= Solution
With the <Fourier transform> convention of 5A, use the printed <Fourier inversion> formula directly. The possible choice of values at $s=\pm1$ has no effect on the integral. Integration over the support gives
$$
g(x)=\frac1{2\pi}\int_{-1}^1(e^s-e^{-s})e^{-isx}ds
=\frac1\pi\left[\frac{\sinh(1-ix)}{1-ix}-\frac{\sinh(1+ix)}{1+ix}\right].
$$
Combining the two fractions yields
$$
\boxed{g(x)=\frac{2i}{\pi(1+x^2)}\bigl(x\sinh1\cos x-\cosh1\sin x\bigr).}
$$
In particular $g$ is imaginary-valued and odd, as expected from the real odd transformed function with this inverse-transform sign.
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