DC value of $x(t)=\frac{1}{t}$
You need to be careful with the definition of the DC value of a signal. The actual time average, which is often called DC value is given by

$$\overline{s(t)}=\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T/2}^{T/2}s(t)dt\tag{1}$$

whereas the value of the signal's Fourier transform at $\omega=0$ is given by

$$S(0)=\int_{-\infty}^{\infty}s(t)dt\tag{2}$$

If $(2)$ is finite, $(1)$ equals zero, and if $(1)$ is finite but non-zero, $(2)$ doesn't exist, and the Fourier transform $S(\omega)$ has a Dirac impulse at $\omega=0$. See also this related answer.

For the given signal $s(t)=1/t$ both integrals in $(1)$ and $(2)$ do not converge in the conventional sense. However, their Cauchy principal value exists and equals zero, as mentioned in a comment by Robert Bristow-Johnson. Hence, the DC-value of $s(t)=1/t$ equals zero, regardless whether you define it by $(1)$ or by $(2)$.

Note that the Fourier transform of $s(t)=1/t$ equals

$$S(\omega)=-j\pi\,\textrm{sgn}(\omega)\tag{3}$$

which is just the frequency response of a Hilbert transformer (scaled by $\pi$).

From $(3)$, $S(0)$ doesn't exist, but the average of the left-sided and right-sided limits $\frac12 (S(0^-)+S(0^+))$ equals zero, and - using the Cauchy principal value according to the definition of the Hilbert transform - the Hilbert transform of a constant is indeed zero.