Assume $\phi$ is smooth in $[-1,1]$, for each $x \in (-1,1)$ and $\epsilon > 0$ consider
$$F_\epsilon(x) = \int_{-1}^{x-\epsilon} \frac{\phi(y) \text{d} y}{y-x} + \int_{x+\epsilon}^1 \frac{\phi(y) \text{d} y}{y-x}$$
It is known that, as $\epsilon \to 0$
$$F_\epsilon(x) \to \text{p.v.} \int_{-1}^1 \frac{\phi(y) \text{d} y}{y-x}$$
In other words we have pointwise convergence. Is this convergence uniform in $x$? Or do we have convergence in $L^p(-1,1)$?
Hint: Take $x=0.$ Note that
$$F_\epsilon(0)= \int_\epsilon^1\frac{\phi(y)-\phi(-y)}{y}\,dy.$$
So if $\epsilon_1<\epsilon_2,$ then
$$F_{\epsilon_2}(0) - F_{\epsilon_1}(0) = \int_{\epsilon_1}^{\epsilon_2}\frac{\phi(y)-\phi(-y)}{y}\,dy.$$
This integral, by the MVT, is bounded above in absolute value by $2\|\phi'\|_\infty(\epsilon_2-\epsilon_1).$ That's just for $x=0,$ but you can do the same for other $x,$ although the end points may cause trouble. I'd guess we get uniform convergence on $[-a,a]$ for $0<a<1.$