Space of all improper Riemann-integrable functions not closed under products and other operations

Question

If $R[a,b]$ denotes the space of all Riemann-integrable functions in the closed interval $[a,b]$, then this space is closed under taking linear combinations, product of functions, powers of functions and absolute value. Now if $R[a,\infty)$ denotes the space of improper Riemann-integrable functions, then it is closed under linear combinations, right? But not under powers or products. Is this correct? What would be a nice example of an $f$, such that the improper integral in $[a,\infty)$ exists but that of $f^2$ doesnt? Also, why is it not closed under taking the positive values: for $f \in R[a,\infty)$, we do not have necessarily $f^+ \in R[a,\infty)$? What is a nice example for that?

Answer 1

If there is a blowup somewhere, then $f$ may be integrable when $f^2$ is not, because squaring "makes the blowup worse". For instance $f(x)=1/\sqrt{x}$ on $(0,1]$ is improperly integrable but its square is not.

If there is cancellation between positive and negative pieces, then $f$ may be integrable when $f^+$ is not, because the cancellation was preventing the integral from blowing up. For instance, define the sequence of numbers $H_n=\sum_{k=1}^n \frac{1}{k}$. A function $f$ which is $1$ on $[H_k,H_{k+1}]$ for odd $k$ and $-1$ on $[H_k,H_{k+1}]$ for even $k$ will be improperly integrable but $f^+$ will not be. (This is essentially just the fact that $\sum_{k=1}^\infty \frac{(-1)^k}{k}$ is conditionally but not absolutely convergent.)

Answer 2

For example, $\displaystyle\int_0^\infty \dfrac{\sin(x)}{\sqrt{x}}\; dx$ exists but $\displaystyle\int_0^\infty \left(\dfrac{\sin(x)}{\sqrt{x}}\right)^2\; dx$ and $\displaystyle\int_0^\infty \left(\dfrac{\sin(x)}{\sqrt{x}}\right)^+\; dx$ do not.