Lorentzian kernels converge to delta and principal-value distributions

With $x=\eta y$, the first integral is ${\int }_{\mathbb{R}}\psi (\eta y)\mathrm{/}(1+y^2)dy$. Its integrand converges pointwise to $\psi (0)\mathrm{/}(1+y^2)$ and is bounded in modulus by $\mathrm{\parallel }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{/}(1+y^2)$, an integrable function. By [F1] the limit is $\psi (0){\int }_{\mathbb{R}}(1+y^2{)}^{-1}dy=\pi \psi (0)$; the last integral is $\arctan (y){\mathrm{\mid }}_{-\mathrm{\infty }}^{\mathrm{\infty }}=\pi$.

Let $L={\sup }_x\mathrm{\mid }{\psi }^{\mathrm{\prime }}(x)\mathrm{\mid }<\mathrm{\infty }$ and set $q(x)=\psi (x)-\psi (0){\mathbf{1}}_{[-1,1]}(x)$. The integral of $Q_{\eta }(x)\psi (0){\mathbf{1}}_{[-1,1]}(x)$ is zero because its integrand is odd on a symmetric interval. For $0<\mathrm{\mid }x\mathrm{\mid }\le 1$, $\mathrm{\mid }{Q}_{\eta }(x)q(x)\mathrm{\mid }\le L{x}^2\mathrm{/}(x^2+{\eta }^2)\le L$ by the mean-value inequality; for $\mathrm{\mid }x\mathrm{\mid }>1$, $\mathrm{\mid }{Q}_{\eta }(x)q(x)\mathrm{\mid }\le \mathrm{\mid }\psi (x)\mathrm{\mid }$. This gives an integrable majorant independent of $\eta$, and for $x\mathrm{\ne }0$ the integrand tends to $q(x)\mathrm{/}x$. Applying [F1] yields ${\int }_{\mathbb{R}}q(x)\mathrm{/}xdx$.

For $0<\delta <1$, the constant part $\psi (0)\mathrm{/}x$ integrates to zero on $\{\delta <\mathrm{\mid }x\mathrm{\mid }\le 1\}$ by oddness. Therefore $\int q(x)\mathrm{/}xdx={\lim }_{\delta \downarrow 0}{\int }_{\mathrm{\mid }x\mathrm{\mid }>\delta }\psi (x)\mathrm{/}xdx$, proving existence of the symmetric principal value and the second limit. The two distributional conclusions are precisely the limits already established in steps 1.1 and 1.2.