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.

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    $\begingroup$ If $deg(p)=deg(q)-1$, your function is not integrable at $\infty$. $\endgroup$ – Jean-Claude Arbaut Mar 21 '13 at 6:29
  • $\begingroup$ Are you assuming that $q$ has no zeros? $\endgroup$ – Julien Mar 21 '13 at 6:37
  • $\begingroup$ I'm assuming $q$ has no real roots, but it may have complex zeroes. $\endgroup$ – Student1 Mar 21 '13 at 6:38
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    $\begingroup$ The function decays too slowly, for large $x$ it behaves like $\frac{c}{x}$. $\endgroup$ – André Nicolas Mar 21 '13 at 6:42
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    $\begingroup$ Well,for instance take $f(x)=2x/(1+x^2)$. Then $\int_{0}^Bf(x)dx=\ln(1+B^2)\longrightarrow +\infty$ as $B\longrightarrow +\infty$. What you are talking about here, I guess, is Cauchy Pincipal Value, which is not the Lebesgue integral over $(-\infty,+\infty)$. That's $\lim_{B\rightarrow+\infty}\int_{-B}^Bf(x)dx$. $\endgroup$ – Julien Mar 21 '13 at 7:18

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.


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.

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    $\begingroup$ To save it, we would need for instance to assume that $p/q$ is odd. Which makes it a highly noninteresting fact wich has nothing to do with rational functions... $\endgroup$ – Julien Mar 21 '13 at 21:55
  • $\begingroup$ @julien Indeed. $\endgroup$ – Did Mar 21 '13 at 22:12

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