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

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 $\infty$?

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You may be able to compare the two to see how they work. Consider what happens as you go away from the origin. The function $1/x$ is fairly considerable in size - it takes roughly $x$ units of space to add 1 to its sum. On the other hand, $1/x^2$ is only $1/x$ the size of the former function. It diminishes fast enough so that 1 can never be added to the sum. Another way to see this is to break the function into chunks which each add 1 unit to the total sum. The first occurs at $e$, the second at $e^2$, then $e^3$, and so on. – Matt Groff Mar 30 '11 at 7:29
@Matt Groff: Your last method helped, thanks. – pimvdb Mar 30 '11 at 7:44
You're welcome! The key was analyzing the antiderivative. – Matt Groff Mar 30 '11 at 7:59

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.

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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.

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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 $f(xy)=f(x)+f(y)$)

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