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Re: Re: does it make sense ? Provisio,

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  • Subject: [mg107734] Re: [mg107624] Re: [mg107608] does it make sense ? Provisio,
  • From: michael partensky <partensky at gmail.com>
  • Date: Thu, 25 Feb 2010 01:51:26 -0500 (EST)

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 at 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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