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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
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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.

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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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