When are two commuting linear operators functions of each other I've computed that the following is valid for certain functions but I've hit a slight bump in my proof. I was wondering if someone could clear it up.
If we formally consider the integral operators $$E f(s) = \frac{1}{\Gamma(-s)} \int_{0}^{\infty} f(-y) y^{-s-1} dy$$
and $$Y f(s) = \int_{-\infty}^{\infty} f(y) \frac{s^y}{\Gamma(y+1)} dy$$
I've shown that if $Q f(s) = s f(s)$ then
$$Q Y E f = Y E Q f$$ and $$\frac{d}{ds} Y E f = Y E \frac{df}{ds}$$

Q: What step should I use to show that, because they commute, $YE$ is a
  function of $Q$ and $ \frac{d}{ds}$ and therefore the constant linear
  operator; $\alpha = \alpha Q^0 = \alpha \frac{d^0}{ds^0}$?

I mean that $Y = E^{-1}$, which I have verified for a few functions. Considering only such functions that converge I won't go into that here.
 A: Your question is going to be difficult to answer in a general way. It's going to have to depend on the specifics of your problem. For example, consider the 3x3 matrix
$$
A = \left[\begin{array}{ccc}
   1 & 0 & 0 \\
   0 & 2 & 0 \\
   0 & 0 & 2
\end{array}\right].
$$
The following matrix P commutes with A
$$
P = \left[\begin{array}{ccc}
   0 & 0 & 0 \\
   0 & 0 & 0 \\
   0 & 0 & 1
\end{array}\right].
$$
P is a projection because $P^{2}=P$. There is no function $f$ for which $P=f(A)$ because of the degeneracy in the eigenspace for $A$ corresponding to eigenvalue 2. All polynomials in $A$ leave the lower $2\times 2$ block unchanged, up to a scalar; so there is no way to get $P$ out of polynomial functions of $A$. Even a power series in $A$ won't give you want you want because such a power series always reduces to a polynomial through the minimal polynomial $p(\lambda)=(\lambda-1)(\lambda-2)$ for $A$. Things couldn't be much nicer: low-dimension, symmetric matrices, diagonal.
One has the following interesting way of characterizing the situation for an $N\times N$ complex matrix $A$. The matrix $A$ has the property that $AB=BA$ implies $B=p(A)$ iff there is a vector $x \in \mathbb{C}^{N}$ such that $\{ x, Ax, A^{2}x,\ldots, A^{N-1}x\}$ is a basis for $\mathbb{C}^{N}$. (That is, only polynomials in $A$ commute with $A$ iff $A$ has a cyclic vector $x$.) There is an equivalent condition in terms of the Jordan canonical form for $A$: $A$ has a cyclic vector iff there is only one Jordan block for each eigenvalue.
For general operators, especially unbounded ones, it is not going to be easy to say what you want, even for selfadjoint operators. If $A$ and $B$ are bounded, selfadjoint, and commute, then each is a function of a third which commutes with both, but $A$ may not be a function of $B$ or $B$ may not be a function of $A$. You can already see that for $3\times 3$ matrices.
