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Re: Re: Re: Incorrect symbolic improper

  • To: mathgroup at smc.vnet.net
  • Subject: [mg103672] Re: [mg103661] Re: [mg103636] Re: Incorrect symbolic improper
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Fri, 2 Oct 2009 08:22:18 -0400 (EDT)
  • References: <200909301141.HAA14962@smc.vnet.net>
  • Reply-to: drmajorbob at yahoo.com

On the other hand, M7 knows that the following diverges:

Integrate[1/(1 + x^a), {x, 0, Infinity},
  Assumptions -> {0 < Re[a] < 1}]

It's common (and inevitable) for general methods to fail for certain  
values of a parameter.

Even the quadratic formula:

Solve[a x^2 + b x + c == 0, x]

{{x -> (-b - Sqrt[b^2 - 4 a c])/(
    2 a)}, {x -> (-b + Sqrt[b^2 - 4 a c])/(2 a)}}

fails utterly for a == 0, and it gives complex roots when the discriminant  
is negative.

To make a long story short, we have to check Mathematica's results using  
other methods or more specific conditions, before betting the farm on them.

Bobby

On Thu, 01 Oct 2009 05:41:53 -0500, Dan Dubin <ddubin 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}]
>
> The result given in v. 7.0.1 is
>
> If[Re[a] > 0, (\[Pi] Csc[\[Pi]/a])/a,
>   Integrate[1/(1 + x^a), {x, 0, \[Infinity]},
>    Assumptions -> Re[a] <= 0]]
>
> 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.
>
>> The integral you tried is a classical one. It is always calculated
>> in the textbooks on application of
>> complex variables to calculation of integrals. Its exact value is
>> therefore, well-known. Evaluate this please:
>>
>> HoldForm[\!\(
>> \*SubsuperscriptBox[\(\[Integral]\), \(-\[Infinity]\), \
>> \(\[Infinity]\)]\(
>> FractionBox[\(Cos[a\ x]\), \(
>> \*SuperscriptBox[\(b\), \(2\)] +
>> \*SuperscriptBox[\(x\), \(2\)]\)] \[DifferentialD]x\)\) = \[Pi]/
>>    b Exp[-a b]]
>>
>> assuming a>0 and b>0. Its evaluation at a=b=1 yields:
>>
>> In[5]:= \[Pi]/b Exp[-a b] /. {a -> 1, b -> 1}
>>
>> Out[5]= \[Pi]/\[ExponentialE]
>>
>> which is obviously the same as the solution returned by Mathematica
>> that you showed. Another point that when I evaluated your
>> integral with parameter
>>
>> In[6]:= Integrate[Cos[a x]/(1 + x^2), {x, -\[Infinity], \[Infinity]},
>>  Assumptions -> a \[Element] Reals]
>>
>>
>> Out[6]= \[ExponentialE]^-Abs[a] \[Pi]
>>
>>  it returned (by my machine Math 6.0, Windows XP)
>>
>> \[ExponentialE]^-Abs[a] \[Pi]
>>
>> that is in line with the above exact solution, rather than with the  
>> value
>>
>> \[Pi] Cosh[a]
>>
>> that you report. So may be you still have a problem.
>>
>> Alexei
>>
>>
>> Below is a definite integral that Mathematica does incorrectly.
>> Thought someone might like to know:
>>
>> In[62]:= Integrate[Cos[x]/(1 + x^2), {x, -\[Infinity], \[Infinity]}]
>>
>> Out[62]= \[Pi]/E
>>
>> What a pretty result--if it were true. The correct answer is \[Pi]*Cosh
>> [1], which can be checked by adding a new parameter inside the
>> argument of Cos and setting it to 1 at the end:
>>
>> In[61]:= Integrate[Cos[a x]/(1 + x^2), {x, -\[Infinity], \[Infinity]},
>>   Assumptions -> a \[Element] Reals]
>>
>> Out[61]= \[Pi] Cosh[a]
>>
>> Regards,
>>
>> Jason Merrill
>>
>> --
>> Alexei Boulbitch, Dr., habil.
>> Senior Scientist
>>
>> IEE S.A.
>> ZAE Weiergewan
>> 11, rue Edmond Reuter
>> L-5326 Contern
>> Luxembourg
>>
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>> Fax:   +352 2454 3566
>>
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>>
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>


-- 
DrMajorBob at yahoo.com


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