Calculating integral $\int_{-\infty}^{\infty}\frac{1}{(1+y^2)^2}dy$ using Residue theorem

Question

In one of the questions about applying the residue theorem in order to solve an integral:

$$\int_{-\infty}^{\infty}\frac{1}{(1+y^2)^2}dy$$

The user (@user438666) answered the question and everything is clear to but one part:

Take the semicircle $\Gamma=[-R,R] \cup \{z:|z|=R, \mathrm{Im}z>0\}$. The integral on this path is

$$I=\oint_\Gamma \frac{\mathrm{d}z}{(1+z^2)^2}=\int_{-R}^R\frac{1}{(1+t^2)^2}\mathrm{d}t +\int_0^\pi\frac{Rie^{i\theta}}{(1+Re^{2i\theta})^2}\mathrm{d}\theta$$

Let us call the second term $I_2$ and notice $$ |I_2|\leq\int_0^\pi\frac{R}{|1+Re^{2i\theta}|^2}\mathrm{d}\theta\rightarrow0$$

as $R\rightarrow \infty$. This implies $$ \lim\limits_{R\rightarrow\infty} I=\int_{-\infty}^{\infty}\frac{1}{(1+t^2)^2}\mathrm{d}t$$ On the other hand by the residue theorem

$$ I=2\pi i\mathrm{Res}\left( \frac{1}{(1+z^2)^2}, z=i\right)=2\pi i \left[\frac{\mathrm{d}}{\mathrm{d}z}\frac{1}{(z+i)^2}\right]_{z=i}=2\pi i \frac{-2}{(2i)^3}=2\pi i \frac{1}{4i}=\frac{\pi}{2}$$

Because $z=i$ is the only singularity in the semicircle, the other singularity in $z=-i$ is on the other half of the plane! Do you understand the reasoning?

MY QUESTION: Can someone explain to me how did the user calculate in the last part the Residue: $\mathrm{Res}\left( \frac{1}{(1+z^2)^2}, z=i\right)$. I see that he has calculated the derivative, but I'm not sure whether I can use it since it is not L'Hospital's rule.

Answer 1

If $f$ has a pole of order $2$ at $a$ then the residue of $f$ at $a $ is $\lim_{z \to a} \frac d {dz} [(z-a)^{2} f(z)]$. This result can be found in almost any book which discusses Residue Theorem.