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From: Arthur Norman <acn1@ca...>  20120929 12:39:36

I note that int(exp(x^2),x,y,infinity) seems to yield sqrt(pi)*(erf(y)+1)/2 for me when I have specfn loaded, and then when I substitute y=0 it tidies up nicely. So I guess it will be possible but Term is just about to start here and I will be rather loaded for the coming week  I hoep to look into it after that. Arthur On Sat, 29 Sep 2012, Andrey G. Grozin wrote: > In the current reduce (svn 1759), if I say > > load_package specfn; int(exp(x^2),x,0,infinity); > > this produces > > sqrt(pi)*erf(infinity) >  > 2 > > If I don't load specfn, I get > > sqrt(pi) >  > 2 > > So, loading specfn reduces the quality :( Can this be avoided? > > Andrey > 
From: Rainer Schöpf <rainer.schoepf@gm...>  20120929 12:11:18

On Sat, 29 Sep 2012 at 18:30 +0700, Andrey G. Grozin wrote: > In the current reduce (svn 1759), if I say > > load_package specfn; int(exp(x^2),x,0,infinity); > > this produces > > sqrt(pi)*erf(infinity) >  > 2 > > If I don't load specfn, I get > > sqrt(pi) >  > 2 > > So, loading specfn reduces the quality :( Can this be avoided? Yes. The problem are these two rules in the specfn package: int(1/e^(~tt^2),~tt,0,~z) => erf(z)/2*sqrt(pi), int(1/e^(~tt^2),~tt,~z,infinity) => erfc(z)/2*sqrt(pi), which are used even if z is infinity. The obvious solution is to replace them by int(1/e^(~tt^2),~tt,0,~z) => erf(z)/2*sqrt(pi) when z freeof infinity, int(1/e^(~tt^2),~tt,~z,infinity) => erfc(z)/2*sqrt(pi) when z freeof infinity I'll make the change later today. Rainer 
From: Andrey G. Grozin <A.G.Grozin@in...>  20120929 11:47:40

In the current reduce (svn 1759), if I say load_package specfn; int(exp(x^2),x,0,infinity); this produces sqrt(pi)*erf(infinity)  2 If I don't load specfn, I get sqrt(pi)  2 So, loading specfn reduces the quality :( Can this be avoided? Andrey 