In this set of lecture notes by Terry Tao on the Hilbert transform, the following statement is made:
Suppose $f\in C^1({\bf R})$ obeys a qualitative decay bound $f(x)=O_f(\langle x\rangle^{-1})$. Also suppose $f$ extends holomorphically to the upper half-plane and obeys the decay bound $f(z)=O_f(\langle z\rangle^{-1})$ in this region. Then Cauchy's theorem gives $$ \frac{1}{2\pi i}\lim_{\varepsilon\to 0}\int_{\bf R} \frac{f(y)}{y-(x+i\varepsilon)}=f(x) $$ and $$ \frac{1}{2\pi i}\lim_{\varepsilon\to 0}\int_{\bf R} \frac{f(y)}{y-(x-i\varepsilon)}=0. $$
In the statement above, $\langle z\rangle:=\sqrt{1+|z|^2}$ denotes the Japanese bracket of $x$ (cf. Tao's notes, corollary 6.7) and $O$ denotes the usual Big O notation (cf. Tao's notes, section 3, Landau notation)
Question:
Would anyone elaborate how Cauchy's theorem (I think the "Cauchy integral formula" would be more precise here) is applied here?
I attempted to use Jordan's lemma with the semi-circle contour to show that $$ \frac{1}{2\pi i}\int_{\bf R}\frac{f(y)}{y-(x+i\varepsilon)}\ dy=f(x+i\varepsilon),\quad \varepsilon>0. $$ But with the decay assumption of $f$, I don't see how to get $$ \lim_{R\to\infty}\max_{\theta\in[0,\pi]} \left|\frac{f(Re^{i\theta})}{Re^{i\theta}-(x+i\varepsilon)}\right|=0 $$
Here is a related excerpt from the notes (regarding the Plemelj formula and the discussion afterwards):
[...]
(The RHS of the first identity in the original notes should be $f(x)$ instead of $f(x+i\varepsilon)$.)
With a semi-circular path, consisting of $\ell_1=[-R,R]$ and $\ell_2=\{Re^{it}: 0\le t\le\pi\}$, we have:
$$\left|\int_{\ell_2}\frac{f(y)}{y-(x\pm i\varepsilon)}dy\right|\le C\frac{1}{\sqrt{1+R^2}}\times\frac{1}{|R-|x\pm\varepsilon||}\times \pi R$$
the factor $C\frac{1}{\sqrt{1+R^2}}$ coming from the estimate of $f$, the factor $\frac{1}{|R-|x\pm\varepsilon||}$ from the triangle inequality (estimate of $\frac{1}{y-(x\pm\varepsilon)}$), and the third factor is the length of the semicircle. The RHS $\to 0$ when $R\to\infty$.
On the other side,
$$\int_{\ell_1}\frac{f(y)}{y-(x\pm i\varepsilon)}dy=\int_{-R}^R \frac{f(y)}{y-(x\pm i\varepsilon)}dy$$
As the sum of those is a contour integral to which you can apply Cauchy's formula, the formulas from the red rectangle directly follow by letting $R\to\infty$.