# Commutativity of Fourier sine/cosine transforms

In an earlier post I asked if the Fourier sine and cosine transforms were commutative, answers to which were given in the affirmative. However, in Mathematica I've taken a closer look to find this:

FourierSinTransform[FourierCosTransform[x/(x^2 + 1), x, t], t, x]


gives the answer $$\frac{2x\log(x)}{\pi(1+x^2)},$$ and

FourierCosTransform[FourierSinTransform[x/(x^2 + 1), x, t], t, x]


gives the answer $$\frac{1}{1+x^2},$$

which are clearly not the same functions. Are the Fourier sine/cosine transforms really commutative then, or maybe I'm doing something wrong in Mathematica?

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It looks like Mathematica is computing this incorrectly. The function $x/(x^2+1)$ is odd, so its Fourier cosine transform should be identically zero (hence the Fourier sine transform of its Fourier cosine transform should be zero). Likewise, since the Fourier sine transform is always odd, the Fourier cosine transform of the Fourier sine transform of $x/(x^2+1)$ should be identically zero as well. – Nick Strehlke Jul 6 '13 at 16:19
It seems that Mathematica defines the Fourier cosine (resp. sine) transform by $$\mathcal{F}_c(f(x))(t)=\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos(xt)\text{d}x,$$ where the limits of integration are from $0$ to $\infty$, and not $-\infty$ to $\infty$. – poirot Oct 12 '13 at 9:02