When doing a Hilbert-transformation, why not simply multiplying by an exponential?
In contrast to Jason R's answer I claim that the Hilbert transform is a phase shift by $-\pi/2$ for real-valued signals. By definition, a phase shifter shifts the phase of a sinusoidal signal by some given phase $\phi$:

$$x(t)=\cos(\omega_0t)\quad\Longrightarrow\quad y(t)=\cos(\omega_0t+\phi)\tag{1}$$

Since

$$\cos(\omega_0t)=\frac{e^{j\omega_0t}+e^{-j\omega_0t}}{2}\tag{2}$$

and

$$\cos(\omega_0t+\phi)=\frac{e^{j\omega_0t+\phi}+e^{-j\omega_0t-\phi}}{2}\tag{3}$$

it is clear that in order for the signal $x(t)$ to experience a phase shift by $\phi$, the positive frequencies need to be shifted by $\phi$ and the negative frequencies by $-\phi$. For $\phi=-\pi/2$ that's exactly what a Hilbert transformer does.

A constant transfer function like $H(f)=e^{-j\pi/2}=-j$ doesn't really do much apart from scaling the input signal, in this case by the imaginary unit. So the signal $x(t)=\cos(\omega_0t)$ is "transformed" into $y(t)=-j\cos(\omega_0t)$. Of course, for the special case of the complex-valued input signal $x(t)=e^{j\omega_0t}$ multiplication by $-j$ achieves a phase shift by $-\pi/2$, but usually we want to shift the phase of real-valued signals, and in that case we need a frequency-dependent transfer function.