Variant of the Sokhotski–Plemelj theorem I am aware of the Sokhotski–Plemelj theorem (I have also heard people referring to it as the "Dirac identity") which states that in the limit $\eta\rightarrow 0^+$
$$\frac{1}{x\pm i\eta}=\mathcal P\frac{1}{x}\mp i\pi\delta(x) \, .$$
Now, I am reading the book "Solid State Physics" by G. Grosso and G. Pastori Parravicini which states on page 430 that using the above formula it can "easily be proved" that
$$\frac{\hbar\omega}{E_j-E_i-\hbar\omega-i\eta}=
\frac{E_j-E_i}{E_j-E_i-\hbar\omega-i\eta}-1\, .$$
However, I fail to see how the latter formula follows from the former. Is there a trick that I am missing here?
 A: I don't think we need Sokhotski-Plemelj for this.
Think of $E_j - E_i$ as a fixed value $E$.
Then the formula is re-written as
$$\frac{\hbar \omega}{E - \hbar \omega - i \eta}\, .$$
Now let $x \equiv \hbar \omega$ and you get
$$\frac{x}{E -x - i \eta} \, .$$
This integral is dominated by the part where $x \approx E$ so let's try shifting the variables $y \equiv E - x$,
$$\frac{E - y}{y - i \eta}$$
and then expand the numerator and put the original variables back in:
$$\frac{E_j - E_i}{E_j - E_i - \hbar \omega - i \eta} - \frac{E_j - E_i - \hbar \omega}{E_j - E_i - \hbar \omega - i \eta} \, .$$
The first term already matches the first term in the target expression, so we only need to worry about the second term.
The second term is 1 because, well, in the limit $\eta \rightarrow 0$ it's identically 1.
A: Noting the simple relation 
$$
\frac{\Delta E_{i,j}}{\Delta E_{i,j}-\hbar\omega-i\eta}-\frac{\Delta E_{i,j}-\hbar\omega-i\eta}{\Delta E_{i,j}-\hbar\omega-i\eta}=\frac{\hbar\omega+i\eta}{\Delta E_{i,j}-\hbar\omega-i\eta}
$$
and by Sokhotski-Plemelj theorem
$$
\lim_{\eta\rightarrow 0^+} \frac{i\eta}{\Delta E_{i,j}-\hbar\omega-i\eta}=0
$$
because, if the limit 
$$
\lim_{\eta\rightarrow0^+}\int_{a}^{b}\frac{f(x)}{x\pm i\eta}dx
$$
exist, that's what the theorem says, then
$$
\lim_{\eta\rightarrow0^+}\int_{a}^{b}\frac{i\eta f(x)}{x\pm i\eta}dx= \left(\lim_{\eta\rightarrow0^+} i\eta\right)\left( \lim_{\eta\rightarrow0^+}\int_{a}^{b}\frac{f(x)}{x\pm i\eta}dx\right)=0
$$
for any target function $f(x)$.
