Why does $\int_1^\infty\frac1xdx=\infty$ but $\int_1^\infty\frac1{x^2}dx=1$ when their graphs are the same shape? Why is it that $$\int_{1}^\infty \frac{1}{x}dx = \infty,$$ but$$\int_{1}^\infty \frac{1}{x^2}dx = 1,$$ when their graphs are the same shape?
They are basically the same shape. Even if $\frac{1}{x^2}$ approaches zero faster than $\frac{1}{x}$, that should not matter, right?
 A: I think some calculations can be useful in understanding how $f(x) = 1/x$ and $g(x) = 1/x^2$ are different.
If I choose $x = 10$, then $f(10) = 0.1$ but $g(10) = 0.01$.  For $f(x)$ to equal $0.01$, we would need to choose $x = 100$, but at this $x$-value, $g(100) = 0.0001$.  And for $f(x)$ to equal $0.0001$, we'd need to choose $x = 10000$, but $g(10000) = 0.00000001$.
In general, for a given fixed $x > 1$, $g(x)$ is $x$ times smaller than the corresponding $f(x)$.  So if $x = 10^6$,  $g$ is already a million times smaller than $f$ at that value.  The extent to which $g$ is smaller than $f$ is exactly proportional to the value of $x$, so the farther out along the positive $x$-axis you go, the more dramatic the ratio of $g$ to $f$.
There are other more familiar pairs of functions that also behave in this way.  For instance, compare $f(x) = x$ to $g(x) = x^2$.  Here, $g$ is $x$ times larger than $f$, and a quick sketch of the graphs of $f$ and $g$ show just how dramatically differently these behave.  Moreover, you can observe that $x$ and $x^2$ are the reciprocals of $1/x$ and $1/x^2$--so if you remark how much more rapidly $x^2$ grows compared to $x$, you must also see that $1/x^2$ decreases that much more rapidly to $0$ compared to $1/x$.
A: They may look similar on $[1,\infty)$, but with a change of perspective they look rather different. Let $z=1/x$, or $x=1/z$. Then $dx=-1/z^2$ and the limits of integration become $1\mapsto 1$ and $+\infty\mapsto 0^+$, or just $0$. Then
$$
\int_1^{\infty} x^{-2}\,dx = \int _{1}^{0}(-dz) = \int _0^1 \,dz = 1 
$$So, under this substitution, the new integrand is a constant and the range of integration is $[0,1]$: this is well-behaved. But the other one looks very different:
$$
\int _1^{\infty} x^{-1}\,dx = \int_{1}^{0} z^{-1}(-dz ) = \int _0^1 z^{-1}\,dz
$$Post-transformation, the integrand still has a singularity (of course, a priori it could still be improperly integrable, but...) and the FTC shows that the value of the integral is unbounded:
$$
=\left. \ln(z)\right|_0^1 = \ln(1)-\ln(0) = -(-\infty)=\infty
$$Strictly speaking, I should be taking the lower limit as $a\to 0+$ for some positive $a$, but you get the idea.
