Using a trivial example to illustrate the question -
$$\int_0^\infty 2 dx$$
$=2x \mid_0^\infty = 2(\infty) - 2(0)$
Can we actually say $2(\infty)$? It doesn't seem valid me.
Using a trivial example to illustrate the question -
$$\int_0^\infty 2 dx$$
$=2x \mid_0^\infty = 2(\infty) - 2(0)$
Can we actually say $2(\infty)$? It doesn't seem valid me.
An integral, one (or both) of limits of which is infinite, is an improper integral. For such type of integral you can't take Newton-Leibniz formula as is, you'll have to use the definition of improper integral:
$$I=\int_0^\infty 2dx\equiv\lim_{t\to\infty}\int_0^t2dx.$$
Now $\int_0^t2dx=2t$, and, following the definition,
$$I=\lim_{t\to\infty} 2t=\infty,$$
so the integral diverges.
The expression $2 \cdot (+\infty) - 2 \cdot 0$ does indeed make sense. It is rather useful to do arithmetic/analysis with $\pm \infty$ alongside the other real numbers; e.g $\lim_{x \to +\infty}$ has all of the same properties that $\lim_{x \to a}$ does, and $\lim_{x \to a} f(x) = +\infty$ behaves much more like a convergent limit than other sorts of divergence (e.g. that of $\lim_{x \to 0} \sin(1/x)$), and $[-\infty, +\infty]$ is a compact interval.
The name of the structure that includes $\pm \infty$ at the endpoints of the real line is the "extended real numbers".