PING: Dave Cantrell


The advice you gave me about how to do my integral was wrong. 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. 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, I have to go to press with the glitch in my paper.
If I do that, scientists everywhere will be spitting mad at both of us.
Please dig out your file on this topic.

.

Relevant Pages

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  • Re: PING: Dave Cantrell
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  • Re: PING: Dave Cantrell
    ... I tweaked Mathematicas output on your integrals into a form, ... give real (I assume that is the 'glitch' you refer to) results for a certain domain of x and a certain ordering of the parameters: ...
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