integrate and assume

[rtoy]

Imported from SourceForge on 2024-07-01 18:06:10 Created by fabienamiot on 2017-05-02 13:22:42 Original: https://sourceforge.net/p/maxima/bugs/3308

---

(%i1) bug_report();

Please report bugs to:
    http://sourceforge.net/p/maxima/bugs
To report a bug, you must have a Sourceforge account.
Please include the following information with your bug report:
-------------------------------------------------------------
Maxima version: "5.34.1"
Maxima build date: "2014-10-24 16:12:54"
Host type: "i686-pc-linux-gnu"
Lisp implementation type: "GNU Common Lisp (GCL)"
Lisp implementation version: "GCL 2.6.12"
-------------------------------------------------------------
The above information is also reported by the function 'build_info'.

(%o1) 
(%i2) assume(L>0);
(%o2)                               [L > 0]
(%i3) assume(notequal((1/l1+1/l3)/(1/l1+1/l2),-1));
                                                 1    1
                                                -- + --
                                                 l3   l1
(%o3)                      [notequal(-------, - 1)]
                                                 1    1
                                                -- + --
                                                 l2   l1
(%i4) assume(notequal((1/l2+1/l3)/(1/l1+1/l2),-1));
                                                 1    1
                                                -- + --
                                                 l3   l2
(%o4)                      [notequal(-------, - 1)]
                                                1    1
                                               -- + --
                                                l2   l1
(%i5) is(equal((1/l1+1/l3)/(1/l1+1/l2),-1));
(%o5)                                false
(%i6) integrate((exp(x/l1+x/l2)+exp(x/l3+x/l1)+exp(x/l3+x/l2))^2,x,0,L);
   1    1
   -- + --
   l3   l1
Is ------- equal to - 1?
   1    1
   -- + --
   l2   l1

Looks like this is an (other) example of a integrate/assume problem, maybe similar to those in bug 2829.

It should be outlined that :

(%i1) assume(L>0);
(%o1)                               [L > 0]
(%i2) assume(notequal((1/l2)/(1/l1),-1));
                                        l1
(%o2)                         [notequal(--, - 1)]
                                        l2
(%i3) assume(notequal((1/l3)/(1/l1),-1));
                                        l1
(%o3)                         [notequal(--, - 1)]
                                        l3
(%i4) integrate((exp(x/l1)+exp(x/l2)+exp(x/l3))^2,x,0,L);

does not raise any question, even though the problem is very similar.

[rtoy]

Imported from SourceForge on 2024-07-01 18:06:11 Created by robert_dodier on 2017-06-13 04:56:39 Original: https://sourceforge.net/p/maxima/bugs/3308/#6d5f

---

• Description has changed:

Diff:

--- old
+++ new
@@ -1,3 +1,4 @@
+~~~~
 (%i1) bug_report();

 Please report bugs to:
@@ -42,10 +43,11 @@
    1    1
    -- + --
    l2   l1
-   
+~~~~
 Looks like this is an (other) example of a integrate/assume problem, maybe similar to those in bug 2829.

 It should be outlined that :
+~~~~
 (%i1) assume(L>0);
 (%o1)                               [L > 0]
 (%i2) assume(notequal((1/l2)/(1/l1),-1));
@@ -57,5 +59,5 @@
 (%o3)                         [notequal(--, - 1)]
                                         l3
 (%i4) integrate((exp(x/l1)+exp(x/l2)+exp(x/l3))^2,x,0,L);
-
+~~~~
 does not raise any question, even though the problem is very similar.