Define $f$ by $$f(z) = \frac{z^{1/2}}{z^2+z+1}$$ where $z \mapsto z^{1/2}$ is the holomorphic function on $\mathbb{C} \setminus (-\infty,0] $ and $z^{1/2}$ is the regular square root for real $z>0$.
Compute the integral $$\int^{\infty}_0\frac{x^{1/2}}{x^2+x+1}dx.$$
I thought of computing the residues of $f$ and then doing a contour integration over the upper half circle going from $\theta = 0$ to $\theta = 3\pi/4$. But in my calculations I always get stuck. Am I doing something wrong here?
If you do the substitution $x=y^2$ and $\mathrm dx=2y\,\mathrm dy$, your integral becomes$$\int_0^\infty\frac{2y^2}{y^4+y^2+1}\,\mathrm dy=\int_{-\infty}^\infty\frac{y^2}{y^4+y^2+1}\,\mathrm dy$$The roots of $y^4+y^2+1$ are $\pm\frac12\pm\frac{\sqrt3}2i$. Among these, those with imaginary part greater than $0$ are $\pm\frac12+\frac{\sqrt3}2i$. So, by the residue theorem,\begin{align}\int_{-\infty}^\infty\frac{y^2}{y^4+y^2+1}\,\mathrm dy&=2\pi i\left(\operatorname{res}_{z=\frac12+\frac{\sqrt3}2i}\frac{z^2}{z^4+z^2+1}+\operatorname{res}_{z=-\frac12+\frac{\sqrt3}2i}\frac{z^2}{z^4+z^2+1}\right)\\&=2\pi i\left(\frac{\sqrt3-i}{4\sqrt3}-\frac{\sqrt3+i}{4\sqrt3}\right)\\&=\frac\pi{\sqrt3}.\end{align}