Pure Imaginary Poles (Why is it problematic?)

Question

Consider the LCCDE : \begin{equation} \sum_{k=0}^{N}\alpha_{k}\frac{d^{k}f(t)}{dt^{k}}=\sum_{m=0}^{M}\beta_{m}\frac{d^{k}g(t)}{dt^{k}} \end{equation} Taking the Fourier transform on both sides, we get : \begin{equation} \sum_{k=0}^{N}\alpha_{k}(2j\pi\lambda)^{k}\widehat{f}(\lambda)=\sum_{m=0}^{M}\beta_{m}(2j\pi\lambda)^{m}\widehat{g}(\lambda) \tag{1}\end{equation} Consider the two polynomials : $$ P(x):=\sum_{k=0}^{N}\alpha_{k}x^{k}\qquad\text{and}\qquad Q(x):=\sum_{m=0}^{M}\beta_{m}x^{m} $$ Then since $\displaystyle\frac{P(x)}{Q(x)}$ has no poles then $\displaystyle\frac{P(2j\pi\lambda)}{Q(2j\pi\lambda)}$ has no poles for real $\lambda$ and thus $(1)$ becomes : \begin{equation} \widehat{g}(\lambda)=\frac{P(2j\pi\lambda)}{Q(2j\pi\lambda)}\widehat{f}(\lambda) \end{equation}

The impulse response $h=\mathcal{F}^{-1}\{H\}$ is computed by decomposing $H$ into partial fractions. The poles $\displaystyle\frac{P}{Q}$ are assumed to lie off the imaginary axis. There are thus two cases to consider :

1. ($\frac{P}{Q}$ has only simple poles) : In this case, $H$ can be decomposed in the form : \begin{equation} H(\lambda)=\sum_{m=0}^{M}\frac{\beta_{m}}{2j\pi\lambda-z_{m}} \end{equation} where $z_{1},z_{2},\cdots,z_{M}$ are the poles. Therefore : \begin{equation} h(t)=\left(\sum_{m\in \mathcal{M}_{-}}\beta_{m}e^{z_{m}t}\right)u(t)-\left(\sum_{m\in \mathcal{M}_{+}}\beta_{m}e^{z_{m}t}\right)u(-t) \end{equation} where we define : \begin{align*} &\mathcal{M}_{-}:=\{m\in\{1,2,\cdots,M\}\;|\;\Re(z_{m})<0\}\\ &\mathcal{M}_{+}:=\{m\in\{1,2,\cdots,M\}\;|\;\Re(z_{m})>0\} \end{align*}
2. ($\frac{P}{Q}$ has multiple poles) : In this case, let $z_{1},z_{2},\cdots,z_{p}$ the poles and let $\gamma_{1},\gamma_{2},\cdots,\gamma_{p}$ be their multiplicites. Then we can write $H$ as : \begin{equation} H(\lambda)=\sum_{m=1}^{p}\sum_{\gamma=1}^{\gamma_{m}}\frac{\beta_{m,\gamma}}{(2j\pi\lambda-z_{p})^{\gamma}} \end{equation} We can see that : \begin{equation} h(t)=\left(\sum_{m\in \mathcal{M}_{-}}P_{m}e^{z_{m}t}\right)u(t)-\left(\sum_{m\in \mathcal{M}_{+}}P_{m}e^{z_{m}t}\right)u(-t) \end{equation} where : $$ P_{m}:=\sum_{\gamma=1}^{\gamma_{m}}\beta_{m,\gamma}\frac{t^{\gamma-1}}{(\gamma-1)!} $$

Remark : The case where we have pure imaginary poles, can not be solved using the two cases above, for a differential equation of the form : $$ g''+\omega^{2} g=f $$ where $\displaystyle\frac{P(x)}{Q(x)}=\frac{1}{x^{2}+\omega^{2}}$ has two poles are on the imaginary axis. In this case, $h$ is a sinusoid and the Fourier transform of $H$ (when $H$ is considered to be a function) is no longer defined

I didn't understand why we would be concerned in finding Fourier transform of $H$ as I didn't understand what the author meant by saying when $H$ is considered to be a function.

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[Reference] : C. Gasquet, P. Witomski, Fourier Analysis and Applications: Filtering, Numerical Computations, Wavelets. Translated by R. Ryan