Find an expression for $\displaystyle \int_0^{\infty} \frac{1}{p(x)} \, dx$ where $\displaystyle p(x)$ is a real polynomial having no nonnegative real zeroes.

Ok. So I think it's clear that I'm supposed to set this up for an application of the residue theorem by considering $\displaystyle \int_C \frac{\log z}{p(z)}dz$ where $\displaystyle C$ is the cut-annulus contour (an annulus missing a band about the real line). Then the residue theorem says that this complex line integral has the value $\displaystyle 2 \pi i \cdot \sum_{j=1}^n \left( Res _{1/p(x)}(z_j) \cdot Ind _{C}(z_j) \right) $, where $\displaystyle n$ is the number of zeroes of $\displaystyle p(z)$ and each $\displaystyle z_j$ is one of these zeroes. We can ignore the index for each zero given the contour because for any zero (which produces an isolated singularity) the index will be one.

But now what? All I've really done is write down the residue theorem. I don't know the zeroes of the polynomial, so no further calculation can be done with the above. Am I supposed to break the integral up now and look at the integral over each piece? If I let $\displaystyle C1$ be the staight-line piece just above the real axis, $\displaystyle C2$ be the outer ring, $\displaystyle C3$ be the straight-line piece just below the real axis, and $\displaystyle C4$ be the inner ring, would that go something like:

$\displaystyle \int_{C1}\frac{\log z}{p(z)}dz + \int_{C2}\frac{\log z}{p(z)}dz + \int_{C3}\frac{\log z}{p(z)}dz + \int_{C4}\frac{\log z}{p(z)}dz$ ?

But then I have to fix the inner radius, the outer radius, and the width of the deleted band of the annulus, and then take limits as the width and inner radius go to zero and the outer radius goes to infinity? I think what happens is that the integrals over the rings vanish, but the other 2 integrals do not...but I have no information really about the polynomial so I'm lost. How should I proceed (especially formally)?