## #1140 integrate loops

closed
nobody
5
2007-07-06
2007-03-01
Anonymous
No

Hello

I had to interrupt maxima when it was at calculating the following integral:

(%i101) m(r,th);

2 2
(%o101) SQRT(z + 2 r rM (1 - COS(th)) + (rM - r) )
(%i102) F(d);

2
2 d0
4 A d0 (epsi - --- + 1) w
2
d
(%o102) - --------------------------
3
d
(%i103) integrate(1/m(th,r)^2*F(m(th,r))*z^2,th,0,%PI);

2 2 2 2
Is z - COS (r) rM + rM positive, negative, or zero?

pos;
2 2 2
Is z + (1 - COS (r)) rM positive, negative, or zero?

pos;
Maxima encountered a Lisp error:

Console interrupt.

Best wishes,
Jocelyn

## Discussion

• Nobody/Anonymous - 2007-03-01

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Version is Maxima 5.9.1

• Nobody/Anonymous - 2007-03-01

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the functions are:
m(r,th):=sqrt(z^2+(rM-r)^2+2*rM*r*(1-cos(th)));

and
F(d):=-4*w/d*(d0/d)^2*(1+epsi-(d0/d)^2)*A;

• Raymond Toy - 2007-03-03

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It does seem to loop. Not sure if it really does, because I didn't wait for very long.

Note, however, that all of the integrals are of the form 1/(th^2+a*th*b)^(n/2) where n is odd (5 or 7 here). Maxima can evaluate all of these integrals quite quickly. I think, but I'm not sure, that maxima is busy simplifying complicated intermediate expressions because the coefficients are not simple.

• Raymond Toy - 2007-03-07

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What is the integrand? You have the function m(r,th), but the integral says m(th,r). Given the limits of 0 and %pi, I think your integrand is wrong. Did you really want m(r,th)?

• Raymond Toy - 2007-06-21

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Marking as Pending

• Raymond Toy - 2007-06-21
• status: open --> pending

• SourceForge Robot - 2007-07-06
• status: pending --> closed

• SourceForge Robot - 2007-07-06

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This Tracker item was closed automatically by the system. It was
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