Corrections to the $\delta$ limit of the Cauchy distribution The Cauchy distribution
$$
C(x) = \frac{1}{\pi} \frac{\gamma}{(x-\omega)^2 + \gamma^2}
$$
approaches the Dirac delta distibution as $\gamma \to 0$. What is the $O(\gamma)$ correction to this? I am thinking it should involve a Cauchy principal value, based on considering two cases:
Case 1: $x \ne \omega$
Off resonance, $C(x)$ has a convergent series expansion for $|\gamma| < |x - \omega|$
$$
C(x) = \frac{1}{\pi} \frac{\gamma}{(x-\omega)^2} + O(\gamma^3)
$$
Case 2: $x = \omega$
On resonance, $C(x)$ diverges as $\gamma \to 0$ as
$$
C(x) \sim \frac{1}{\pi\gamma}
$$
But this is the same divergent behavior that leads us to identify the Dirac delta distribution as the $\gamma \to 0$ limiting behavior.
These two cases lead me to want to write
$$
C(x) \sim \delta(x-\omega) + \frac{\gamma}{\pi} P \frac{1}{(x-\omega)^2}
$$
where $P$ means to take the Cauchy principal value. But if I use this in expansion in real calculations, will I recover correct $O(\gamma)$ behavior? Or have I made some subtle mistake?
 A: Let $\varphi$ be a test function. Then
$$
\int_{-\infty}^{\infty} \frac{1}{\pi} \frac{\gamma}{(x-\omega)^2 + \gamma^2} \varphi(x) \, dx
= \{ x=y+\omega \}
= \int_{-\infty}^{\infty} \frac{1}{\pi} \frac{\gamma}{y^2 + \gamma^2} \varphi(y+\omega) \, dy
\\= \{ y=\gamma z \}
= \int_{-\infty}^{\infty} \frac{1}{\pi} \frac{\gamma}{\gamma^2z^2 + \gamma^2} \varphi(\gamma z+\omega) \, \gamma\,dz
= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \varphi(\gamma z+\omega) \,dz
\\= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \left(\varphi(\omega)+\gamma z\varphi'(\omega)+O((\gamma z)^2)\right) \,dz
\\= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \varphi(\omega) \, dz 
+ \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \gamma z\varphi'(\omega)\,dz +O(\gamma^2)
.
$$
The first term is the $\delta$ term:
$$
\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \varphi(\omega) \, dz
= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \, dz \, \varphi(\omega)
= \varphi(\omega)
$$
since $\int_{-\infty}^{\infty} \frac{1}{z^2+1} \, dz = \pi.$
The term linear in $\gamma$ is
$$
\frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{z^2+1} \gamma z\varphi'(\omega)\,dz
= \frac{\gamma}{\pi} \int_{-\infty}^{\infty} \frac{z}{z^2+1}\,dz\, \varphi'(\omega)
= 0
$$
since $\int_{-\infty}^{\infty} \frac{z}{z^2+1}\,dz=0.$
