$$\begin{aligned}\int_{0}^{\infty}\frac{\cos(2\pi x)}{x^{4}+x^{2}+1}\mathrm{d}x &=\Re\left(\int_{0}^{\infty}\frac{e^{2\pi xi}}{x^{4}+x^{2}+1}\mathrm{d}x\right)\\ &=\Re\left(\lim_{R\to \infty}\int_{0}^{R}\frac{e^{2\pi xi}}{x^{4}+x^{2}+1}\mathrm{d}x\right)\\ &=\Re\left(-i\lim_{R\to \infty}\int_{0}^{R}\frac{1}{\gamma(t)^{4}+\gamma a(t)^{2}+1}\mathrm{d}t\right) \end{aligned}$$

Now we have $2$ poles with the order of $2$ so: $$ \text{Res}(f,-i)=(t+i)^{2}\frac{4x^{3}+2x}{(t-i)^{2}(t+i)^{2}}=\frac{4i-2i}{-4}\\ \text{Res}(f,i)=(t-i)^{2}\frac{4x^{3}+2x}{(t-i)^{2}(t+i)^{2}}=\frac{4i-2i}{-4} $$ Therefore they cancel each out, so I get that the integral should be $0$. Can someone spot my mistakes? and provide me solution?

2nd attempt: $$ \int_{0}^{\infty}\frac{\cos(2\pi x)}{x^{4}+x^{2}+1}\mathrm{d}x=\Re\left(\int_{0}^{\infty}\frac{e^{2\pi xi}}{x^{4}+x^{2}+1}\mathrm{d}x\right)\\\int_{0}^{\infty}\frac{\cos(2\pi x)}{x^{4}+x^{2}+1}\mathrm{d}x=\Re\left(\int_{0}^{\infty}\frac{e^{2\pi xi}}{(x\pm(-1)^{\frac{2}{3}})(x\pm\sqrt[3]{-1})}\mathrm{d}x\right)\\\Re\left(\underset{R\rightarrow\infty}{\lim}\int_{0}^{R}\frac{e^{2\pi xi}}{(x\pm(-1)^{\frac{2}{3}})(x\pm\sqrt[3]{-1})}\mathrm{d}x\right)=\\\Re\left(\underset{R\rightarrow\infty}{\lim}\frac{\oint_{c}\frac{e^{2\pi xi}}{(x\pm(-1)^{\frac{2}{3}})(x\pm\sqrt[3]{-1})}\mathrm{d}x}{2}\right)=\pi i(Res(f,\sqrt[3]{-1})+Res(f,-\sqrt[3]{-1}))\\Res(f,\sqrt[3]{-1}))=\frac{e^{2\pi\sqrt[3]{-1}i}}{(\sqrt[3]{-1}+(-1)^{\frac{2}{3}})((\sqrt[3]{-1}-(-1)^{\frac{2}{3}}))(\sqrt[3]{-1})+\sqrt[3]{-1})}\\Res(f,-\sqrt[3]{-1}))=\frac{e^{2\pi\sqrt[3]{-1}i}}{(\sqrt[-3]{-1}+(-1)^{\frac{2}{3}})((\sqrt[-3]{-1}-(-1)^{\frac{2}{3}}))(\sqrt[3]{-1})+-\sqrt[3]{-1})}\\\Re\left(\underset{R\rightarrow\infty}{\lim}\frac{\oint_{c}\frac{e^{2\pi xi}}{(x\pm(-1)^{\frac{2}{3}})(x\pm\sqrt[3]{-1})}\mathrm{d}x}{2}\right)=\pi i(\frac{e^{2\pi\sqrt[3]{-1}i}}{(\sqrt[3]{-1}+(-1)^{\frac{2}{3}})((\sqrt[3]{-1}-(-1)^{\frac{2}{3}}))(\sqrt[3]{-1})+\sqrt[3]{-1})}+\frac{e^{2\pi\sqrt[3]{-1}i}}{(\sqrt[-3]{-1}+(-1)^{\frac{2}{3}})((\sqrt[-3]{-1}-(-1)^{\frac{2}{3}}))(\sqrt[3]{-1})+-\sqrt[3]{-1})}) $$

  • $\begingroup$ How do you know you have a mistake? $\endgroup$ – Eric Wofsey Aug 1 '20 at 16:42
  • $\begingroup$ because it seems really off. $\endgroup$ – hash man Aug 1 '20 at 16:42
  • $\begingroup$ The poles are the four complex sixth roots of 1 $\endgroup$ – Empy2 Aug 1 '20 at 17:06

There are several issues.

  • Starting from the second line, the cosine is the real part, not the imaginary of $e^{i2\pi x}$.
  • Then, which one is your closed contour? Do you draw the semicircle in the upper half of the complex plane, or in the lower?.
  • Also, in that case, what's happening to the integral from $-R$ to $0$?
  • If you have a semicircle, you might capture only one of the poles inside.


Based on your second attempt, here are some suggestions:

  • Starting at the beginning, $$ \int_{0}^{\infty}\frac{\cos(2\pi x)}{x^{4}+x^{2}+1}\mathrm{d}x=\frac12\int_{-\infty}^{\infty}\frac{\cos(2\pi x)}{x^{4}+x^{2}+1}\mathrm{d}x=\frac12\Re\left(\int_{-\infty}^{\infty}\frac{e^{2\pi xi}}{x^{4}+x^{2}+1}\mathrm{d}x\right)$$ If you use instead $\lim_{R\to\infty}\int_{-R}^R...$, you will notice that for your integral the sine part is $0$, which is the imaginary part, so you don't need to take the real part.
  • The notation $\sqrt[3]{-1}$ is confusing. Use $e^{i\pi/3}$ instead. Alternatively, this is $\frac 12+i\frac{\sqrt 3}2$.
  • The above notation (especially the one with the exponential) allows you to quickly determine which poles are inside your contour
  • The above notation will also allow you to simplify your final expression
  • $\begingroup$ thank you! regarding the 2nd comment why half a circle and not a full circle? $\endgroup$ – hash man Aug 1 '20 at 17:06
  • $\begingroup$ if you can show a full solution I'll be delighted . $\endgroup$ – hash man Aug 1 '20 at 17:07
  • $\begingroup$ Because a full circle does not contain, as part of the contour, the real line $\endgroup$ – Andrei Aug 1 '20 at 17:08
  • $\begingroup$ so why a half a circle and not a quarter of a circle? as we begin with 0 and not -R $\endgroup$ – hash man Aug 1 '20 at 17:09
  • $\begingroup$ Because then you need to calculate the integral on the imaginary axis. But your function is even in $x$, so the integral should be easy to express $\endgroup$ – Andrei Aug 1 '20 at 17:14

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