Fourier transform of $f(x)=\frac{1}{1+x+x^2}$

Question

I'm trying to find the fourier transform of $f(x)=\frac{1}{1+x+x^2}$ with the fourier transform given as $F(n) = \frac{1}{\sqrt{2\pi}}\int_{-\pi}^{\pi}f(x)e^{-inx}dx$.

I guess I have to find some other representation for $f$ in order to be able to calculate this but I have no clue where to start.

Thanks!

Answer 1

You might use G. Sassatelli good hint, and compute the easier calculation then.

Your result will be:

$$\frac{\sqrt[6]{-1} \sqrt{2 \pi } e^{-(-1)^{5/6} n} \left(\theta (-n)+e^{-\sqrt{3} n} \theta (n)\right)}{1+\sqrt[3]{-1}}$$

Answer 2

Using the contour $\gamma=[-R,R]\cup Re^{-i[0,\pi]}$, as $R\to\infty$, which circles the singularity at $x=-\frac12-i\frac{\sqrt3}2$ clockwise, $$ \begin{align} \frac1{\sqrt{2\pi}}\int_{-\infty}^\infty\frac{e^{-inx}}{1+x+x^2}\,\mathrm{d}x &=\frac1{\sqrt{2\pi}}\int_\gamma\frac{e^{-inx}}{\left(x-\left(-\frac12-i\frac{\sqrt3}2\right)\right)\left(x-\left(-\frac12+i\frac{\sqrt3}2\right)\right)}\,\mathrm{d}x\\ &=\frac1{\sqrt{2\pi}}\underbrace{\ \ (-2\pi i)\ \ \vphantom{\frac1{\sqrt3}}}_\text{clockwise}\underbrace{\frac{e^{-in\left(-\frac12-i\frac{\sqrt3}2\right)}}{-i\sqrt3}}_\text{residue}\\[6pt] &=\sqrt{\frac{2\pi}3}\,e^{-n\sqrt3/2}e^{in/2} \end{align} $$