calculate: $\int_{0}^{\infty}\frac{\cos(2\pi x)}{x^{4}+x^{2}+1}dx$; find my mistake

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

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

EDIT

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
• thank you! regarding the 2nd comment why half a circle and not a full circle? – hash man Aug 1 '20 at 17:06
• if you can show a full solution I'll be delighted . – hash man Aug 1 '20 at 17:07
• Because a full circle does not contain, as part of the contour, the real line – Andrei Aug 1 '20 at 17:08
• so why a half a circle and not a quarter of a circle? as we begin with 0 and not -R – hash man Aug 1 '20 at 17:09
• 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 – Andrei Aug 1 '20 at 17:14