Let us apply this method to the integral $$ \begin{equation} I(x)={\cal P}\int_{-1}^{+1}dt\frac{e^t}{t}. \tag{23} \end{equation} $$ The integrand diverges at \( x=t=0 \). We rewrite it using Eq. (21) as $$ \begin{equation} {\cal P}\int_{-1}^{+1}dt\frac{e^t}{t}=\int_{-1}^{+1}\frac{e^t-1}{t}, \tag{24} \end{equation} $$ since \( e^x=e^0=1 \). With Eq. (22) we have then $$ \begin{equation} \int_{-1}^{+1}\frac{e^t-1}{t}\approx \sum_{i=1}^{N}\omega_i\frac{e^{t_i}-1}{t_i}. \tag{25} \end{equation} $$