Yesterday I was dealing with this problem: Let $f : \mathbb{R}-\{0\} \rightarrow \mathbb{R}$ continuos and nonnegative and:
$$F(x)= \int_{1}^{x} \frac{f(s)}{s} ds \hspace{5mm}x \gt 0 $$ $$ F(x) = \int_{-1}^{x} \frac{f(s)}{s}ds \hspace{5mm} x \lt 0$$
-For which $f$ we have $F \in BV(\mathbb{R})$?
-For these $f$ give a formula for the total variation of $F$.
Taking the case $x>0$, I was thinking that because of $f$ is continuos by the fundamental theorem of calculus it should be $F'(t) = \frac{f(t)}{t} \hspace{3mm} \forall t \in [1,x]$. So $F'$ is continuous (because is equal to a composition of continuos functions) and we can use the formula:
$$T_{F}(x) = \int_{1}^{x} |F'(t)| dt $$
At this point we must choose $f$ such that $$\int_{1}^{+\infty} \left|\frac{f(s)}{s}\right| ds \lt \infty$$
Maybe I'm totally wrong.