       • To: mathgroup at smc.vnet.net
• From: "David Park" <djmp at earthlink.net>
• Date: Sat, 16 Dec 2006 05:18:24 -0500 (EST)

```Aaron,

I'm not certain why you insist on working with the general indefinite
integral. The indefinite integral always has a possible implied constant
that must be added. Furthermore the result contains an ArcTan function which
is a multifunction. The general result for ArcTan[x] should be ArcTan[x] + n
Pi where n is an integer.

If you look at your specific case you will see that you have the term

-0.463512 ArcTan[0.0224733 (102 - 96 x)]

If you use Chop to get rid of the imaginary part of the first answer then
you can compare the two answers with the n Pi term, and calculate the value
of n by...

1.6552 == -1.25713 + n*Pi*-0.46351240544347894;
Solve[%]
{{n -> -2.}}

And you see the two answers are the same if you take the proper multivalue
for ArcTan.

Either go with integrating specific cases, or try to find some method of
picking the correct multivalue for ArcTan, which may be extremely difficult.

David Park

From: aaronfude at gmail.com [mailto:aaronfude at gmail.com]

Hi,

This is similar to my previous posts, but I have worked hard to come up
with the simplest possible example which highlights what ought to be
considered in bug in Mathematica. Please comment and suggest a work
around.

F = Log[(-1 + d x + e Sqrt[f + 2g x + h  x^2])];
st = Integrate[F, x];

specific = st /. { d -> -2, e -> 10, f -> 1.3, g -> -1, h -> 1 };
(specific /. x -> 1 ) - (specific /. x -> 0)
NIntegrate[F /.  { d -> -2, e -> 10, f -> 1.3, g -> -1, h -> 1 }, {x,
0, 1}]

The results are:

-1.25713 - 4.44^-16*I

and

1.6552

Thank you very much in advance!

Aaron Fude

```

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