Are compositions of the Fourier sine and cosine transforms commutative? That is to say, is it true or false that
$$\mathcal{F}_c(\mathcal{F}_s(f(x)))(\xi)\equiv\mathcal{F}_s(\mathcal{F}_c(f(x)))(\xi),$$
and if they are not then are there any conditions on $f$ for which they might be?
I can't seem to find any documents online about general properties of the Fourier sine and cosine transforms (so far). My 7-th edition Table of Integrals, Series and Products only states the basic properties.
 A: Yes, they do commute. Define $\mathcal F_-$ by $\mathcal F_-(f)(\xi) = \mathcal F(f)(-\xi)$, where $\mathcal F$ is the ordinary Fourier transform, $\mathcal F = \mathcal F_c + i\mathcal F_s$. Then
$$
\mathcal F_c = {\mathcal F + \mathcal F_-\over 2},\quad \mathcal F_s = {\mathcal F - \mathcal F_-\over 2i}. \tag{1}
$$
The Fourier inversion formula says that $\mathcal F \mathcal F_- = \mathcal F_- \mathcal F$ is a constant multiple of the identity. So, from $(1)$ and the fact that $\mathcal F$ and $\mathcal F_-$ commute, it follows that
$$
\mathcal F_c\mathcal F_s = {\mathcal F^2 - \mathcal F_-^2\over4i} = \mathcal F_s\mathcal F_c.
$$
A: Recall that sine and cosine transforms are defined by
\begin{eqnarray}
\mathscr{F}_c f(\omega) & = & \frac{2}{\pi} \int_0^\infty P_e f(t) \cos(\omega t) dt \\
\mathscr{F}_s f(\omega) & = & \frac{2}{\pi} \int_0^\infty P_o f(t) \sin(\omega t) dt \ ,
\end{eqnarray}
where
\begin{eqnarray}
P_e f(t) & = & \frac{1}{2}(f(t)+f(-t)) \\
P_o f(t) & = & \frac{1}{2}(f(t)-f(-t)) \ .
\end{eqnarray}
Now assume that $f$ is even. Then $\mathscr{F}_c f$ is even and $\mathscr{F}_s \mathscr{F}_c f$ is $0$ because the projection $P_o$ in the expression of $\mathscr{F}_s$ maps it to $0$. Then assume that $f$ is odd. Then $\mathscr{F}_c f$ is $0$ because the projection $P_e$ in the expression of $\mathscr{F}_c$ maps it to $0$. Hence also $\mathscr{F}_s \mathscr{F}_c f$ is $0$. Now linearity shows that $\mathscr{F}_s \mathscr{F}_c f = 0$ for any $f$ whenever both integrals exist. Similar reasoning shows that $\mathscr{F}_c \mathscr{F}_s f = 0$ for any $f$ whenever both integrals exist. But nevertheless, the equation $(\mathscr{F}_s \mathscr{F}_c f)(t) = (\mathscr{F}_c \mathscr{F}_s f)(t)$ holds for every $t \in \mathbb{R}$ whenever all necessary integrals exist.
