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Topic: are the commercial results continuous?
Replies: 59   Last Post: Jul 30, 2012 4:28 PM

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 clicliclic@freenet.de Posts: 940 Registered: 4/26/08
Re: are the commercial results continuous?
Posted: Jul 16, 2012 1:33 AM

Axel Vogt schrieb:
>
> Maple 15 finds them (in terms of arctan and ln) and using
> 15 Digits 'confirms' numerically:
>
> J:=Int(1/((x^2 + x + 1)*sqrt(x^2 - x + 1)), x=-1 .. 0);
> value(J): #combine(%, ln): evala(%);
> evalf(J), evalf(%);
>
> 0.927208724125717, 0.927208724125708
>
> J:=Int( (2+3*x-2*x^2)/((1+x+x^2)^2*sqrt(1-x+x^2)), x=-1 .. 0);
> value(J): #combine(%, ln): evala(%);
> evalf(J), evalf(%);
>
> 0.0614319468787267, 0.0614319468787171
>
> J:=Int( (1-x+3*x^2)/((1+x+x^2)^2*sqrt(1-x+x^2)), x=-1 .. 0);
> value(J): #combine(%, ln): evala(%);
> evalf(J), evalf(%);
>
> 2.60271958298176, 2.60271958298176

Oops. I overlooked that jumps in the Maple antiderivatives do not have
to be inside x = -1 .. 0, of course. So:

Assuming that the Maple antiderivatives for the above integrands show
jumps, are definite integrals whose range includes such a jump position
handled correctly, and do (simplified) symbolic results for them contain
no stuff like Mathematica's Piecewise constructs?

Martin.

PS: This is what I am up to: I want to discuss one integrand of this
type and briefly remark whether Maple/Mathematica produce discontinuous
antiderivatives and are able to correct for the jumps in definite
integrals. My remark applies to the particular integrand discussed, but
it had better reflect the generic behavior of the systems rather than
some exceptional situation. Unfortunately, Mathematica turns out to
sometimes correct for jumps (by adding unsimplifyable Piecewise stuff)
and sometimes not ...