Raymond Toy wrote:
>>>>>> "maryam" == maryam shamssolary <shamssolary at gmail.com> writes:
>
> maryam> Dear Master
>
>
> maryam> I am a young researcher in numerical linear algebra. For some
calculations, I used your "Maxima"
> maryam> software. But I have some problems with some commands such as:
> maryam> ?integrate(1/(r+s*x+t*x^2)*1/sqrt(1-x^2), x, -1, 1);
>
>
> maryam> give me this message:?
> maryam> "Is? "4*r*t-s^2"? positive, negative, or zero?"
>
>
> maryam> I tried different answers but did not?! What do you suggest?
>
> Maxima cannot do all integrals of this form. It appears that maxima
> can only do this if t*x^2+s*x+r has real roots. The case for complex
> roots is not handled. This is why maxima is asking for the sign of
> 4*r*t-s^2. The roots are real if 4*r*t-s^2 is zero or negative.
>
> After this maxima may ask many more questions as it breaks the
> original integral into two new integrals. I find the questions hard to
> answer. :-(
>
> You might get a feel for what is going on by using some numerical
> values for r, s, and t.
>
> Sorry that maxima can't easily produce an answer for you.
>
> Ray
By the way i have tried with maple and mathematica which all give very
complicated answers. So i have tried to find something simpler by hand.
Suppose you know the roots of r+s*x+t*x^2. Then you can as well replace
that by (x-a)*(x-b).
The inverse of which can be written as
1/(a-b)*(1/(x-a)-1/(x-b)).
This reduces the problem to one that maxima easily handles:
(%i3) integrate(1/(x-a)*1/sqrt(1-x^2), x, -1, 1);
Is a - 1 positive, negative, or zero?
p;
2
2 %pi sqrt(a - 1)
(%o3) - ------------------
2
2 a - 2
If you answer n; you have some more sign questions, for example:
(%i2) integrate(1/(x-a)*1/sqrt(1-x^2), x, -1, 1);
Is a - 1 positive, negative, or zero?
n;
Is a + 1 positive, negative, or zero?
p;
Principal Value
2 2
log(2) sqrt(1 - a ) log(2 - 4 a) sqrt(1 - a )
(%o2) ------------------- - -------------------------
2 2
a - 1 a - 1
As you can see these formulas require appropriate signs for the arguments
of the logs to be "correct" in an elementary sense. Are they true? I
don't know. For exemple in the second case one has -1<a<1 so for
1/2<a<1 one has 2-4*a <0 which renders the formula highly suspect.
--
Michel Talon