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Re: Integrate question

"Oskar Itzinger" <oskar at> wrote:
> Mathematica 5.2 under IRIX complains that
> Integrate[x/(3 x^2 - 1)^3,{x,0,1}]
> doesn't converge on [0,1].

Actually, that's good! The integral doesn't converge. The integrand has a
pole at x = 1/Sqrt[3].

> However, Mathematica 2.1 under Windows gives the corrrect answer, (1/16).

The answer 1/16 is correct in the sense of Cauchy principal value. But I
think that the behavior of version 5.2 is better, that is, if an integral
doesn't converge in the usual sense, complain that it doesn't converge,
rather than giving a Cauchy principal value which was not specifically

> When did Mathematica lose the ability to do said integral?

Of course you can get the Cauchy principal value easily in version 5.2.
The quick and dirty method is to get the antiderivative and then just use

In[3]:= Integrate[x/(3 x^2 - 1)^3, x]

Out[3]= -(1/(12*(-1 + 3*x^2)^2))

In[4]:= (% /. x - > 1) - (% /. x -> 0)

Out[4]= 1/16

However, it would have been reasonable to expect that setting the option
PrincipalValue to True would even more easily give the desired value. But
that would be "reasonable to expect" only if one had not carefully read the
documentation for PrincipalValue:

"Setting PrincipalValue->True gives finite answers for integrals that had
simple pole divergences with PrincipalValue->False"

Note the crucial word "simple". The pole at x = 1/Sqrt[3] in not simple,
and thus, after having read the documentation, it should not be a surprise

Integrate[x/(3 x^2 - 1)^3, {x, 0, 1}, PrincipalValue -> True]

also causes version 5.2 to complain that the integral diverges.

I consider the restriction of PrincipalValue to simple poles to be
unfortunate. Is it still that way in version 6? If so, I hope that the
utility of PrincipalValue will be broadened in some future version.

David W. Cantrell

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