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Edit detail for #440 cannot take first of empty list in integral revision 4 of 4

1 2 3 4
Editor: kratt6
Time: 2009/01/16 05:07:16 GMT-8
Note: fixed in FriCAS revision 486

added:

From kratt6 Fri Jan 16 05:07:16 -0800 2009
From: kratt6
Date: Fri, 16 Jan 2009 05:07:16 -0800
Subject: fixed in FriCAS revision 486
Message-ID: <20090116050716-0800@axiom-wiki.newsynthesis.org>

Status: open => fixed somewhere 


Submitted by : kratt6 at: 2009-01-13T01:50:49-08:00 (10 years ago)
Name :
Axiom Version :
Category : Severity : Status :
Optional subject :  
Optional comment :

axiom
integrate(log(1-z^3)*(%i*z)^(1/2), z)

\label{eq1}{\left(
\begin{array}{@{}l}
\displaystyle
{{\sqrt{{16}\  i}}\ {\log \left({{{4 \  i \  z \ {\sqrt{i \  z}}}+{i \ {\sqrt{{16}\  i}}}}\over 4}\right)}}- 
\
\
\displaystyle
{{\sqrt{{16}\  i}}\ {\log \left({{{4 \  i \  z \ {\sqrt{i \  z}}}-{i \ {\sqrt{{16}\  i}}}}\over 4}\right)}}+ 
\
\
\displaystyle
{{\left({4 \  z \ {\log \left({-{z^3}+ 1}\right)}}-{8 \  z}\right)}\ {\sqrt{i \  z}}}
(1)
Type: Union(Expression(Complex(Integer)),...)

another instance --kratt6, Tue, 13 Jan 2009 02:09:57 -0800 reply
axiom
integrate(sin(z)*csc(z)*(1-1/(%i*z)^(1/2))^(1/2), z)

\label{eq2}{\left(
\begin{array}{@{}l}
\displaystyle
{i \ {\sqrt{i \  z}}\ {\log \left({{\sqrt{{{\sqrt{i \  z}}- 1}\over{\sqrt{i \  z}}}}+ 1}\right)}}- 
\
\
\displaystyle
{i \ {\sqrt{i \  z}}\ {\log \left({{\sqrt{{{\sqrt{i \  z}}- 1}\over{\sqrt{i \  z}}}}- 1}\right)}}+ 
\
\
\displaystyle
{{\left({4 \  z \ {\sqrt{i \  z}}}-{2 \  z}\right)}\ {\sqrt{{{\sqrt{i \  z}}- 1}\over{\sqrt{i \  z}}}}}
(2)
Type: Union(Expression(Complex(Integer)),...)

simpler instance --kratt6, Tue, 13 Jan 2009 02:12:25 -0800 reply
axiom
integrate((1-1/(%i*z)^(1/2))^(1/2), z)

\label{eq3}{\left(
\begin{array}{@{}l}
\displaystyle
{i \ {\sqrt{i \  z}}\ {\log \left({{\sqrt{{{\sqrt{i \  z}}- 1}\over{\sqrt{i \  z}}}}+ 1}\right)}}- 
\
\
\displaystyle
{i \ {\sqrt{i \  z}}\ {\log \left({{\sqrt{{{\sqrt{i \  z}}- 1}\over{\sqrt{i \  z}}}}- 1}\right)}}+ 
\
\
\displaystyle
{{\left({4 \  z \ {\sqrt{i \  z}}}-{2 \  z}\right)}\ {\sqrt{{{\sqrt{i \  z}}- 1}\over{\sqrt{i \  z}}}}}
(3)
Type: Union(Expression(Complex(Integer)),...)

fixed in FriCAS revision 486 --kratt6, Fri, 16 Jan 2009 05:07:16 -0800 reply
Status: open => fixed somewhere