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Re: Re: definite integration of 1/a
*To*: mathgroup at smc.vnet.net
*Subject*: [mg86453] Re: [mg86443] Re: definite integration of 1/a
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Tue, 11 Mar 2008 05:33:22 -0500 (EST)
*References*: <200803110800.DAA24976@smc.vnet.net>
On 11 Mar 2008, at 09:00, Bill Rowe wrote:
> On 3/10/08 at 2:04 AM, chris at chiasson.name (Chris Chiasson) wrote:
>
>> In[1]:= Integrate[1/a,{a,1,ablah},Assumptions\[RuleDelayed]ablah>0]
>
>> Out[1]= \!\(If[ablah > 1, Log[ablah],
>> Integrate[1\/a, {a, 1, ablah}, Assumptions \[Rule] ablah
>> \[LessEqual] 1]]\)
>
>> I was expecting 1/a. Is there something I can do to get that? Thank
>> you.
>
> If you were expecting a function of a, then you should be doing
> an indefinite integral rather than a definite integral, i.e.,
>
> Integrate[1/a, a, Assumptions :> a > 0]
>
> log(a)
>
It seems to me that this is only an illusion that Assumptions do
anything in the case of indefinite integration. I can't see how they
could really have any relevance since the Risch algorithm cannot make
any use of them. Indeed note that
In[79]:= Integrate[1/a, a]
Out[79]= Log[a]
In[80]:= Integrate[1/a, a, Assumptions :> a < 0]
Out[80]= Log[a]
and even
Integrate[1/a, a, Assumptions :>
Re[a] == 0 && Im[a] == 0]
Log[a]
So one would assume that they are being completely ignored. However,
this is not quite so:
In[85]:= Integrate[1/a, a, Assumptions :> a == 0]
Out[85]= ComplexInfinity
However, I am sure that this is not really intended to mean anything
and is just a side-effect of something else. Assumptions are not meant
to work with indefinite integration and it probably should be
considered if not exactly a bug then slghly careless design that they
actually work instead of producing some syntax error message.
Andrzej Kozlowski
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