Determining the contour to use during contour integration

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.

• $z^4=1$ and so $z^4+1=1+1=2$ and so they are not poles. Commented Jan 28, 2021 at 3:26
• I’m voting to close this question because I made a mistake in the question which makes the question pointless.
– DJA
Commented Jan 28, 2021 at 3:30

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.
$$z \in \left\{\frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} i \;,\; \frac{\sqrt 2}{2} - \frac{\sqrt 2}{2} i \;,\; -\frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} i \;,\; -\frac{\sqrt 2}{2} - \frac{\sqrt 2}{2} i\right\}$$
• should all the $\sqrt{2}$ be $\frac{1}{\sqrt{2}}$ Commented Jan 28, 2021 at 3:27