Integral related to Poisson kernel $\textbf{Problem}$: Find the value of the integral 
$$I=\int_{-\infty}^0 P.V.\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\partial f}{\partial y} \frac{(x-y)}{(x-y)^2+z^2} \ dy \ dz,$$
with $f$ a sufficiently well behaved function of $y$ alone, and $P.V.$ meaning we are to take the principal value of the integral. 
$\textbf{Part of a solution}$: Exchanging the order of integration, we have 
$$I = P.V.\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{\partial f}{\partial y} \int_{-\infty}^0  \frac{(x-y)}{(x-y)^2+z^2} \ dz \ dy.$$
and we let 
$$J=\frac{1}{\pi}\int_{-\infty}^0  \frac{(x-y)}{(x-y)^2+z^2} \ dz$$
Now, when $x\neq y$ this integrates to $1/2$, so that $I=0$ if we assume $f$ is compact. 
If $x-y$ is small, say $\epsilon$, then we have 
$\pi \delta(z) = \lim_{\epsilon \to 0} \frac{\epsilon}{\epsilon^2 +z^2}$, so that $J=1/2$. 
However, I do not know how to simultaneously take the limit of the other part of the integral, which I believe is something of the form: 
$$K= \lim_{y\to x}P.V.\int_{-\infty}^{\infty}\frac{\partial f}{\partial y} \ dy$$
I know the solution should be $-f(x) +c_o$ for $c_o$ some constant, as the physics of the problem yields $H(x,z)_{xx}+H(x,z)_{zz}=0$, with the boundary condition $\partial H /\partial z|_{z=0} = \partial f/\partial x$.
This makes solving the integral explicitly ancillary, however, I want to solve more complicated versions of this integral, so I'd like to understand the analysis.
Any tips on how to solve the remaining part of the integral are appreciated.
 A: We have the integral of interest $I(x)$ given by 
$$I(x)=\frac{1}{\pi}\int_{-\infty}^{0} \int_{-\infty}^{\infty}\frac{df(y)}{dy}\frac{x-y}{(x-y)^2+z^2}\,dy\,dz$$
Integrating the inner integral by parts and exploiting that $f$ is a test function gives
$$\begin{align}
I(x)&=-\frac{1}{\pi}\int_{-\infty}^{0} \int_{-\infty}^{\infty}f(y)\frac{\partial}{\partial y}\left(\frac{x-y}{(x-y)^2+z^2}\right)\,dy\,dz\\\\
&=\frac{1}{\pi}\frac{d}{dx}\int_{-\infty}^{0} \int_{-\infty}^{\infty}f(y)\left(\frac{x-y}{(x-y)^2+z^2}\right)\,dy\,dz
\end{align}$$
Interchanging the order of integration yields
$$\begin{align}
I(x)&=\frac{1}{\pi}\frac{d}{dx} \int_{-\infty}^{\infty}f(y)\left(\int_{-\infty}^{0}\frac{x-y}{(x-y)^2+z^2}\,dz\right)\,dy\\\\
&=\frac{1}{\pi}\frac{d}{dx} \int_{-\infty}^{\infty}f(y)\left(\frac{x-y}{|x-y|}\left.\arctan\left(\frac{z}{|x-y|}\right)\right|_{z\to-\infty}^{z=0}\right)\,dy\\\\
&=\frac12 \frac{d}{dx} \int_{-\infty}^{\infty}f(y)\left(\frac{x-y}{|x-y|}\right)\,dy\\\\
&=\frac12 \int_{-\infty}^{\infty}f(y)\frac{d}{dx}\left(\frac{x-y}{|x-y|}\right)\,dy\\\\
&=\frac12 \int_{-\infty}^{\infty}f(y)2\delta (x-y)\,dy\\\\
&=f(x)
\end{align}$$
as expected!
