let be the integra
$$ I(a)=\int_{0}^{b} dx \frac{f(x)}{(x+ie-a)} $$ with 'e' an small quantity going to 0
from Shothotsky's formula $$ (x+ie-a)^{-1}= P(1/(x-a))-i\pi \delta (x-a) $$
so $$ I(a)=\int_{0}^{b} dx \frac{f(x)-f(a)}{(x+ie-a)}-i\pi f(a) $$
however . what would happen if $ a \to 0 $ ??
i believe the integral $$ I(0)=\int_{0}^{b} dx \frac{f(x)}{(x+ie)}$$ should be equal to
$$ I(0)= \int_{0}^{b} dx \frac{f(x)-f(0)}{(x+ie)}-i\pi f(0)$$ but since the lower limit is also zero perhaps we have a problem there