Is 1/x, integrated from 1 to infinity, considered "integrable"? $\large \int\limits_1^\infty \frac{dx}{x} $ diverges by comparison with the infinite sum $\sum\limits^\infty\frac{1}{n}$.
But $\frac{1}{x}$ is continuous at all non-zero values of $x$.
So is $\frac{1}{x}$ "integrable" from 1 to infinity, or does the fact that the integral diverges mean it is not integrable?
I'm just trying to make sure I am using the dominated convergence theorem correctly.  DCT states that if the integrand is bounded by some "integrable" function, then the limit of the integrals, is the integral of the limit, roughly speaking.

It is necessary that the upper bound in the dominated convergence theorem has a finite integral. To illustrate this, we can construct a sequence of functions converging point wise to and bounded above by $\frac{1}x$, but where each function has an integral of $0$. In particular, define $k_n$ to be the solution to:
$$\int_{1}^{k_n}x^{-1-1/n}=\int_{k_n}^{\infty}x^{-1-1/n}.$$
Noting that $$\lim_{n\rightarrow \infty}\int_{1}^{c}x^{-1-1/n}<\infty$$
and
$$\lim_{n\rightarrow \infty}\int_{c}^{\infty}x^{-1-1/n}=\infty$$
implies that $\lim_{n\rightarrow\infty}k_n=\infty$, because the above establishes that no finite bound "splits" the value of the integral equally for all $n$ - as the tail end will get continually larger where the other end will not.
If we define
$$f_n(x)=\begin{cases}x^{-1-1/n}&&\text{if }x<k_n\\-x^{-1-1/n}&&\text{if }x\geq k_n\end{cases}$$
then it is clear that $\int_{1}^{\infty}f_n(x)=0$ and $\lim_{n\rightarrow\infty}f_n(x)=\frac{1}x$ and $|f_n(x)|\leq \frac{1}x$. If we could apply the dominated convergence theorem using $\frac{1}x$ as the bound, we would conclude that $\int_{1}^{\infty}\frac{1}x=0$.
