How correct is the statement $\int_1^∞ (\frac{1}{x^2}) \,dx= 1$? In Calculus 2, our professor always writes something along the lines of this for improper integrals:
$$
\int_1^∞ \frac{1}{x^2} \,dx= \lim_{b\to ∞}\int_1^b \frac{1}{x^2} \,dx=\lim_{b\to ∞}\left(-\frac{1}{b}+\frac{1}{1}\right)=1
$$
But...this isn't technically true at all, right?$\int_1^∞ \frac{1}{x^2} \,dx$ is not even Riemann-Integrable to begin with because the domain of integration is unbounded. What I think is really happening is that
$$
\int_1^∞ \frac{1}{x^2} \,dx=DNE
$$
but...
$$
\lim_{b\to ∞}\int_1^b \frac{1}{x^2} \,dx=1
$$
Just like $\frac{e^0-1}{0}=DNE$ but $\lim_{x\to 0}\frac{e^x-1}{x}$ converges to $1$. You wouldn't say, given a function $f(x)=\frac{e^x-1}{x}$, that
$$
f(0)=1
$$
...would you?

I suppose you could say (if you are pedantic), that the first equality
$$
\int_1^\infty \frac{1}{x^2} \,dx
\color{red}{=} \lim_{b\to \infty}\int_1^b \frac{1}{x^2} \,dx
\tag1$$
is only provisional: namely, the equality holds provided the limit exists (which we do not yet know).  So keep that in mind until the computation finishes,
$$
\lim_{b\to \infty}\int_1^b \frac{1}{x^2} \,dx=\lim_{b\to ∞}\left(-\frac{1}{b}+\frac{1}{1}\right)=1
$$
and we see that the limit does, indeed, exist.  Then we are OK, and everything is correct.

I would say it is incorrect to write
$$
\int_1^\infty \frac{1}{x^2} \,dx=DNE
\tag2$$
when you mean
$$
\int_1^\infty \frac{1}{x^2} \,dx\quad\text{does not exist}.
$$
But $(2)$ is a short way of saying something, just as $(1)$ is.
