Find $$ \int_0^\infty \frac{\sqrt{x}\ dx}{x^2+5x+6}$$
This problem is slightly unusual as the poles are on the real axis, + it has the square root so we will need to deal with a log branch.
Here a terrible sketch of my contour.
I know that the integral over $z_2$ and $z_4$ goes to $0$ regardless of how much of an arch we make. $z_1$ gives us the desired integral letting $\epsilon \to 0$. Now $z_3$ is parameterized so that it is a straight line a little bit below the real axis $z_3(t)=t+i\delta, t\in [-\sqrt{R^2-\delta^2},-\sqrt{\epsilon^2-\delta^2}]$. Now by definition of contour integral over $z_3$ we get $$\int_{-\sqrt{R^2-\delta^2}}^{-\sqrt{\epsilon^2-\delta^2}}\frac{\sqrt{t+i\delta}}{(t+i\delta)^2+5(t+i\delta)+6}$$ letting $\delta \to 0$ ( I am having trouble justifying moving the limit inside) we get $\int_{-R}^{-\epsilon}\frac{\sqrt{t}}{t^2+5t+6}$ which is imaginary, so we just have to take the real part of the residue and we are done.
Is this correct? It gives me the right answer but i am not sure if all i did was valid.
Best Answer
If you do $x=y^2$ and $\mathrm dx=2y\,\mathrm dy$, then your integral becomes$$2\int_0^\infty\frac{y^2}{y^4+5y^2+6}\,\mathrm dy=\int_{-\infty}^\infty\frac{y^2}{y^4+5y^2+6}\,\mathrm dy.$$Now, you are integrating a function which has poles at $\pm\sqrt2\,i$ and $\pm\sqrt3\,i$. So, your integral is equal to$$2\pi i\left(\operatorname{res}\left(\sqrt 2\,i,\frac{z^2}{z^4+5z^2+6}\right)+\operatorname{res}\left(\sqrt 3\,i,\frac{z^2}{z^4+5z^2+6}\right)\right).$$These residues are equal to $\frac{\sqrt2}2i$ and to $-\frac{\sqrt3}2i$ respectively. So, your integral is equal to $\pi\left(\sqrt3-\sqrt2\right)$.