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

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.

• Note that if $f(z)=\frac{a_{-2}}{(z-i)^2}+\frac{a_{-1}}{z-i}+a_0+a_1(z-i)+...$, then $(z-i)^2f(z)=a_{-2}+a_{-1}(z-i)+a_0(z-i)^2+...$ and $((z-i)^2f(z))'=a_{-1}+a_0(z-i)+...$. Therefore, if you have a pole of order $2$ at $i$, you multiply by $(z-i)^2$, take derivative and evaluate the result at $z=i$, you get $a_{-1}$, which is the residue. Commented Jan 26, 2020 at 23:58
• "In one of the questions..." what question? Link? Commented Jan 27, 2020 at 0:13

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.
• I have totally forgotten the square of $(z-a)$. Thank you very much! Commented Jan 27, 2020 at 0:03