How correct is the statement $\int_1^∞ (\frac{1}{x^2}) \,dx= 1$?

Question

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?

Answer 1

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 $a=0.99$.

Answer 2

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.

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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.