Logarithm of the Fourier transform? I've found this paper on the logarithm of the discrete fourier transform which proves that 
$$
log F = 1/4 i \pi (I - (1 +i)F + F^2 - (1 - i)F^3)
$$
where $F$ is the unitary discrete Fourier transform operator.
Is there a similar analogous version known for the infinite dimensional Fourier transform?
I am asking this from the point of view of trying to better understand cumulants, whose generating function is the logarithm of the characteristic function. The characteristic function is the fourier transform of the PDF of a random variable, thus motivating this question.
EDIT: Please take as many "niceness" assumptions as needed to answer the question regarding convergence.
 A: Let us start with the identity $F^4=I$. As a result, we have formally
\begin{align}
\ln F&=\ln(I+F-I)=\sum_{k\ge 1}\frac{(-1)^{k-1}(F-I)^k}{k}
\\&=
\sum_{1\le k\le 3}\frac{(-1)^{k-1}(F-I)^k}{k}+
\sum_{k\ge 4}\frac{(-1)^{k-1}}{k}\sum_{0\le l\le 3}\binom{k}{l} F^l(-1)^{k-l}
\\&\hskip94pt+
\sum_{k\ge 4}\frac{(-1)^{k-1}}{k}\sum_{k\ge l\ge 4}\binom{k}{l} F^l(-1)^{k-l},
\end{align}
and since in the last sum $F^l=F^j$ with $j\equiv l\mod 4$, we get a polynomial in $F$ with degree 3, that is
$$
\ln F= a_0+a_1 F+a_2 F^2
+a_3 F^3.$$
I guess that it is straightforward to get the expressions of the $a_j$ from the above identity. Also you can use the fonction $g(x)=e^{-π x^2}$ which is such that $(F-I) g=0$ implying that
$$
0=(\ln F)(g)=(a_0+a_1 +a_2 
+a_3)(g)\Longrightarrow \sum_{0\le j\le 3}a_j=0,
$$
and in fine you will get the same coefficients as yours.
A: $$F=e^{\frac{1}{4} \pi  i \left(D^2-x^2+1\right)}$$
This is for unitary case.
A: For $F$ any linear map such that $F^4=Id$ then  $F=  \sum_{k=0}^3 i^k  T_k$ where $T_k =\frac14 \sum_{m=0}^3 i^{-mk} F^m $, $T_k^2=T_k,T_k T_l = 0$ which gives that $(\sum_{k=0}^3 c_k  T_k)^n = \sum_{k=0}^3 c_k^n  T_k$.
With $L=\sum_{k=0}^3 \frac{ki\pi}{2} T_k$ we get that $\exp(L)=\sum_{n=0}^\infty  \sum_{k=0}^3 \frac{(\frac{ki\pi}{2})^n}{n!}  T_k=\sum_{k=0}^3 \exp(\frac{ki\pi}{2}) T_k= F$.
