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>