There are many questions about assumptions here. Many of them are problem specific. That is why I want to ask a question in the most general way. Why does Mathematica solves this
Assuming[{x < 0}, Integrate[1/x, x]]
as
Log[x]
and not
-Log[-x]
?
-Log[-x] is not a correct result, but Log[-x] is.
In fact the expressions Log[-x] and Log[x] differ only in a constant I Pi, so both are correct antiderivatives for all $x \in \mathbb{C}$.
While the result given by Mathematica is correct, it is complex valued for x < 0. I think you are looking for a real valued result. I do not think it is possible to ask Integrate to automatically provide one.
For definite integrals this won't be a problem though as that complex constant is cancelled:
Integrate[1/x, {x, a, -1}, Assumptions -> a < -1]
(* ==> -Log[-a] *)
dx. However, I have one more question about the different part of your answer. Here (mathematica.stackexchange.com/a/41112/12223) you said "x > 0 is sufficient and (in Mathematica) implies that x is also real". So, does Mathematica assume x to be real or not?
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Commented
Feb 7, 2014 at 13:18
x<0 is equivalent to x < 0 && Element[x, Reals]. It doesn't influence the result of the integral though. Log[x] is a correct result for all complex (real or not real) x.
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Assuming[x \[Element] Reals, Re[Integrate[1/x, x]]] give Re[Log[x]] instead of Log[-x]? (sorry for asking these follow up questions, I'm trying to get an intuition of how Mathematica works)
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Commented
Feb 7, 2014 at 16:13
Re[Log[x]] is a correct answer to the question, right? If you'd like to simplify Re[Log[x]] to Log[-x] with the assumption that x<0, you can use FullSimplify[ComplexExpand@Re[Log[x]], x < 0].
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Log[-x], not-Log[-x]. $\endgroup$