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Given $f(x)$ continuous for all $x\in \mathbb R$, and $f(x)$ nonzero on $\mathbb R$, $0 such that $|f(x)|\leq e^{a|x|}$ for all $x\in \mathbb R$. What conditions should $f$ have so that the integral

$\int_{-\infty}^{\infty}\bigg|\frac{e^{b|x|}}{f(x)}\bigg|^{2}\,dx$ be finite?

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    I think there's only so much you can say (in terms of the sandwich theorem and Taylor series) about f, because the "borderline" of functions with infinite area (between -inf and +inf) is not really a solid border of one function, but a range of functions, e.g. 1/x, 1/(xlogx), 1/(xloglogx). And given any one of these functions, we can always find another function. So there is only so much you can say about f. But by the Sandwich theorem, f cannot eventually be greater than any of these functions. e.g. 1/x^2 is eventually less than all of these functions.2012-05-09

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If $a=b$ then $e^{b|x|}/|f(x)|\ge 1$, and the integral diverges. In general, the upper bound on $|f(x)|$ works against you because $f$ is in the denominator. To prove convergence, you need a lower bound on $|f|$. For example, if $|f(x)|\ge c(|x|+1)e^{b|x|}$ for some $c>0$, then $e^{b|x|}/|f(x)| = c^{-1}/(|x|+1)$, which is square integrable.