Let us say we want to integrate
$$\int_{-\infty}^{\infty} \frac{dx}{1+x^4}$$
We do this by c contour integration of the form:
$$\oint_{-\infty}^{\infty} \frac{dz}{1+z^4}$$ However my question concerns the type of contour we use. Now everywhere that I have seen do this online do it via taking the semicircle counter. However, we know we have poles at $$z = 1,-1,i,-i$$
So my question is since we have a singularity at -1, 1 then surely these would lie on the path integral if we choose a semicircle. Surely a more appropriate contour would be a semi circle with 2 small semi circles of very small (tend to 0) radius $\epsilon$ as shown below.
The poles are not at $z \in \{1,-1,i,-i\}$. Note that, for all of these, $z^4 = 1$ and thus the denominator is nonzero.
Rather, the poles are at
$$z \in \left\{\frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} i \;,\; \frac{\sqrt 2}{2} - \frac{\sqrt 2}{2} i \;,\; -\frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} i \;,\; -\frac{\sqrt 2}{2} - \frac{\sqrt 2}{2} i\right\}$$