Suppose $f:{\bf R}\to{\bf R}$ is sufficiently smooth and supported on $[-R,R]$. Then $$ \int_{|t|>\epsilon}\frac{f(x-t)}{t}\ dt=\int_{\epsilon<|t|<R}\frac{f(x-t)-f(x)}{t}\ dt\ .\tag{*}$$
In his lecture notes on Hilbert transform, Tao makes a one-line calculation as follows:
Writing $\{|t|>\epsilon\}$ as a union of $(-\infty,-\epsilon]$ and $[\epsilon,\infty)$, playing around with the change of variable $u=x-t$ back and forth on the left hand side of ($*$), I got $$ \int_{\epsilon<|t|<R}\frac{f(x-t)}{t}\ dt +\int_{R}^{R+x}\frac{f(x-t)}{t}\ dt -\int_{-R}^{-R+\epsilon}\frac{f(x-t)}{t}\ dt, $$ which is not quite the right hand side yet. It seems that the calculation would be very simple by using "symmetry" as indicated in the notes. But I don't know how. Would anyone elaborate how it is done (by symmetry)?
This is actually quite simple. You first take $R$ large enough so that $f(x-t) = 0$ if $|t| > R$, which is possible, since $f$ has compact support. So the integral becomes $\int_{\epsilon < |t| < R} \frac{f(x-t)}{t}dt$. Now note $\int_{\epsilon < |t| < R} \frac{f(x)}{t}dt = 0$ (by symmetry) since $f(x)$ is just a constant. You're done.