The docs state
The Fourier sine transform of a function f(t) is by default defined to be Sqrt[2/\[Pi]]\!\(
\*SubsuperscriptBox[\(\[Integral]\), \(0\), \(\[Infinity]\)]\(\(f(t)\)\ \(sin(\[Omega] t)\) d t\)\).
It's easier viewed here: http://reference.wolfram.com/language/ref/FourierSinTransform.html
On the right side of the definition, the variable omega appears only as a factor inside Sin, and Sin is an odd function, so any function returned by FourierSinTransform should be an odd function.
However, Mathematica returns a non-odd function on FourierSinTransform of ArcTan:
In:= FourierSinTransform[ArcTan[t], t, \[Omega]]
Out= (E^-\[Omega] Sqrt[\[Pi]/2])/\[Omega]
Also easier to view in the docs, where it's given as an example.
The docs also state "Results from FourierSinTransform and FourierTransform differ by a factor of I for odd functions". That's how it should be, but that's not how it really is when they're applied to ArcTan.
If, as is sometimes done, you restrict the domain to ω > 0 (or >= 0), it all seems fine to me.
Thanks, I guess you are right. I didn't know what the full spectrum (-infinity, +infinity) was supposed to be, so all I noticed was that the full answer wasn't odd. I hadn't realized the upper half was correct.