Let $\hat{O}$ be an operator, what is 1/$\hat{O}$? I would like to know the meaning of an operator to the power of minus 1. I often see it in papers, but I do not know how it is defined. For example, in this paper on page 6 (of the pdf), the following has been written:
\begin{align}
*F^{\mu}&=\frac{\mu}{\square+\mu^2}(g^{\mu\nu}-\varepsilon^{\mu\nu\alpha}\frac{\partial_{\alpha}}{\mu})J_{\nu} \\
&=\frac{1}{\mu}(g^{\mu\nu}+\varepsilon^{\mu\nu\alpha}\frac{\partial_{\alpha}}{\mu})^{-1}J_{\nu}.
\end{align}
There are two things here that I am unsure of. The first is the factor $\frac{\mu}{\square+\mu^2}$, how is this things action on functions $J_{\mu}$ defined? The second is the term in the second line which has exponent -1,I just can't see how these operators, which are in the denominator, can act on functions which are in the numerator. The only way I can make sense of this is if it is interpreted as a Taylor series, but then I think things would be very ugly. 
One more thing, if I have the following equation
\begin{align}
(\square+\mu^2)\square \psi_{\alpha\beta\gamma}=0,
\end{align}
can I somehow conclude that this implies $(\square+\mu^2) \psi_{\alpha\beta\gamma}=0$ by applying $\square^{-1}$, or something along these lines?
 A: These operators are usually defined in momentum space, where $\partial_\mu \to i p_\mu$. For instance
$$
\frac{1}{\Box + m^2} \phi(x) = \frac{1}{\Box + m^2} \int \frac{d^4 p}{ (2\pi)^4 } e^{i p \cdot x } {\tilde \phi}(p) = \int \frac{d^4 p}{ (2\pi)^4 } \frac{ e^{i p \cdot x } }{ - p^2 + m^2 }  {\tilde \phi}(p) 
$$
Going back to position space, we then have
$$
\frac{1}{\Box + m^2} \phi(x) =  \int \frac{d^4 p}{ (2\pi)^4 } \frac{ e^{i p \cdot x } }{ - p^2 + m^2 } \int d^4 y e^{- i p \cdot y} \phi(y) =\int d^4 y G(x-y) \phi(y)
$$
where
$$
G(x-y) =   \int \frac{d^4 p}{ (2\pi)^4 } \frac{ e^{i p \cdot ( x - y )  } }{ - p^2 + m^2 } ~. 
$$
We can proceed in a similar way with all other "inverse" derivative operators. 
A: Let $T:X\to Y$ be an operator (linear map) between Banach (also locally convex) spaces, with domain $D(T)\subseteq X$. If (and only if) $T$ is injective (i.e. if and only if $Tx=0\Leftrightarrow x=0$), then its inverse $T^{-1}$ is defined as the operator with domain $\mathrm{Ran}(T)\subseteq Y$, that maps $(Tx)\mapsto x$.
Of course $T^{-1}$ satisfies:
$$D(T^{-1})=\mathrm{Ran}(T)\; ,\; \mathrm{Ran}(T^{-1})=D(T)$$
$$\forall x\in D(T)\; ,\; T^{-1}(Tx)=x\; ;\; \forall y\in \mathrm{Ran}(T)\; ,\; T(T^{-1}y)=y\; .$$
In the theory of operators from a Hilbert space to itself, usually the inverse is required to be bounded, i.e. $T$ has to be surjective as well as injective.
I am not sure that the operators in OP's examples are injective and/or surjective, since not much information is provided and it depends from which spaces they are acting upon.
