So, I'm working on the following problem.
Compute $ \int_{-\infty}^{\infty} \frac{x^2}{x^4 +4}dx $
I've chosen the following contours: $$ C_1: z(t)=Rt, \ -1 \le t \le 1 \\ C_2: z(\theta)=Re^{i\theta}, \ 0\ \le \theta \le \pi \\ C=C_1 \, + \, C_2 $$
Now, to find the singularities.
$$ x^4+4=0 \Rightarrow x=\sqrt[4]{-4}=\sqrt[4]{4e^\pi}=\sqrt2 \sqrt[4]{e^\pi}=\sqrt2 \, \exp[i(\frac{\pi}{4}+\frac{2\pi k}{4})] \, \, \text{for} \, \, k=0,1,2,3 $$
Let $ z_1=\sqrt2 e^{\pi i/4}, \ z_2= \sqrt2 e^{3\pi i/4}, \ z_3=\sqrt2 e^{-3\pi i/4}, \ z_4=\sqrt2 e^{-\pi i/4} $
So, my question is regarding the choice for which singularity to choose in the CIF. I understand that $z_3$ and $z_4$ are outside of the contour and therefore irrelevant. But now I'm left with $z_1$ and $z_2$. In my professor's solution to this problem, he says
Notice that for $R >\sqrt2$, $C$ may be deformed into a contour which is positively oriented and contains a small circle around one relevant singularity, $1 +i$, which I’ll call $C_3$, the other singularity $−1 +i$, which I’ll call $C_4$, the line connecting them $C_5$ and $−C_5$ to close the contour.
I don't really follow his logic here, why is $z_1$ the only relevant singularity when $z_2$ is also on the same line? Could someone please elaborate on this?
Thank you.