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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.

pshmath0
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2 Answers2

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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 $$ \mathcal F_c\mathcal F_s = {\mathcal F^2 - \mathcal F_-^2\over4i} = \mathcal F_s\mathcal F_c. $$

Nick Strehlke
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  • By $\mathcal{F}_c\mathcal{F}_s$ do you mean $\mathcal{F}_c(\mathcal{F}_s(f(x)))$ ? – pshmath0 Jun 12 '13 at 07:26
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    @pbs Yes! That's exactly what I meant. – Nick Strehlke Jun 12 '13 at 08:49
  • I may have found a counter example: http://math.stackexchange.com/questions/435171/commutativity-of-fourier-sine-cosine-transforms (or else I've erred!) – pshmath0 Jul 03 '13 at 12:33
  • @pbs: See my comment on your question. By the way (to see that my answer agrees with Vihtanen's), the expression $\mathcal F^2 - \mathcal F_-^2$ is identically zero because $\mathcal F^2 = \mathcal F_-^2$ is a constant multiple of the operator that sends a function $f(x)$ to the function $f(-x)$. – Nick Strehlke Jul 06 '13 at 16:24
  • Mathematica appears to define the Fourier sine and cosine functions by $\mathcal{F}_s(f(x))(t) = 2\int_0^\infty f(x)\sin(x t)\text{d} x$ and $\mathcal{F}_c(f(x))(t) = 2\int_0^\infty f(x)\cos(x t)\text{d} x$, where the limits of integration are from $0$ to $\infty$, not $-\infty$ to $\infty$, which would explain the problem. I have also seen these exact definitions in many text books on the subject (see "Integral Transforms and their Applications" by Davies (2002), for example). Does commutativity hold in this case? Most likely not given the above example. – pshmath0 Jul 07 '13 at 17:52
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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.

Juha-Matti Vihtanen
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