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Re: [Axiom-developer] please check your Schaums

From: Raymond E. Rogers
Subject: Re: [Axiom-developer] please check your Schaums
Date: Mon, 28 Apr 2008 08:45:45 -0400
User-agent: Thunderbird (X11/20080226)

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?


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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.

(%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.


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