I am trying to solve an ODE via Fourier transforms,
$$ f^{\prime \prime} = -f $$
Naively following the usual approach I get
$$- \xi ^2 \hat {f} = - \hat{f} \\ (\xi^2 -1 ) \hat{f} = 0 \\ \hat{f} = 0$$ which seems a dead end.
The solution is $\sin (x) $ and is not integrable and its Fourier transform exists in the distributional sense
$$ \widehat{\sin(x)} = \sqrt{2 \pi} \,\frac{\delta (\xi - 1 )- \delta (\xi + 1)}{2i} $$ and it seems the relationship
$$ \hat {f ^{\prime}} = i \xi \hat{f} $$ does apply, if I consider the Fourier Transforms of $\sin$ and $\cos$. Probably taking excessive freedoms, the numerator equals $0$ unless $\xi = 1,-1$, and then the $i$ factor makes just things right with respect to $ \widehat{\cos} (x) $.
So in the "naive"calculation above it is the division by the factor $1-\xi^2$ that is to blame: nonsense comes from the division by $0$.
However, I would like to see how this specific calculation is to be rigorously handled.
The distributional solution to $(\xi^2 -1 ) \hat{f} = 0$ is not $\hat{f}\equiv 0$ but $\hat{f}(\xi) = A\,\delta(\xi+1)+B\,\delta(\xi-1),$ where $A$ and $B$ are constants.