at gmail.com>*Date*: Wed, 23 Nov 2005 01:12:11 -0500 (EST)*References*: <dlupfj$nb2$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

These settings seem to work: In[8]:= NIntegrate[chi[foo[x], y1, y2], {x, 0, 20}, MinRecursion -> 4, MaxRecursion -> 200, SingularityDepth -> 1000] NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration being 0, oscillatory integrand, or insufficient WorkingPrecision. If your integrand is oscillatory try using the option Method->Oscillatory in NIntegrate. Out[8]= 1.33977 As you said, the reason for NIntegrate to stop with the default settings is because the sampling points of the first integration step are where the integrand is 0. Increasing the number of initial sampling points with the option MinRecursion allows NIntegrate to see that the integrand is non-zero, and on what areas the computational efforts should be concentrated. These plots with different values for the parameter minrec (eg, 0, 2, 4) might clarify my explanation further: In[43]:= minrec = 4; In[44]:=pnts = Reap[NIntegrate[chi[foo[x], y1, y2], {x, 0, 20}, MinRecursion -> minrec, MaxRecursion -> 200, SingularityDepth -> 1000, EvaluationMonitor :> Sow[x]]][[2,1]]; In[45]:= Length[pnts] Out[45]= 4972 In[46]:=ListPlot[Transpose[{pnts, Range[Length[pnts]]}]] Further, if the zeroes of your function foo are known (e.g. foo[x_]:=Sin[x]) the piecewise handling of NIntegrate will simplify the integrand to the non-zero parts of it only (and the integration will be much faster). Again you can see this with the plots above, and using Length[pnts]. In[58]:=pnts = Reap[NIntegrate[chi[Sin[x], y1, y2], {x, 0, 20},EvaluationMonitor :> Sow[x]]][[2,1]]; In[59]:=Length[pnts] Out[59]=77 In[60]:=ListPlot[Transpose[{pnts,Range[Length[pnts]]}]] Anton Antonov, Wolfram Research, Inc.