-3

Here is the integration I want to calculate (complex analysis):

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

But I do not know how to factorize the following equation $x^4 + x^2 + 1$ to get the singularities. I first let $y = x^2$ and I used the quadratic formula to get $y = \frac{-1 \pm \sqrt{3}i}{2}$ but then what should I do to get $x$? Could someone help me please?

EDIT:

I think my professor did it using that any complex number can be written as $r e^{i \theta}$ but I do not know how he got the $\theta$, could anyone show me the solution by this method please?

Intuition
  • 3,269
  • If $y=x^2$, $x=\sqrt{y}$. Probably easier by using de Moivre on the polar form, $2 \text{cis} \dots$ whatever the angle is. – Nij Sep 30 '22 at 11:01
  • 1
    You can compute it using polar coordinates. Singularities are at $y = \exp\left(\frac{2 \pi}{3}\right)$ and $y = \exp\left(\frac{4 \pi}{3}\right)$, thus $x = e^{i \phi}$ where $\phi \in \left{\frac{\pi}{3}, \frac{5 \pi}{3}, \frac{2 \pi }{3}, \frac{4 \pi }{3}\right}$. – MrGhost Sep 30 '22 at 11:07
  • @AnneBauval no it does not, I think my professor did it using that any complex number can be written as $r e^{i \theta}$ but I do not know how he got the $\theta $ – Intuition Sep 30 '22 at 11:41
  • Then, use this one. – Anne Bauval Sep 30 '22 at 11:58
  • @MrGhost How did you know these values? – Intuition Sep 30 '22 at 12:35
  • 1
    @Secretly since $e^{i \phi} = \cos \phi + i \sin \phi$ I observed that $- \frac{1}{2} = \cos \left(\frac{2 \pi}{3}\right)$ and also $\frac{\sqrt{3}}{2} = \sin \left(\frac{2 \pi}{3}\right)$. Thus $\phi = \frac{2\pi}{3}$. Similarly for $-\frac{1}{2} - \frac{\sqrt{3}}{2}$ you get that $\phi = \frac{4\pi}{3}$. – MrGhost Sep 30 '22 at 13:17

3 Answers3

1

You can multuply $x^4+x^2+1$ with $x^2 - 1$ and solve equation $x^6=1$ then remember that extrude root $x = \pm 1$.

1

You can compute the square roots of your $y_\pm:=\frac{-1\pm i\sqrt3}2$ directly, by the classical method indicated in the proposed duplicate. E.g. for $y_+$: let $z=a+ib$ with $a,b\in\mathbb R$. $$\begin{align}z^2=\frac{-1+i\sqrt3}2&\Leftrightarrow a^2-b^2=-\frac12,\;a^2+b^2=1,\;2ab>0\\ &\Leftrightarrow a^2=\frac14,\;b^2=\frac34,\;ab>0\\ &\Leftrightarrow z=\pm\frac{1+i\sqrt3}2.\end{align}$$

Anne Bauval
  • 34,650
0

Method I

$x^4+x^2+1=x^4+2x^2+1-x^2=(x^2+1)^2-x^2=(x^2+x+1)(x^2-x+1)$ $=(x-a_1)(x-a_2)(x-a_3)(x-a_4)$

where $a_1,a_2,a_3,a_4$ are roots of those two quadratic equations.

Method II

Start from your $x^2=y= \frac{-1 \pm \sqrt{3}i}{2}=cos\pm\frac{2\pi}{3}+isin\pm\frac{2\pi}{3}=cos\pm\frac{4\pi}{3}+isin\pm\frac{4\pi}{3}$

Then you can get 4 $x$'s like

$x=cos\pm\frac{\pi}{3}+isin\pm\frac{\pi}{3}, cos\pm\frac{2\pi}{3}+isin\pm\frac{2\pi}{3}$

Abel Wong
  • 1,173