Hilbert transform properties
HINT:


Prove that
$$y(t)=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{x(\tau)}{t-\tau}d\tau=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{x(t-\tau)}{\tau}d\tau\tag{1}$$
Use the right-most expression in $(1)$ to show that the system is time-invariant.


Equivalently, you can directly show that the original input-output relation is a convolution integral, from which it follows that the system must be LTI. Then find the expression for the impulse response $h(t)$ and show that $h(t)\neq 0$ for $t&lt;0$, which means that the system is not causal.

EDIT:

The same variable substitution that can be used to show $(1)$ can also be used to directly show the equivalence of the response to $x(t-t_0)$ and the delayed response to $x(t)$:

$$y_2(t)=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{x(\tau-t_o)}{t-\tau}d\tau$$

$$\begin{align}y(t-t_o)&amp;=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{x(\tau)}{(t-t_o)-\tau}d\tau{\Huge|}_{\tau=u-t_0}\\&amp;=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{x(u-t_0)}{(t-t_o)-(u-t_0)}du\\&amp;=\frac{1}{\pi}\int_{-\infty}^{+\infty}\frac{x(u-t_0)}{t-u}du\\&amp;=y_2(t)\end{align}$$