Let $f(x) = p(x)/q(x)$ and $deg(p) = deg(q) - 1$. Show that $\int_{-\infty}^\infty f(x) = 0$

Question

I would like to show that $\int_{-\infty}^\infty f(x)dx = 0$ where $f(x) = \dfrac{p(x)}{q(x)}$ and $deg(p) = deg(q) - 1$. Also, $q(x)$ has no real roots. I was considering integrating along the contour $C_R$, where $C_R$ is the real line segment from $-R$ to $R$ and the upper semi circle, in which case

$$\lim_{R \to \infty} \int_{C_R} f(z)dz = \lim_{R \to \infty} \int_{-R}^R f(x) dx+\int_{\Gamma_R}f(z)dz = 2\pi i\sum_{k}Res(f, z_k)$$

where $z_k$ are the zeroes of $q(x)$ in the upper half plane, and $\Gamma_R$ is the upper semicircle. However, I'm not sure where to proceed from here

Any help would be appreciated.

Answer 1

It's not true in general. For example, the Cauchy principal value integral $$\int_{-\infty}^\infty \frac{dx}{x-i} = \pi i$$

EDIT: More generally, the Cauchy principal value

$$ \int_{-\infty}^\infty \frac{dx}{x-r} = \int_{-\infty}^\infty \frac{dx}{x-{\text Im}(r)} = \cases{\pi i & if $\text{Im}(r)>0$\cr -\pi i & if $\text{Im}(r)<0$}$$

The Cauchy principal value of $\int_{-\infty}^\infty f(x)\ dz$ is $\pi i$ times the difference between the sum of the residues of $f$ in the upper half plane and the sum of the residues in the lower half plane.

Answer 2

The result is not true, and it is hard to see how to "save" it. If $p(x)=x+a$ and $q(x)=x^2+1$, then $\lim\limits_{R\to+\infty}\int\limits_{-R}^{+R}f(x)\mathrm dx=a\pi$ hence the Cauchy principal value exists but is not zero in general.