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I always thought that $y=\frac{1}{x^2}$ was symmetric across the line $y=x$. However, when I computed the area between the function and the $x$-axis from $x=1$ to infinity, I got 1

$\int_1^\infty \frac{1}{x^2} = lim_{b\rightarrow\infty} \int_1^b \frac{1}{x^2} = lim_{b\rightarrow\infty} 1 - \frac{1}{b} = 1$.

Also, I computed the area between the y-axis and the reciprocal function from $x=0$ to $x=1$, I got an infinite area.

$\int_0^1 \frac{1}{x^2} = lim_{b\rightarrow0} \int_b^1 \frac{1}{x^2} = lim_{b\rightarrow0} \frac{1}{b} - 1 = \infty$.

However, when inspecting the graph of $y=\frac{1}{x^2}$, assuming that $y=\frac{1}{x^2}$ is symmetric over the line $y=x$, the difference between $\int_1^\infty \frac{1}{x^2}$ and $\int_0^1 \frac{1}{x^2}$ should be the area of the square with vertices of the origin, (1,0), (1,1) and (0,1), not $\infty$.

Why then, is the difference between the two integrals infinite, instead of 1, the area of the square? Is $y=\frac{1}{x^2}$ not really symmetric over $y=x$, or is there another insight I am missing?

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    $\begingroup$ Just a quick observation: A graph $y = f(x)$ is symmetric across the line $y = x$ if and only if $x = f(y)$, i.e., exchanging $x$ and $y$ preserves the graph. $y = 1/x$ has this property, and $y = 1/x^{2}$ doesn't. :) $\endgroup$ – Andrew D. Hwang Mar 20 '15 at 16:59
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FOR THE ORIGINAL FUNCTION $\frac 1x$

$\int_b^1 \frac 1x\,dx$ is $-\ln b$, not $\frac 1{b^2}-1$, as you wrote in your second gray box. However, with that correction, you still get an integral of $\infty$.

More important, in your first grey box,

$$\int_1^\infty \frac{1}{x}\,dx = \lim_{b\rightarrow\infty} \int_1^b \frac{1}{x}\,dx = \lim_{b\to\infty}(\ln b-\ln 1)=\infty$$

FOR THE EDITED FUNCTION $\frac 1{x^2}$

As you can see in your edited graph, the function $y=\frac 1{x^2}$ is not symmetric across the line $y=x$, it is just nearly so. For example, the point $(2,\frac 14)$ is on the graph but not $(\frac 14,2)$. Looking at the graph, the right area is "thinner" than the left area. The difference in symmetry is enough to cause what you see with the integrals, with the right one finite but the left one infinite.

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  • $\begingroup$ Thanks for the feedback. I actually meant $y=\frac{1}{x^2}$ instead of $y=\frac{1}{x}$, and have edited the question accordingly. $\endgroup$ – user207766 Mar 20 '15 at 16:48
  • $\begingroup$ @Andy: You are welcome! $\endgroup$ – Rory Daulton Mar 20 '15 at 16:56

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