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Let $R(z)=\displaystyle \frac{P(z)}{Q(z)}$ be a rational function of order(Q) $\geq \mathrm{order}(P)+2$ and $Q(x)\neq 0$ for all $x\in \mathbb{R}$. Then we have: $$ \int_{-\infty}^{\infty}R(x)\mathrm{d}x=2\pi i\sum_{z:\ \mathrm{Im} \ z>0}{\rm Res}(R,z) $$

Why is

order(Q) $\geq \mathrm{order}(P)+2$ and $Q(x)\neq 0$ for all $x\in \mathbb{R}$


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I guess you meant "Let $\,R(z)\,$ be a rational function with rational coefficients..." , i.e. $\,P(z), Q(z)\in\mathbb{Q}[z]\,$ ....? – DonAntonio Jun 6 '12 at 23:46
I fixed it. The coefficients are complex. – Chris Jun 7 '12 at 0:15
up vote 1 down vote accepted

The basic reason is that you need a bounded growth condition on the function. In order to evaluate this integral, you're actually doing a complex line integral over a half circular region, but you need the complex part to go to zero as the radius goes to infinity. If your function doesn't have bounded growth, the residue formula will retain some of the complex line integral, not just the real line.

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