# [Maxima] tellsimp problem

Richard Hennessy rvh2007 at comcast.net
Sun Oct 26 10:37:34 CDT 2008

```Wait a minute.  The result of an integration of a function f on an interval [a,b] such that f is finite everywhere in
the interval and is defined everywhere is always continuos.  If f goes to inf or minf then you may have a problem like
that.  In this case f is discontinuos but integrate(f(x), x) is not because f never goes to inf or minf anywhere in the
range [a,b] which in this case is true.  There is no other choice that makes f infinite.  I think the answer %o45 is
wrong and that explains it.

Rich

----- Original Message -----
From: "Barton Willis" <willisb at unk.edu>
To: "Richard Hennessy" <rvh2007 at comcast.net>
Cc: "Maxima List" <maxima at math.utexas.edu>
Sent: Sunday, October 26, 2008 8:46 AM
Subject: Re: [Maxima] tellsimp problem

If your goal is to compute definite integrals, be careful:

(%i40) subst('unit_step = lambda([s], (s + abs(s))/(2*s)), unit_step(x));
(%o40) (abs(x)+x)/(2*x)

(%i41) pmint(%,x);
(%o41) (x^3*abs(x)+x^4)/(2*x^2*abs(x))

The expression (%o41) has a removable discontinuity at 0; otherwise it is
continuous on the reals.
Except for the troublemaker at zero, (%o41) valid on the reals.

(%i44) subst('unit_step = lambda([s], (s + abs(s))/(2*s)), unit_step
(1-x^2));
(%o44) (abs(x^2-1)-x^2+1)/(2*(1-x^2))

(%i45) pmint(%,x);
(%o45) (x^5*abs(x^2-1)-2*x^3*abs(x^2-1)+x*abs(x^2-1)-x^3*(x^2-1)^2
+x*(x^2-1)^2)/(2*(x^2-1)^2*abs(x^2-1))

The expression (%o45) has nonremovable discontinuities, so (%o45) isn't
valid on the reals (but
it is valid piecewise, I think).

Barton

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