Inconsistent Integrate behavior?

Question

Evaluating the following integral with Mathematica

Integrate[q0/(-1 + q0^2), {q0, 0, A}]

gives a conditional expression

ConditionalExpression[
    -((I π)/2) + 1/2 Log[-1 + A^2], -1 < Re[A] <= 1 || A ∉ Reals]

However, evaluating

Integrate[q0/(-1 + 12 q0^2), {q0, 0, A}]

Mathematica returns

-(1/24) I (π + I Log[-1 + 12 A^2])

Is this a bug or is there a reason for this behavior?

(Mathematica Ver.10.3.1)

Answer 1

Lets start with a simpler case

Integrate[q0/(-1 + a q^2), q]

$\frac{\log \left(1-a \text{q}^2\right)}{2 a}$

When you put limit [0,A], it has no problem with q=0. But it is not defined when $aA^2>1$. So you always have to obey that condition. You can check that by

Integrate[q0/(-1 + a q0^2), {q0, 0, A}]

In your second case Integrate[q0/(-1 + 12. q0^2), {q0, 0, A}] does not show the condition but you can see it in the answer. To avoid this you can use floating number like

Integrate[q0/(-1 + 12. q0^2), {q0, 0, A}]

ConditionalExpression[ 0.103538 + 0.0416667 Log[0.288675 - 1. A] + 0.0416667 Log[0.288675 + 1. A], -0.288675 < Re[A] <= 0.288675 || A [NotElement] Reals]

If you want to see the condition, then you might want to give mathematica a heads up with an Assumption like

Integrate[q0/(-1 + 12 q0^2), {q0, 0, A},Assumptions -> A \[Element] Reals]

$\text{ConditionalExpression}\left[\frac{1}{24} \log \left(1-12 A^2\right),A>0 ~ \&\& ~ 6 A\leq \sqrt{3}\right]$

Otherwise you may not get a condition, specially while dealing with exact numbers.