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.
12. A^2
and you will see it. $\endgroup$Integrate[q0/(-1 + b q0^2), {q0, 0, A}] /. b -> 12
to get your conditional expression. I agree this is an inconsistency. You can also see the issue if you doIntegrate[q0/(-1 + 12 q0^2), {q0, 0, 1}]
you get a correct message that the integral does not converge whileIntegrate[q0/(-1 + 12 q0^2), {q0, 0, A}] /. A -> 1
gives an erroneous result. $\endgroup$