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.