I have to calculate the integral $$I =\int_{-\infty}^{\infty} \frac{\sin (ax)}{x^2 + x +1}dx$$.
I know that $I = \Im(\int_{-\infty}^{\infty} \frac{e^{iax}}{x^2 + x +1}dx)$. Then I substitute $x=z$ and get $f(z) = \frac{e^{iax}}{z^2 + z + 1} = \frac{e^{iax}}{(z-i)(z+i)+ z}$
I know I can use the contour $C = C_1 + C_R$ where $C_1$ is a straight line on the real axis from $-R$ to $R$ and $C_R$ is a semicircle from $-R$ to $R$. So I get $\int_{C_1 + C_R} f(z)dz = 2\pi i \sum Res(f,z_0)$.
The integral over the contour $C_R$ is for $R \to \infty$ equal to $0$. So we are left with the integral over $C_1$. We have then $\lim_{R \to \infty} \int_{C_1} f(z)dz = \lim_{R \to \infty} \int_{-R}^{R}f(x)dx = \int_{-\infty}^{\infty}f(x)dx = 2\pi i \sum Res(f,z_0)$
The residues can be determined by the following equation: $$Res(f,z_0) = \frac{1}{(N-1)!}\lim_{z\to z_0}(\frac{d}{dz})^{N-1} (z-z_0)^{N}f(z)$$ where $N$ is the order of the poles. Since I have $f(z) = \frac{e^{iax}}{(z-i)(z+i)+ z}$, the poles are at $z_0 = -\frac{1}{2}\pm i\frac{\sqrt{3}}{2} = -e^{\mp i\frac{\pi}{3}}$ but in my case I need only the $z_0 = -e^{-i\frac{\pi}{3}}$.
Here I'm stuck because I'm not sure what the order is and how to use it on the equation for calculating the residues