How to obtain an inverse integral transform for $F(q) = \int_0^R \chi(t) \sin qt \, \mathrm{d}t$ provided that $\chi$ is an odd function?

Question

For an odd function, the sine Fourier transform is defined as $$ F(q) = \int_0^\infty \chi(t) \sin qt \, \mathrm{d} t \, , $$ and the back transform is obtained as $$ \chi(t) = \frac{2}{\pi} \int_0^\infty F(q) \sin qt \, \mathrm{d}q \, . $$

By defining a transform law as

$$ F_R(q) = \int_0^R \chi_R(t) \sin qt \, \mathrm{d}t \, , $$

where $0<R<\infty$, I was wondering whether an inverse formulation can be obtained. I.e. expressing $\chi_R(t)$ in term of $F_R(q)$ via an integral equation and that for finite $R$.

Hints and ideas welcome.

Thanks

R