I am considering an integral around the path $ \Gamma = C_1 \cup C_{\varepsilon_{1}} \cup C_2 \cup C_{\varepsilon_{2}}$ of a function $f(z)$ that has a pole in every cross in the images below. In particular, this function has infinitely many simple poles in the imaginary axis.
The contours $C_1$ and $C_2$ vary from $-\infty$ to $\infty$ and $C_{\varepsilon_{1}}$ and $C_{\varepsilon_{2}}$ can be arbitrarily close to the imaginary axis without changing the value of the integration, a result assumed to be known. The expression for the contour integral is
$\int_{\Gamma}f(z)dz = \int_{C_1}f(z)dz + \int_{C_2}f(z)dz + \int_{C_{\varepsilon_{1}}}f(z)dz + \int_{C_{\varepsilon_{2}}} f(z)dz. \tag{1}$
To be able to do the integration I need to change the contour without changing the value of the integral. I can do the following
In the above icontour, I know how to show that the integrals through $C_{R_1}$ and $C_{R_2}$ go to zero using Jordan's lemma, the remaining integrals I can do using the residue theorem for the four poles inside the contour. So that, the resulting general expression after the vanishing of the integrals through $C_{R_1}$ and $C_{R_2}$
$\int_{\Gamma}=\int_{C_1}f(z)dz + \int_{C_2}f(z)dz.\tag{2}$
To complete my demonstration I need to show that the integrals through $C_{\varepsilon_{1}}$ and $C_{\varepsilon_{2}}$ go to zero, with that, equation (1) is equal to (2) and I can do the integration.
I think they go to zero because the path can be arbitrarily small, but I am not sure of this argument. I would appreciate if someone could give an argument on how I can show that the expressions for $C_{\varepsilon_{1}}$ and $C_{\varepsilon_{2}}$ vanish in equation (1).
Edit: $f(z)=\dfrac{1}{(1+e^z)(z^2 + a^2)((z-c)^2 - b^2)},$ where $a,b$ and $c$ are constants.