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Edit detail for #292 integrate(sin(x)**2) problem revision 2 of 10

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Editor: kratt6
Time: 2007/12/20 02:58:10 GMT-8
Note: more details

changed:
-integrate(sin(x)**2,x=0..1) returns "potentialPole".   This doesn't look right.  We have integrate(sin(x)**2,x)=(t-cos(t)sin(t))/2; no pole.
\begin{axiom}
integrate(sin(x)**2,x=0..1)
\end{axiom}
returns "potentialPole". This doesn't look right.  We have 
\begin{axiom}
integrate(sin(x)**2,x)
\end{axiom}
and
\begin{axiom}
integrate(sin(x)**2,x=0..1, "noPole")
\end{axiom}


There is no pole.

Submitted by : (unknown) at: 2007-11-17T22:22:16-08:00 (16 years ago)
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Axiom Version :
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axiom
integrate(sin(x)**2,x=0..1)
LatexWiki Image(1)
Type: Union(pole: potentialPole,...)

returns "potentialPole". This doesn't look right. We have

axiom
integrate(sin(x)**2,x)
LatexWiki Image(2)
Type: Union(Expression Integer,...)

and

axiom
integrate(sin(x)**2,x=0..1, "noPole")
LatexWiki Image(3)
Type: Union(f1: OrderedCompletion? Expression Integer,...)

There is no pole.