Given $f(x)$ continuous for all $x\in \mathbb R$, and $f(x)$ nonzero on $\mathbb R$, $0
$$\int_{-\infty}^{\infty}\bigg|\frac{e^{b|x|}}{f(x)}\bigg|^{2}\,dx$$ be finite?
Given $f(x)$ continuous for all $x\in \mathbb R$, and $f(x)$ nonzero on $\mathbb R$, $0
$$\int_{-\infty}^{\infty}\bigg|\frac{e^{b|x|}}{f(x)}\bigg|^{2}\,dx$$ be finite?
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.