If we then note that $f(x)$ is just a constant, we have also $$\begin{equation*} f(x)\int_{-1}^{+1} \frac{ds}{s}=\int_{-1}^{+1}f(x) \frac{ds}{s} =0. \end{equation*}$$

Subtracting this equation from Eq. (20) yields $$\begin{equation} I_{\Delta}(x)={\cal P}\int_{-1}^{+1}ds\frac{f(\Delta s+x)}{s}=\int_{-1}^{+1}ds\frac{f(\Delta s+x)-f(x)}{s}, \tag{21} \end{equation}$$ and the integrand is no longer singular since we have that $\lim_{s \rightarrow 0} (f(s+x) -f(x))=0$ and for the particular case $s=0$ the integrand is now finite.