Surface under $\frac{1}{x}$ is $\infty$, while surface under $\frac{1}{x^2}$ is $1$?

Question

Since the antiderivative of $\frac{1}{x}$ is $\ln(|x|)$, the surface under the graph of $\frac{1}{x}$ with $x>1$ is $\infty$.

However, the antiderivative of $\frac{1}{x^2}$ is $-\frac{1}{x}$, so the surface under the graph of $\frac{1}{x^2}$ with $x>1$ is $1$.

Although I understand how one can calculate this, I can't really imagine it. Both graphs have an asymptote, so how can the surface under $\frac{1}{x}$ be $\infty$, whilst the surface under $\frac{1}{x^2}$ is $1$? It doesn't look like the surface under $\frac{1}{x}$ is infinite.

Perhaps my question is a little argumentative but I'm looking for a way to imagine that this is true. So, is there any way to intuitively "see" that the surface under $\frac{1}{x}$ is $\infty$?

Answer 1

Imagine a sequence of rectangles with corners at the origin and at the points $(2^n,2^{-n})$. The upper halves of these rectangles are all disjoint, they all lie under the graph, and they all have the same area -- thus they must add up to an infinite area under the graph.

Answer 2

Unfortunately, I don't have enough rep to post this as a comment, but you can apply the same argument that is used to show the divergence of the series $\frac{1}{n}$ to show that a Riemann sum for $\frac{1}{x}$ diverges as well.

That might be more intuitively satisfying than a formal calculation.

Answer 3

The area above $x\in[1,2]$ is the same as the area above $[2,4]$, is the same as above $[4,8]$, is the same as ... That's because if you multiply $x$ by 2 and divide $y$ be 2, the areas don't change, and the part of the plane above $[1,2]$ and below the graph is mapped to the area above $[2,4]$, which is mapped to ...

(btw this way one can see that the integral of $1/x$ satisfies $f(xy)=f(x)+f(y)$)