Can the area under $\frac{1}{x}, x \in [1,+\infty]$ be finite?

Question

Because of the Riemann integral theory, we know that $\int_{[1,+\infty]}\frac{1}{x}=\lim_{M \rightarrow +\infty}\ln(M) $ and so $\frac{1}{x}$ is not improperly R-integrable in this sense. With another argument, one can show that this holds also with Lebesgue measure.

But, in a naive sense, are there arguments for the improper integrability of $\frac{1}{x}, x \in [1,+\infty]$?

For example (take in mind that this is a naive argument):

I can take the following intervals in $\mathbb{R}^2$: Let $a_n:=\sum_{i=1}^{n} \frac{1}{i}, n >0$ and let $A_n:=[a_n,a_{n+1}]\times [0,\frac{1}{n}] \subset\mathbb{R}^2$. Now, $\cup_n A_n \supset \Gamma(\frac{1}{x}), x\in[1,+\infty]$ (i.e. the area under the graph of $\frac{1}{x}$), and the $A_n$ are disjoint (except for a set of (naive-)"measure" zero in $\mathbb{R}^2$) and so I am naturally led to say that $|\cup A_n|= \sum_n |A_n| = \sum_n (a_{n+1}-a_n)(\frac{1}{n})=\sum_n(\sum_i^{n+1}\frac{1}{i}-\sum_i^n\frac{1}{i})(\frac{1}{n})=\sum_n \frac{1}{n(n+1)}$ which is convergent.

Now, because of the monotonicity of this naive-measure, I would say that $\Gamma(\frac{1}{x})$ is finite.

I know that my argument is wrong somehow, but I neither can't see why nor relates this reasoning to the Riemann measure or the Lebesgue one. Can someone help me clarifying this situation which seems legit in a naive sense, but formally it is not the case?

Answer 1

To illustrate LeBtz's comments: If we set $$ a_{n} = \sum_{i=1}^{n} \frac{1}{i} \sim \log n, $$ the shaded gray rectangles are the first hundred of $[a_{n}, a_{n+1}] \times [0, \frac{1}{n}]$, while the blue rectangles are the first six of $[n, n + 1] \times [0, \frac{1}{n}]$, relevant because if their total area were finite, the improper integral of the reciprocal function would be finite.

By contrast, the finiteness of the total area of the shaded rectangles has no bearing on the finiteness of the area under the graph, since the heights of the gray rectangles drop off exponentially with their horizontal position (i.e., like $\frac{1}{n}$ at position $\sim \log n$).