It is not a homework question. I just want to learn complex integrations.
I want to evaluate the following integral $$ I = \int \frac{dE}{2\pi} E^2 \left(\frac{1}{(E-E_n+iη)^2 (E-E_m+iη) }-\frac{1}{(E-E_n-iη)^2 (E-E_m-iη) }\right) $$
$\eta$ is a real positive number and $E_{n/m}$ are real numbers. I believe that this integral is equal to the following complex integral $$ I = \oint dz f(z) = -\oint \frac{dz}{2\pi} z^2 \left(\frac{1}{(z-E_n-iη)^2 (z-E_m-iη) }\right) $$ where the integral is taken over the upper half of the complex plane. According to Cauchy's theorem, the integral is $$ I = 2i\pi\sum_{i} \text{Res}(f(z),z_i) $$ $f(z)$ has two poles. One first-order pole at $z_1=E_m+i\eta$ and one second-order pole at $z_2=E_n+i\eta$. The residues are (I hope I am doing them right): $$ \text{Res}(f(z),z_1) = -\frac{1}{2π} \frac{E_m^2}{(E_m-E_n )^2} \\ \text{Res}(f(z),z_2) = -\frac{1}{2π} \left( \frac{2E_n}{(E_n-E_m)}-\frac{E_n^2}{(E_n-E_m )^2}\right) $$ Hence, the integral is $$ I = -i\left(\frac{E_m^2}{(E_m-E_n )^2} +\frac{2E_n}{(E_n-E_m)}-\frac{E_n^2}{(E_n-E_m )^2}\right) \\ = -i\left(\frac{E_m^2-E_n^2}{(E_n-E_m )^2} +\frac{2E_n}{(E_n-E_m)}\right)\\ = -i\frac{1}{(E_n-E_m )^2}\left(E_m^2-E_n^2 +2E_n(E_n-E_m )\right) = -i $$
The value is simply $-i$, but during numerical integration, I consistently observe a real part of 0, as expected. However, the imaginary part equals -1 only as $\eta$ approaches 0. I am curious about the reason behind this phenomenon. Why does the residue I calculate show no dependence on $\eta$? Is there something I might be overlooking?