Re: Re: Incorrect symbolic improper integral
- To: mathgroup at smc.vnet.net
- Subject: [mg103739] Re: [mg103715] Re: Incorrect symbolic improper integral
- From: DrMajorBob <btreat1 at austin.rr.com>
- Date: Sun, 4 Oct 2009 05:36:55 -0400 (EDT)
- References: <200909301141.HAA14962@smc.vnet.net> <ha212q$n8u$1@smc.vnet.net>
- Reply-to: drmajorbob at yahoo.com
> I do not think that DrMajorBob comments about Solve are applicable in
> the cases below:
It looks to me as if you've repeatedly proved my point, which was:
Check ALL Mathematica's results with another method and/or more specific
conditions, before betting the farm on them.
Using the very same Integrate function on the same integrand with "a"
symbolic every time (though varying its range) isn't a sufficient check,
as you've demonstrated.
I'm thinking more along the lines of
NIntegrate[1/(1 + x^(1/2 + I)) // Im, {x, 0, Infinity}]
That's both a specific value of "a" and a different method, and the result
strongly indicates divergence for a value that, according to some of
Mathematica's results, should give convergence.
So again... Mathematica isn't always right. So check it, if the result is
important.
Bobby
On Sat, 03 Oct 2009 08:04:04 -0500, ADL <alberto.dilullo at tiscali.it> wrote:
> Following Dan Dubin's comment, I made some tests, reported below with
> some visual simplification, which show that there are some troubles in
> assumptions management in Mathematica.
> Note that, when I gave the parameter a specific value, real or
> complex, I found no problems with this integral.
>
> My version is 7.0.1 on Windows.
> I do not think that DrMajorBob comments about Solve are applicable in
> the cases below:
>
> Assuming[a > 0, Integrate[1/(1 + x^a), {x, 0, Infinity}]]
> ==> If[a > 1, (Pi*Csc[Pi/a])/a, (*otherwise...*)]
> -OK-
>
> Assuming[a < 0, Integrate[1/(1 + x^a), {x, 0, Infinity}]]
> ==> Message: Integrate::idiv:Integral does not converge on {0,
> Infinity}
> ==> Integrate[(1 + x^a)^(-1), {x, 0, Infinity}]
> -OK-
>
> Assuming[a != 0, Integrate[1/(1 + x^a), {x, 0, Infinity}]]
> ==> If[Re[a] > 0, (Pi*Csc[Pi/a])/a, (*otherwise...*)]
> -WRONG-
>
> Assuming[Element[a, Reals], Integrate[1/(1 + x^a), {x, 0, Infinity}]]
> ==> If[a > 0, (Pi*Csc[Pi/a])/a, (*otherwise...*)]
> -WRONG-
>
> Assuming[a < 0, Integrate[1/(1 + x^a), {x, 0, Infinity}]]
> ==> Message: Integrate::idiv:Integral does not converge on {0,
> Infinity}
> ==> Integrate[(1 + x^a)^(-1), {x, 0, Infinity}]
> -OK-
>
> Assuming[a < 1, Integrate[1/(1 + x^a), {x, 0, Infinity}]]
> ==> If[a > 0, (Pi*Csc[Pi/a])/a, (*otherwise...*)]
> -WRONG-
>
> Assuming[a < -1, Integrate[1/(1 + x^a), {x, 0, Infinity}]]
> ==> Message: Integrate::idiv:Integral does not converge on {0,
> Infinity}
> ==> Integrate[(1 + x^a)^(-1), {x, 0, Infinity}]
> -OK-
>
>
> Regards
> ADL
>
> On Oct 1, 12:42 pm, Dan Dubin <ddu... at ucsd.edu> wrote:
>> OK, many people have replied that the given integral was in fact done
>> correctly by Mathematica. Here's a related integral that is not done
>> correctly:
>>
>> Integrate[1/(1 + x^a),{x,0,Infinity}]
>> ...
>> This result is incorrect in the range 0<Re[a]<1. In this range the
>> integral diverges, and is not given by the above cosecant expression.
>> | Professor Dan Dubin
>> | Dept of Physics , Mayer Hall Rm 3531,
>> | UC San Diego La Jolla CA 92093-0319
>> | phone (858) - 534-4174 fax: (858)-534-0173
>> | ddu... at ucsd.edu
>
>
--
DrMajorBob at yahoo.com