Recent changes to 2487: integrate loops forever with simple expressionhttps://sourceforge.net/p/maxima/bugs/2487/Recent changes to 2487: integrate loops forever with simple expressionenFri, 27 Mar 2015 15:34:31 -0000#2487 integrate loops forever with simple expressionhttps://sourceforge.net/p/maxima/bugs/2487/?limit=25#9e5d<div class="markdown_content"><p>limit(-log(1+x^(2/3)) , x, 0, plus),algebraic:true; also fails. replace 2/3 by 1/3 and it succeeds (gives 0). <br />
bprog is part of the program that computes partial fractions.</p>
<p>There is no problem if algebraic:false .<br />
Sorry, not a bug fix.<br />
Rjf</p></div>Richard FatemanFri, 27 Mar 2015 15:34:31 -0000https://sourceforge.netca8aa77b1d30f849e27e418b888e92b83adb20d6integrate loops forever with simple expressionhttps://sourceforge.net/p/maxima/bugs/2487/if you calculate this:
integrate\(x^\(1/3\)/\(x^\(2/3\)+1\), x, 0, 8\);
Maxima will compute forever and never return.
However, the indefinite integral works:
integrate\(x^\(1/3\)/\(x^\(2/3\)+1\), x\);
gives you the antiderivative:
F\(x\) := \(3\*\(x^\(2/3\)+1\)\)/2-\(3\*log\(x^\(2/3\)+1\)\)/2;
Now the original definite integral can be computed as F\(8\) - F\(0\).
This is OK, because F\(x\) doesn't change sign or shows any other notable behavior between 0 and 8.
The question is: why doesn't Maxima do this? Instead it loops forever ...David ScherfgenThu, 08 Nov 2012 15:45:50 -0000https://sourceforge.net89953baa4e7c13cb7bde0c07de75764e60827c41