Re: PING: Dave Cantrell

John Schutkeker <jschutkeker@xxxxxxxxxxxxxxxxxxxx> wrote:
The advice you gave me about how to do my integral was wrong.

No, it wasn't. If you think otherwise, please _quote_ the advice which you
think was wrong.

Anyone interested may see for themselves exactly what was said in the
thread
<http://groups.google.com/group/sci.math/browse_frm/thread/a91c5237d5d724bf>.

My integral was

Int a*u(b-c*u)/{(1-u^2)sqrt[(u-u1)(u-u2)(u-u3)]} du

It turns out that partial fractions are not the way to get the solution,
as you said.

I never said any such thing. Again, if you think otherwise, please give a
specific quotation.

Partial fractions are used to factor the polynomial in the
denominator. I'm trying to put a polynomial into the numerator. So
it's not a factoring problem at all, but just the question of doing a
simple integral with a quadratic in the numerator, rather than the
linear.

My integral,

Int a*u(b-c*u)/{(1-u^2)sqrt[(u-u1)(u-u2)(u-u3)]} du

breaks into two integrals

Int a*b*u/{(1-u^2)sqrt[(u-u1)(u-u2)(u-u3)]} du

- Int a*c*u^2/{(1-u^2)sqrt[(u-u1)(u-u2)(u-u3)]} du,

the first of which we already know how to do. The second one is the key
to solving the problem

Thus it reduces to

Int a*c*u^2/{(1-u^2)sqrt[(u-u1)(u-u2)(u-u3)]} du,

which I need to learn the method for solving.

Wolfram's new on-line integrator can do the solution, and they've almost
worked out all the bugs, but there's still one little glitch that looks
fishy.

If you can't give me the references that you used to teach yourself how
to do this integral,

Someone else in this thread has already given you an excellent reference.

I have to go to press with the glitch in my paper.

What an absurd thing to say! Of course, you don't _have_ to go to press
with a glitch. Indeed, you never _have_ to go to press at all.

If I do that, scientists everywhere will be spitting mad at both of us.

Not at me. I never gave you any incorrect advice.

David Cantrell
.

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