= Solution
The reciprocal in this transform must mean the <principal-value reciprocal distribution>. Its inverse transform is
$$
\begin{aligned}
\mathcal F^{-1}\!\left(\operatorname{PV}\frac1\omega\right)(t)
&=\frac{i}{\pi}\int_0^\infty\frac{\sin(\omega t)}\omega\,d\omega\\
&=\frac i2\operatorname{sgn}t,
\end{aligned}
$$
using the given <Dirichlet integral> and the sign of $t$. Also $\mathcal F^{-1}[\delta](t)=1/(2\pi)$. The inverse of the proposed expression is consequently $(iA/2)\operatorname{sgn}t+B/(2\pi)$. Comparison with $H(t)=[1+\operatorname{sgn}t]/2$ for $t\ne0$ gives
$$
\boxed{A=-i,\qquad B=\pi,\qquad\widetilde H(\omega)=-i\operatorname{PV}\frac1\omega+\pi\delta(\omega).}
$$
The value assigned at the single point $t=0$ does not affect the distribution. This is the <Fourier transform of the Heaviside step function>; treating $1/\omega$ as an ordinary integral would miss the required prescription.
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