       Re: Re: If Integrate returns no result, can we conclude that no closed-form

• To: mathgroup at smc.vnet.net
• Subject: [mg87808] Re: [mg87784] Re: If Integrate returns no result, can we conclude that no closed-form
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Thu, 17 Apr 2008 06:59:18 -0400 (EDT)
• References: <fu4lpc\$sk9\$1@smc.vnet.net> <200804170233.WAA21093@smc.vnet.net>

```On 17 Apr 2008, at 11:33, David W.Cantrell wrote:
> Szabolcs_Horv=E1t <szhorvat at gmail.com> wrote:
>> The documentation says:
>>
>> "In the most convenient cases, integrals can be done purely in
>> terms of
>> elementary functions such as exponentials, logarithms and
>> trigonometric
>> functions. In fact, if you give an integrand that involves only such
>> elementary functions, then one of the important capabilities of
>> Integrate is that if the corresponding integral can be expressed in
>> terms of elementary functions, then Integrate will essentially always
>> succeed in finding it."
>>
>> =
http://reference.wolfram.com/mathematica/tutorial/IntegralsThatCanAndCann
>> otBeDone.html
>>
>> How precise is this?  Can one rely on this information?
>
> I suppose that depends on the definition of "essentially".  ;-)
>
>> Is it really
>> true that if Mathematica cannot integrate an expression made up of
>> elementary functions, then no closed-form result exists?
>
> No. Consider, for example,
>
> In:= Integrate[D[x Sin[x^ArcSin[x]], x], x]
>
> Out= Integrate[x^(1 + ArcSin[x])*Cos[x^ArcSin[x]]*(ArcSin[x]/x +
>          Log[x]/Sqrt[1 - x^2]) + Sin[x^ArcSin[x]], x]
>
> which was done in Versionn 6.0.2 under Windows XP.

This is not an "elementary function" (as far as the Risch algorithm is
concerned).

The class of elementary functions consists of rational functions,
exponentials, logarithms, trigonometric, inverse trigonometric,
hyperbolic, and inverse hyperbolic functions, the solutions of a
polynomial equation whose coefficients are elementary functions) and
any finite nesting (composition) of elementary functions. It does not
include the power of an elementary function to another elementary
function so your example actually shows the opposite of what you seem
to claim, namely that Mathematica has correctly concluded (by using
the Risch algorithm) that the integral cannot be expressed in terms of
elementary functions.

Andrzej Kozlowski

```

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