Re: [Axiom-developer] please check your Schaums

[Raymond E. Rogers]

Doug Stewart wrote:
> address@hidden wrote:
> > In 14.661 Schaums claims:
> >
> > integral acoth(x/a) = x*acoth(x)+a/2*log(x^2-a^2)
> >                         ^^^^^^^^
> >
> > Axiom claims
> >
> > integral acoth(x/a) = x*acoth(x/a)+a/2*log(x^2-a^2)
> >                         ^^^^^^^^^^
> >
> > Is this a Schaums typo?
> >
> > Tim
> >
> >
> >
> > _______________________________________________
> > Axiom-developer mailing list
> > address@hidden
> > http://lists.nongnu.org/mailman/listinfo/axiom-developer
> >
> >   
> My schaums is the same as your Schaums but it is old (not as old as 
> yours) I have a New Schaums at work But I will not Be in to work 
> today  :-)
>
>
> My Maxima agrees with Axium.
>
> integrate(acoth(x/a),x);
> (%o8) (a*log(x^2/a^2-1))/2+x*acoth(x/a)
According to "Table of Integrals, Series, and Products" I.S. 
Gradshteyn/I.M. Ryzhik;
Axiom is right.  In fact a dimensional analysis says that Schaums must 
be wrong.  My personal integration says that Axiom/"Table.."  are off by 
a constant, but it's hard to argue about a constant of integration.  The 
meaning of that statement is: set y=x/a  and evaluate, then back 
substitute and multiply by a to get F(x/a); having done that you would 
get log((x/a)^2+1) as the trailing terms.  But this is the same with a 
constant difference.


RayR