Services & ResourcesWolfram ForumsMathGroup Archive

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

Re: definite integration of 1/a

• To: mathgroup at smc.vnet.net
• Subject: [mg86416] Re: definite integration of 1/a
• From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
• Date: Tue, 11 Mar 2008 02:55:47 -0500 (EST)
• References: <acbec1a40803091930w2e684027l4c0f34d1ad96d2c1@mail.gmail.com> <fr2mik$o6g$1@smc.vnet.net>

"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/

• Follow-Ups:
  • Re: Re: definite integration of 1/a
    • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>