Integrals on unlimited sets How do you evaluate this expression
$$
\left| \int_{1}^{\infty} 1 \; dx - \int_{1}^{\infty} 1 \; dx \right| \quad ?
$$
Using improper integral definition, this should be an indeterminate $\infty - \infty$ form.

In general, if we integrate on a generic set $A$, when evaluating 
$$
\left| \int_{A} f \; dx - \int_{A} g \; dx \right|,
$$
should we make sure that it is not an indeterminate form?

ADDED: probably, in the first case, rules of limits allow to affirm that that difference is $0$. In the second, if improper integrals appears, we can do the same having a well defined result.
You cannot evaluate that expression, for $\infty - \infty$ makes no sense.
Improper integrals of these types are defined in the following way.
$\int_a^{\infty}f(x)dx=\lim\ _{b\rightarrow+\infty}\int_a^bf(x)dx $ assuming $f$ in integrate on $[a,+\infty)$ In your case the answer is zero.if we agree that $\int f+g=\int f+\int g$. Basically infinities cancel each other out.For more about improper integrals check the link out. http://en.wikipedia.org/wiki/Improper_integral
If we wanted to talk generic case weed would have to know about measure theory . Please see the awesome Wikipedia link