What is the operator form of $1/P_x$? I know that the operator form of $1/P_x$ ($P_x$ is the $x$-component of momentum operator $\mathbf{P}$) should have an integral form like: $$\frac{i}{\hbar}\int\,dx,$$ but I'm not sure about the limits of integral. Any help would be appreciated.
 A: The operator
$$\frac{1}{P_x}$$
produces a divergent result if it acts on a wave function with $P_x=0$ (the eigenstate with this eigenvalue looks like a constant function of $x$). The superpositions of a $P_x=0$ states with the eigenstates associated with nonzero values of $P_x$ are still singular.
That's why a well-defined result of $(1/P_x)|\psi\rangle$ is only obtained if $|\psi\rangle$ contains no $P_x=0$ admixture in the superposition. But that's equivalent to 
$$ \int_{-\infty}^{+\infty} dx \,\psi(x) = 0$$
The left hand side is nothing else than the $P_x=0$ Fourier component. For that reason, the prescriptions
$$[\frac{1}{P_x} \psi] (x) \leftrightarrow \frac{i}{\hbar}\int_{-\infty}^x dx'\,\psi(x') $$ 
and
$$[\frac{1}{P_x}\psi](x) \leftrightarrow -\frac{i}{\hbar}\int_{x}^\infty dx'\,\psi(x') $$ 
produce the same result.
If you wanted to generalize the action of an operator $1/P_x$ on a wave function that does contain a $P_x=0$ piece, then you would have to decide whether you mean the "principal value" of $1/P_x$ or $1/(P_x+i\epsilon)$ or something else. This extra refinement could make the operator well-defined for a broader set of test functions $\psi(x)$.
A: One way to treat Quantum mechanics is that writing the operators in terms of matrices.
One can think of operators $\hat{X}$, $\hat{P_x}$ and $\hat{P_x}^{-1}$ in terms of matrix quantum mechanics. In the $|x \rangle$ basis (where $\hat{X} |x \rangle=x|x \rangle$), we can discretize the space up to some overall constant as a unit of spacing, and write the operators as
$$\hat{X}=a{\begin{pmatrix} 
\ddots & 0 & 0 & 0 & 0 & 0&0\\
0 & -2 & 0 & 0 & 0 & 0&0\\  
0 & 0 & -1 & 0 & 0 & 0&0\\  
0 & 0 & 0 & 0 & 0 & 0&0\\
0 & 0 & 0 & 0 & 1 & 0&0 \\
0 & 0 & 0 & 0 & 0 & 2&0 \\
0 & 0 & 0 & 0 & 0 & 0& \ddots \\
\end{pmatrix}}$$
$$\hat{P_x}=-i \hbar \partial_X=i \hbar a^{-1}{\begin{pmatrix} 
\ddots & 0 & 0 & 0 & 0 & 0&0\\
1 & 0 & -1 & 0 & 0 & 0&0\\  
0 & 1 & 0 & -1 & 0 & 0&0\\  
0 & 0 & 1& 0 & -1 & 0&0\\
0 & 0 & 0 & 1 & 0 & -1&0 \\
0 & 0 & 0 & 0 & 1 & 0&-1 \\
0 & 0 & 0 & 0 & 0 & 0& \ddots \\
\end{pmatrix}}$$
Easy to check $$[\hat{X},\hat{P_x}]=i\hbar$$
Thus,
$$\frac{1}{\hat{P_x}}=\hat{P_x}^{-1}$$
Thus, $\frac{1}{\hat{P_x}}=\hat{P_x}^{-1}$ is just the inverse matrix of $\hat{P_x}$. One may need to consider two cases: periodic boundary conditions on a 1D circle $S^1$ or an infinite size 1D system. Treated two cases separately.
