 Solution

Let $\varphi$ be a test function supported in $[-R,R]$. Symmetric cancellation givesThe integrand is bounded by $2\Vert {\varphi }^{\prime }{\Vert }_{\infty }$, so the limit exists andThis is a continuous linear functional on every fixed-support test space, hence a distribution. It also specifies its local order as at most one.

For the boundary value, split the ordinary complex denominator:The real part paired with $\varphi$ isIt converges to the principal value by dominated convergence theorem, using the same bound $2\Vert {\varphi }^{\prime }{\Vert }_{\infty }$. In the second part set $x=\epsilon t$:Here $\Vert \varphi {\Vert }_{\infty }/(1+t^2)$ is an integrable majorant. The correct Sokhotski–Plemelj formula is thereforeThe resulting functional is continuous, with bound $2R\Vert {\varphi }^{\prime }{\Vert }_{\infty }+\pi \Vert \varphi {\Vert }_{\infty }$, so it too is a distribution.

The printed PDF has the wrong sign on its delta term. For a real even nonnegative test function with $\varphi (0)>0$, the principal value is zero while the imaginary part of every regularized integral is negative. Its limit is $-\pi \varphi (0)$, disproving the printed plus sign directly. A plus sign instead belongs to the boundary value $(x-i0{)}^{-1}$.