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Re: Nasty bug in Integrate (version 5.0)


MathUser:

I also run Mathematica 5.0.  I tested all your integrals, and obtained
answers which appear to be correct, except for Out[2] below.  In[2]
doesn't seem to know it should account for the pole at x=0, if a or b
(not both) are negative.  I would have expected a warning message
about the pole, and maybe also about the limit of 1+c as c -> -1. 
Out[2] was not the answer I expected.  I don't see why the Log[ ... ]
expression disappeared.  Maybe an answer to this question can be
posted.

Before running each integral, did you Remove x,a,b to be sure they
don't carry spurious values?

If you have time, try analyzing a Trace on In[2].  I'd be interested
in reading a post of the results.


Mathematica 5.0 for Microsoft Windows
Copyright 1988-2003 Wolfram Research, Inc.
 -- Terminal graphics initialized --

In[1]:= Integrate[1/x + x^c, {x, a, b}]

                                                                      
       Im[b]
Out[1]= (-a + b) If[(Im[a] - Im[b]) (-(Im[b] Re[a]) + Im[a] Re[b]) > 0
|| ------------- >= 0 ||
                                                                      
   Im[a] - Im[b]

                             1 + c    1 + c
           Im[a]            a      - b      + Log[a] + c Log[a] -
Log[b] - c Log[b]
>      -------------- >= 0, -------------------------------------------------------,
       -Im[a] + Im[b]                           (a - b) (1 + c)

                      1                          c
>     Integrate[-------------- + (a + (-a + b) x) , {x, 0, 1},
                a + (-a + b) x

                                                                      
           Im[b]
>      Assumptions -> !((Im[a] - Im[b]) (-(Im[b] Re[a]) + Im[a] Re[b]) > 0 || ------------- >= 0 ||
                                                                      
       Im[a] - Im[b]

               Im[a]
>          -------------- >= 0)]]
           -Im[a] + Im[b]

In[2]:= Integrate[1/x + x^c, {x, a, b}, Assumptions ->
{Element[a,Reals], Element[b, Reals]}]

          1 + c    1 + c
        -a      + b
Out[2]= ----------------
             1 + c

In[3]:= Integrate[1/x + x^c, {x, a, b}, Assumptions -> {a > 0, b > 0}]

          1 + c    1 + c
        -a      + b            b
Out[3]= ---------------- + Log[-]
             1 + c             a


Regards,
Harold


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