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?

This is a convergent improper integral. Note that $\int_{1}^{\infty} 1/x^a dx$ is finite if $a>1$. It will hold even for $a=1.01$ and not hold for $a=0.99$.
