0
$\begingroup$

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.

enter image description here

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

1 Answer 1

3
$\begingroup$

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

$$z \in \left\{\frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} i \;,\; \frac{\sqrt 2}{2} - \frac{\sqrt 2}{2} i \;,\; -\frac{\sqrt 2}{2} + \frac{\sqrt 2}{2} i \;,\; -\frac{\sqrt 2}{2} - \frac{\sqrt 2}{2} i\right\}$$

$\endgroup$
4
  • $\begingroup$ should all the $\sqrt{2}$ be $\frac{1}{\sqrt{2}}$ $\endgroup$ Commented Jan 28, 2021 at 3:27
  • 1
    $\begingroup$ Yeah, that was my bad, got a bit hasty. $\endgroup$ Commented Jan 28, 2021 at 3:27
  • $\begingroup$ oh no... my bad was not paying attention! very silly error from me! $\endgroup$
    – DJA
    Commented Jan 28, 2021 at 3:29
  • 4
    $\begingroup$ All good! It happens to the best of us. :) $\endgroup$ Commented Jan 28, 2021 at 3:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .