Dear Madhusudan,
Many thanks for your message below.
Yes, this is not a so basic question. My brief comments and
some suggestions:
My own computations have as follows:
(C1) display2d : false$
(C2) assume(x<1)$
(C3) assume(x>-1)$
(C4) integrate(sqrt(1-s^2)/(x-s),s,-1,1);
Is x zero or nonzero?
n;
Principal Value
(D4) SQRT(1-x^2)*LOG(-1/(2*x-1))+%PI*x
and agree with yours. My conclusions are:
1. Yes, Maxima does not compute the above Cauchy-type principal
value integral correctly. I think this can be called a bug. This is my
own opinion. I would like to hear further from colleagues, since I
may be wrong.
2. Practically speaking, if I would be interested to compute the
above Cauchy-type principal value integral with Maxima and
correctly, I would simply use the command
(C5) integrate(sqrt(1-s^2)/(z-s),s,-1,1);
Is z-1 positive, negative, or zero?
p;
(D5) %PI*z-%PI*SQRT(z^2-1)
(yes, positive, but valid for complex values of z too) leaving the
variable z outside the critical interval [-1, 1]. This result is
correct. Next, for the computation of the Cauchy-type
principal value integral, I would add (preferably by hand) the
values in (D5) for z = x+0 and z = x-0, getting the average,
i.e. %PI*x taking into account that for these values of z
(i.e. z = x+0 and z = x-0 with -1 < x < 1) the square root
sqrt(z^2-1) takes opposite imaginary values, i.e. %i*sqrt(1-x^2)
and -%i*sqrt(1-x^2) with a sum zero. Therefore, the
contribution is just (1/2)*(%PI*x+%PI*x) = %PI*x. This
is the correct result. This comment takes into account the
second Plemelj formula for Cauchy-type principal value
integrals. (Another way of computation could be a change
of variable.)
When using Maxima, I would prefer the above approach on
the basis of the second Plemelj formula. (D5) is essentially
the correct result, the square root ignored there as I explained
above. A third way would be the epsilon definition of a
Cauchy-type principal value integral. This could succeed too.
(The direct subtraction of the singularity seems impossible.)
In any case, I repeat I believe this is a bug under the assumption
that Maxima is programmed to compute Cauchy-type principal
value integrals, but I doubt if this programming task has been
done in a systematic way.
I am looking forward to hear from the colleagues with much
more experience in Maxima than me. (I am a novice in
Maxima.)
This is the situation according to my opinion. The advice is:
avoid principal value integrals in Maxima and just use the
second Plemelj formula on the related ordinary integrals
instead. I would be glad to hear on a better advice.
Incidentally, I can add that I have recently suggested the
extension of the special functions Maxima package of
Barton with second kind functions corresponding to the
classical orthogonal polynomials. In such a case, perhaps
Maxima could be instructed on the second kind functions
corresponding to the Chebyshev polynomials of the second
kind that these are related to the Chebyshev polynomials
of the first kind as is your correct result %pi*x. (1 means
U[0](x) and x means T[1](x).) Perhaps, you do not like
this approach, but I like it very much.
As far as I am concerned, I am unable to suggest something
more practical for your computation. I do not have more
experience with Maxima so far.
Best regards from Patras,
Nikos
----- Original Message -----
From: "Madhusudan Singh" <chhabra at eecs>
To: <Maxima@www.ma.utexas.edu>
Sent: Sunday, April 13, 2003 4:19 PM
Subject: A not so basic question :)
> Hi
> Thanks to everyone who has been helping me with my basic questions.
> However, I have a question that is not so basic, or at least so it seems
> to me.
>
> Consider the following use of integrate (the integrand is singular
> in the domain) :
>
> assume(x<1);
> assume(x>-1);
> f(x):=integrate(sqrt(1-s^2)/(x-s),s,-1,1);
> f(x);
>
> Maxima returns the Principal Value as :
>
> Principal Value$$\isqrt{1-x^{2}}\*\log
> \left(-\ifracd{1}{2\*x-1}\right)+\pi\*x$$
>
> (what's with the non-tex \ifracd and \isqrt anyway ?)
>
> The correct answer (confirmed with a hand calculation and Mathematica)
> is simply \pi*x.
>
> Is this a bug ?
>
> Thanks,
> MS.