Let $\operatorname{f}: \mathbb{R} \to \mathbb{R}$ be defined s.t. $\operatorname{f}\in\mathcal{C}^{\infty}\left(\mathbb{R},\mathbb{R}\right)$. When the minus one term indicates a reciprocal, how would one evaluate the Fourier transform given by the following integral for $c \in \mathbb{R}\ ?$.
$$ \hat{\operatorname{f}}\left(k\right) = \frac{1}{\,\sqrt{\,{2\pi}\,}\,}\int_{-\infty}^{\infty} {\rm e}^{-{\rm i}kx}\,\, \left(\frac{{\rm d}^{2}}{{\rm d}x^{2}} + c\right)^{-1}\,\operatorname{f}\left(x\right)\,{\rm d}x $$
We can assume that the integrand is finite at every point. Is there any fact that I may exploit about reciprocals when computing these Fourier transforms $?$. If no result holds in the general case, we can assume that $\operatorname{f}\left(x\right)$ is a simple trigonometric polynomial.
Question: Is there any way to represent $\displaystyle\left(\frac{{\rm d}^{2}}{{\rm d}x^{2}} + c\right)^{-1}$ by a Fourier multiplier $?$. If not, are there any other simplifications available $?$.
Let $D$ denote the operator $D=\left(\frac{d^2}{dx^2}+c\right)$ and let $I=DD^{-1}=\left(\frac{d^2}{dx^2}+c\right)\left(\frac{d^2}{dx^2}+c\right)^{-1}$ denote the identity operator.
For any $\phi\in \mathbb{S}$, we have
$$\begin{align} \langle \mathscr{F}\{f\},\phi \rangle&=\langle \mathscr{F}\{DD^{-1}f\},\phi \rangle\\\\ &=\langle DD^{-1}f,\mathscr{F}\{\phi\}\rangle\\\\ &=\langle D^{-1}f,D\mathscr{F}\{\phi\}\rangle\\\\ &=\langle D^{-1}f,\mathscr{F}\{(-k^2+c)\phi\}\rangle\\\\ &=\langle (-k^2+c) \mathscr{F}\{D^{-1}f\},\phi\rangle\\\\ \end{align}$$
Therefore, in distribution we have
$$\mathscr{F}\{f\}=(-k^2+c)\mathscr{F}\{D^{-1}f\}$$
or alternatively
$$\mathscr{F}\{D^{-1}f\}=\frac1{c-k^2}\mathscr{F}\{f\}$$