Construct a function $f$ such that the above limit exists, although it fails to exist with $| f |$ in place of $f$ .

Question

This is related to Baby Rudin, Chapter 6, Exercise 7

Suppose $f$ is a real function on $(0, 1]$ and $f \in \mathscr{R}$ on $[c, 1]$ for every $c>0$. Define \begin{equation}\label{7.0} \int_0^1 f(x)\, dx = \lim_{c \to 0} \int_{c}^1 f(x) \,dx \end{equation} if this limit exists (and is finite). Construct a function $f$ such that the above limit exists, although it fails to exist with $|f|$ in place of $f$.

There are a couple of answers to this question online and they all seem to be related to series. Is it possible to come up with a simple example that does not involve the concept of series? If so, can someone please also provide an explanation as to how their proposed example satisfies the question statement.

Answer 1

The integrand should be highly oscillating.

Try

$$f(u)=\frac{\sin(1/u)}{u}$$ over the interval $[0,1]$. The change of variables

$$u=1/x$$

will give you a well known improper integral that converges conditionally ( converges as an improper Riemann integral, but the the integrand with absolute value is not integrable).

Answer 2

This is just to address a comment from the OP:

For sake of simplicity, consider the integral $\int^\infty_\pi\Big|\frac{\sin x}{x}\Big|\,dx$.

$$ \int^\infty_{\pi}\Big|\frac{\sin x}{x}\Big|\,dx=\sum^\infty_{k=1}\int^{(k+1)\pi}_{k\pi}\Big|\frac{\sin x}{x}\Big|\,dx\geq \frac{1}{\pi}\Big(\int^\pi_0|\sin x|\,dx\Big)\sum^\infty_{k=1}\frac{1}{k+1} $$ Here we have used the $\pi$-periodicity of $|\sin x|$. Furthermore, one also has that

$$ \sum^\infty_{k=1}\int^{(k+1)\pi}_{k\pi}\Big|\frac{\sin x}{x}\Big|\,dx\leq \frac{1}{\pi}\Big(\int^\pi_0|\sin x|\,dx\Big)\sum^\infty_{k=1}\frac{1}{k} $$

Since $\sum^n_{k=1}\frac{1}{k}\sim\log(n)$, we have that $\int^x_1 \Big|\frac{\sin t}{t}\Big|\,dt$ grows logarithmically.