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I'm trying to find the poles/residues of this integral:

$$\int_{-\infty}^\infty \frac{x^2 \, dx}{1 + x^4}$$

I've been given this attempt for a solution, but I don't really understand the procedure for how the poles are found.

EDIT:

From the solution on the bottom of this post:

Specifically I don't see how: $ \frac{p(z)}{q'(z)} = \frac{z^2}{4z^3} = \frac{1}{4z} .$

becomes this:

$ \frac{p(z)}{q'(z)} = - \frac{z^3}{4} $

What exactly is happening here?

END EDIT

As I understand it there should be $2$ poles, each of order $1$.

solution to the problem

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  • 2
    $\begingroup$ To find the poles you need to find the roots of $x^4 +1 = 0$ i.e. you need to compute the fourth roots of $-1$. $\endgroup$ – Joelafrite May 25 '15 at 19:17
  • $\begingroup$ Your user name reminded me of this old song. :-) $\endgroup$ – Lucian May 26 '15 at 1:22
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Hint:

  • Fist find the roots of $z^4 = -1$.
  • Choose a path containg $2$ of those roots. Say a semi-circunference of radius $R > 1$.
  • Use the following Theorem:

If $p$ and $q$ are functions, analytic at $z_0$, with $p(z_0) \neq 0$ and $q(z_0) = 0$ and $q'(z_0) \neq 0$ then $z_0$ is a simple pole of $p(z)/q(z)$ and $$\mathrm{Res}_{z=z_0} \frac{p(z)}{q(z)}= \frac{p(z_0)}{q'(z_0)}$$

what are $p$ and $q$?

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  • 1
    $\begingroup$ Adding to bullet #2, there is a lemma that states that if $f=Q(z)/P(z)$ is a quotient of two polynomials s.t. $degree Q \ge 2+degree P$, then $lim_{\rho \rightarrow \infty} \int_{C^+_\rho} f(z)\ dz = 0$. Meaning, you can take your closing contour to be a semi-circle in the upper half plane in this circumstance. $\endgroup$ – zahbaz May 25 '15 at 19:45
  • $\begingroup$ Ah... I meant if $f=P(z)/Q(z)$. Sorry. $\endgroup$ – zahbaz May 25 '15 at 19:58
  • $\begingroup$ $p(z) = x^2$. $q(z) = 1 + x^4$. so $q'(z) = 4x^3$ $\endgroup$ – jibo May 26 '15 at 8:48
  • $\begingroup$ Thanks! But IMO $ \frac{p(z)}{q'(z)} = \frac{z^2}{4z^3} = \frac{1}{4z} $ and not : $\frac{p(z)}{q'(z)} = - \frac{z^3}{4}$ as the solution claims. What am I missing? $\endgroup$ – jibo May 26 '15 at 9:52
  • $\begingroup$ First, it's not $z$ and rather $z_k$ with $k = 0, 1$. Now multiply your fraction by $\frac{z^4_k}{z^4_k}$ and notice that $z^4_k = -1$. $\endgroup$ – Aaron Maroja May 26 '15 at 13:39

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