does it make sense ? Provisio,
in [Mathematica]
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From: michael partensky on 25 Feb 2010 01:51 Thanks, Daniel. Does it mean that it is worth checking separately all the "unresolved" expressions in a provisio, or it only applies to some special cases (say, Integrals). Best Michael Partenskii On Tue, Feb 23, 2010 at 9:59 AM, Daniel Lichtblau <danl(a)wolfram.com> wrote: > michael partensky wrote: > >> Sorry, Daniel. The Yellow background have been lost. I use Mathematica >> 7.0.1, WinXP. >> Here is the output: >> >> *if[Re(t) < 0, Sqrt[2 \[Pi]] E^(t^2/2) t (erf(t/Sqrt[2]) + 1) + 2, >> Integrate[E^(t Sqrt[u] - u/2), {u, 0, \[Infinity]}, >> Assumptions -> Re(t) >= 0]]* >> >> from evaluating the expression : >> * >> md[t] = Integrate[Exp[t u^(1/2) - u/2], {u, 0, \[Infinity]}]* >> ======================================================================= >> >> And here is the same with two explicitly made assumptions from the If >> statement. The results are analytical in both cases >> * >> In = md[t] = >> Integrate[Exp[t u^(1/2) - u/2], {u, 0, \[Infinity]}, >> Assumptions -> Re[t] < 0] >> Out = Sqrt[2 \[Pi]] E^(t^2/2) t (erf (t/Sqrt[2]) + 1) + 2 >> >> In = md[t] = >> Integrate[Exp[t u^(1/2) - u/2], {u, 0, \[Infinity]}, >> Assumptions -> Re[t] > 0] >> Out = Sqrt[2 \[Pi]] E^(t^2/2) t (erf (t/Sqrt[2]) + 1) + 2* >> >> >> Did you get a different result? >> >> Thanks >> Michael >> > > This looks visually like the result I obtain. You originally had minus > signs where now you have Set (that is, "="), and that had the effect of > altering things a bit. > > *So if I understand correctly, your point is that the proviso (that is, > conditional result) is not needed because the result is actually the same > regardless of sign of Re[t]. That is correct; Mathematica does not realize > the proviso is not in fact needed.* > > Possibly some day we will manage to improve on this. > > Daniel >
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