Hilbert transform from analytic signal
Note that the condition on the relation between the bandwidth of $m(t)$ and the modulation frequency ensures that $h(t)$ is a band pass signal. It can be written as

$$h(t)=\frac12 m(t)e^{j2\pi \nu_c t}+\frac12 m(t)e^{-j2\pi \nu_c t}\tag{1}$$

From our assumption $\nu_m&lt;\nu_c$ it follows that the two expressions on the right-hand side of $(1)$ do not overlap in the frequency domain. Consequently, the analytic signal is simply given by the first term on the right-hand side of $(1)$ (scaled by a factor of $2$), which is the part of the spectrum of $h(t)$ occurring at positive frequencies:

$$h_a(t)=m(t)e^{j2\pi\nu_c t}\tag{2}$$

The Hilbert transform of $h(t)$ is given by the imaginary part of $(2)$:

$$\hat{h}(t)=\text{Im}\{h_a(t)\}=m(t)\sin(2\pi\nu_c t)\tag{3}$$

EDIT: The equation for $h_a(t)$ given in your question is also correct. If you want to continue that way, it's important to see that due to $\nu_c&gt;\nu_m$, $M(\nu+\nu_c)$ equals zero for $\nu&gt;0$. So the integral can be written as

$$h_a(t)=\int_0^{\infty}M(\nu-\nu_c)e^{j2\pi\nu t}d\nu\tag{4}$$

Likewise, $M(\nu-\nu_c)$ is zero for $\nu&lt;0$, so we can extend the lower integration limit in $(4)$ to $-\infty$:

$$h_a(t)=\int_{-\infty}^{\infty}M(\nu-\nu_c)e^{j2\pi\nu t}d\nu=\mathcal{F}^{-1}\{M(\nu-\nu_c)\}=m(t)e^{j2\pi\nu_c t}\tag{5}$$

which is of course identical to $(2)$.