Services & ResourcesWolfram ForumsMathGroup Archive

[Date Index] [Thread Index] [Author Index]

Re: Re: definite integration of 1/a

• To: mathgroup at smc.vnet.net
• Subject: [mg86457] Re: [mg86416] Re: definite integration of 1/a
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Tue, 11 Mar 2008 05:34:08 -0500 (EST)
• References: <acbec1a40803091930w2e684027l4c0f34d1ad96d2c1@mail.gmail.com> <fr2mik$o6g$1@smc.vnet.net> <200803110755.CAA24701@smc.vnet.net>

It seems ot me that your post contains quite an interesting  
observation that has not been made explicit. Note the following, (to  
me) surprising fact:


In[90]:= Simplify[Integrate[1/a, {a, 1, ablah}], ablah > 0]
Out[90]= log(ablah)

but, (here is what I find surprising):

Simplify[Integrate[1/a, {a, 1, ablah},
      Assumptions -> ablah > 0], ablah > 0]

If[ablah > 1, Log[ablah], Integrate[1/a,
      {a, 1, ablah}, Assumptions -> Inequality[0, Less,
          ablah, LessEqual, 1]]]

In other words, in this particular case, providing the Assumption  
ablah>0 within Integrate actually prevents Simplify from getting the  
more general result. We can see how this comes about. Without the  
Assumption ablah>0 Integrate gives a general complex result:

Integrate[1/a, {a, 1, ablah}]
If[Re[ablah] >= 0 || Im[ablah] != 0, Log[ablah],
    Integrate[1/a, {a, 1, ablah}, Assumptions ->
         ! (Re[ablah] >= 0 || Im[ablah] != 0)]]

Note tha condition Re[ablah]>=0 which, of course is satisfied whenver  
ablah >0. But, with the Assumption ablah>0 inIntegrate we get a less  
general real result:

Integrate[1/a, {a, 1, ablah}, Assumptions :> ablah > 0]
If[ablah > 1, Log[ablah], Integrate[1/a,
      {a, 1, ablah}, Assumptions -> Inequality[0, Less,
          ablah, LessEqual, 1]]]

so this is one of the (I think rather rare) cases when Integrate does  
better by working in the complex plane and then interpreting the  
result on the real line than by working on the real line form the start.


Andrzej Kozlowski





On 11 Mar 2008, at 08:55, David W.Cantrell wrote:

> "Chris Chiasson" <chris at chiasson.name> wrote:
>> I meant that I was expecting Log@a
>
> Surely you really meant that you were expecting Log[ablah].
>
> How about the following?:
>
> In[2]:= Simplify[Integrate[1/a, {a, 1, ablah}], ablah > 0]
>
> Out[2]= Log[ablah]
>
> But I must say that I'm slightly surprised that (using version 6.0.2),
>
> In[1]:= Integrate[1/a, {a, 1, ablah}, Assumptions -> ablah > 0]
>
> Out[1]= If[ablah > 1, Log[ablah],
> Integrate[1/a, {a, 1, ablah}, Assumptions -> 0 < ablah <= 1]]
>
> did not give simply Log[ablah].
>
> David
>
>> On Sun, Mar 9, 2008 at 9:30 PM, Chris Chiasson <chris at chiasson.name>
>> 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.
>>>
>>> --
>>> http://chris.chiasson.name/
>

• References:
  • Re: definite integration of 1/a
    • From: "David W.Cantrell" <DWCantrell@sigmaxi.net>