I've recently started to study about the relation between the convergence/divergence of improper integrals and the behaviour of the function as its graph approaches the x-axis. But first, let me be more specific. Here is a useful text I found:
It 's the third example from this page: http://merganser.math.gvsu.edu/david/m202/notes/techniques/improper.html
According to the last sentence, "the convergence of the integral is determined by how fast the graph approaches the x-axis, in other words, how fast the function approaches $\;0\;$"
My Question:
From the above conclusion, is it right or wrong to claim that the divergence of the integral shows that the graph approaches the x-axis slow?
If $\; \int_{a}^{\infty} f(x) dx=\infty\;$ for $\;f \in L_{loc}^1 [a,+\infty)\;$ then it follows $\;f\;$ approaches $\;0\;$ slow. Right or False? For instance, an appropriate function would be $\;f(x)=\frac {1}{x^p} \;$ for $\;0\lt p \lt 1\;$
I want to fully understand this, so any recommendations about related books are -of course- welcome.
Thanks in advance!
If $f(x) \to 0$ sufficiently fast, the integral converges. But conversely, the integral can converge even if the function does not tend to zero: take a function which is $1$ between $n$ and $n+1/2^n$ for integer $n$ and zero elsewhere, for example. This has integral equal to $\sum_{n=1}^{\infty} 2^{-n} $, which is finite, but does not tend to zero. Or one can imagine a function with nonzero parts made out of triangles with area $1/2^n$ at each integer, that become taller and narrower, so in fact $f$ can be unbounded and still have convergent integral.
In general, if $f$ is positive, decreasing for $x>x_0$, and decays fast enough, the integral converges, while if it decays too slowly, the integral diverges. One can prove this using the comparison test, and the examples given are useful for comparison: for example, if $f(x) > k/x$ for $x>x_0$, $\int_{x_0}^{\infty} f(x) \, dx $ diverges. On the other hand, if $f(x) < k/x^2$ for $x>x_0$, the integral converges. (There is no absolute boundary function that always determines if convergence occurs: $1/(x\log{x}$ diverges, $1/(x\log{x}\log\log{x})$ diverges, ..., while $1/(x(\log{x})^2)$ converges, for example.)