Extension: Absolute Convergence of Improper Integrals

Recall from here that a series $$\sum_{i=1}^\infty a_i$$ is absolutely convergent if and only if $$\sum_{i=1}^\infty |a_i|$$ is convergent.

Compare it with this definition:

An improper integral $$\int_0^\infty f(x) dx$$ is absolutely convergent if and only if $$\int_0^\infty |f(x)| dx = L$$ for some number $L$ (i.e. the integral of the "absolute value of $f(x)$ converged)

Ok, we're kind of cheating, in the sense that the example is a bit pathological, but that's ok, since it's a valid example. :D Consider the function $f(x) = \frac{(-1)^{\lfloor x\rfloor}}{1+\lfloor x\rfloor}.$ This is kind of intimidating when you first look at it, but let's draw a picture.

$f(x) = 1$ from $0$ to $1,$

$f(x) = -\frac{1}{2}$ from $1$ to $2,$

$f(x) = \frac{1}{3}$ from $2$ to $3,$ and so on.

Showing that this is convergent but not absolutely convergent is not too difficult and will be similar to the proof in the case of the series. We will leave it as an exercise (or a later editor can fill it in). At least from the picture you can see that for integers $n,$ $$\int_0^n f(x) = 1 -\frac{1}{2} + \cdots +\frac{(-1)^{n-1}}{n}$$ which you should help you get started in dealing with the limits.

Lastly, let's just close this off with one more point.

Well done!