Compute $\int_{0}^{\infty} \frac{x^\alpha}{x^2+x+1}dx$ So I am asked to integrate:
$\int_{0}^{\infty} \frac{x^{\alpha}}{x^{2}+x+1}dx$, where $-1<\alpha<1$
As this is done in the Complex Analysis course I tried to consider the function 
$f(z)=\frac{z^{\alpha}}{z^{2}+z+1}$ 
and to integrate it over the contour:
$\gamma_{1}$: real line from 0 to R, which is basically our integral
$\gamma_{2}$: the quarter of the circle of radius R, with $\theta$ from 0 to $\pi/2$.
$\gamma_{3}$: line from $iR$ to $0.$
Now my argument is that the integral over this contour is $0$ as the are no singularities of $f$ inside the curve.
I managed to prove that the integral over the quarter of the circle tends to $0$ as $R$ tends to infinity but I do not know how i can compute the integral over the imaginary line.

Take the function $$f\left(z\right)=\frac{z^{\alpha}}{z^{2}-z+1}$$ and the classic keyhole contour. Then $$\int_{\Gamma}f\left(z\right)dz=2\pi i\left(\underset{z=\frac{1+i\sqrt{3}}{2}}{\textrm{Res}}f\left(z\right)+\underset{z=\frac{1-i\sqrt{3}}{2}}{\textrm{Res}}f\left(z\right)\right)=2\pi\frac{\left(\frac{1+i\sqrt{3}}{2}\right)^{\alpha}}{\sqrt{3}}-2\pi\frac{\left(\frac{1-i\sqrt{3}}{2}\right)^{\alpha}}{\sqrt{3}}$$ and since over the circumferences we have $$\left|iR\int_{0}^{\pi}\frac{R^{\alpha}e^{i\theta\alpha}}{R^{2}e^{2i\theta}-Re^{i\theta}+1}e^{i\theta}d\theta\right|\leq\pi\frac{R^{1+\alpha}}{R^{2}-R-1}\rightarrow0$$ as $R\rightarrow\infty$ or $R\rightarrow0$ since $-1<\alpha<1$, we have $$\int_{0}^{\infty}\frac{x^{\alpha}}{x^{2}+x+1}dx=\frac{2\pi\left(\left(\frac{1+i\sqrt{3}}{2}\right)^{\alpha}-\left(\frac{1-i\sqrt{3}}{2}\right)^{\alpha}\right)}{\sqrt{3}\left(e^{i\pi a}-e^{-i\pi a}\right)}=\color{red}{\frac{2\pi}{\sqrt{3}\left(1+2\cos\left(\frac{2\pi a}{3}\right)\right)}}.$$
