A particle with a lifetime $\tau$ and rest energy $E_0$ has a wavefunction, in natural units: $$\psi(t) = \psi_0 e^{-iE_0 t} e^{-t/2\tau}.$$

The modulus squares of its Fourier transform $|\hat{\psi(E)}|^2$ gives the probability density of finding the particle at energy $E$:

$$|\hat{\psi(E)}|^2 = \frac{\psi_0}{(E_0-E)^2 + \frac{1}{4\tau^2}}.$$

But if the particle has infinite lifetime, i.e. it is stable with $\tau \rightarrow \inf$, why does it still have a finite spread in energy space?

• You forgot to normalize the function. If you include corresponding coefficient (depending on $\tau$) you would have a consistent limit $|\psi(E)|^2=\delta(E-E_0)$. – A.V.S. Jan 21 '18 at 16:09

If you take the $τ$ to infinity, the $Ψ$ becomes a plane wave solution of the corresponding basic quantum mechanical equation, (no potentials). This cannot be normalized.
For the form you have, the modulus squared will just give $Ψ_0$ a constant saying that there is probability 1 to find the particle anyplace in space. This is solved in quantum field theory, which is based on free particle solutions of the basic quantum mechanical equations, by using wave packets, which are localized.