2 Hilbert transform

Your principal value integral can be expressed in terms has the form of a Hilbert transform. It is probably more helpful to note that it is closely related to the Stieltjes transform of the function $\sqrt{P}$. . The Stieltjes transform of a function $f(y)$ is defined by

$$S(z)=\int_I dy\frac{f(y)}{z-y}$$

for a real interval $I$ and complex $z$ not in $I$. If we take $I=[a,b]$ and $f=\sqrt{P}$, then your integral is given by

$${\cal P}\int_{a}^{b}dy\frac{\sqrt{P(y)}}{x-y}=\frac{1}{2}\lim_{\epsilon\rightarrow 0}[S(x+i\epsilon)+S(x-i\epsilon)]P}\int_{a}^{b}dy\frac{\sqrt{P(y)}}{x-y}=\lim_{\epsilon\rightarrow 0}\Re S(x+i\epsilon).$$

1

Your principal value integral can be expressed in terms of the Stieltjes transform of the function $\sqrt{P}$. The Stieltjes transform of a function $f(y)$ is defined by

$$S(z)=\int_I dy\frac{f(y)}{z-y}$$

for a real interval $I$ and complex $z$ not in $I$. If we take $I=[a,b]$ and $f=\sqrt{P}$, then your integral is given by

$${\cal P}\int_{a}^{b}dy\frac{\sqrt{P(y)}}{x-y}=\frac{1}{2}\lim_{\epsilon\rightarrow 0}[S(x+i\epsilon)+S(x-i\epsilon)].$$