As in textbook:Fourier Analysis (Javier Duoandikoetxea)
In the Schwartz class it is straightforward to characterize the functions whose Hilbert transforms are integrable: for $\phi \in > \mathcal{S}, H \phi \in L^{1}$ if and only if $\int \phi=0$.
I have proved one side of this proposition: (2) $H \phi \in L^1$ then $\|H \phi\|_1<+\infty$ $$ \begin{aligned} \|H \phi\|_1=\int_R|H \phi| d x & \\ \geqslant \int_R H \phi d x & =\int_R \lim _{\epsilon \rightarrow 0} \int_{|y|>\epsilon} \frac{1}{\pi} \frac{\phi(x-y)}{y} d y d x \\ & =\lim _{\epsilon \rightarrow 0} \int_{|y|>\epsilon} \int_R \frac{1}{\pi} \frac{\phi(x-y)}{y} d x d y \\ & = \lim _{\epsilon \rightarrow 0} \int_{|y \mid>\epsilon} \frac{\|\phi(x-y)\|_1}{\pi y} d y \\ & =\|\phi\|_1 \lim _{\epsilon \rightarrow 0} \int_{|y|>\epsilon} \frac{1}{\pi} \cdot \frac{1}{y} d y \end{aligned} $$ $\lim _{\epsilon \rightarrow 0} \int_{| y| >\epsilon} \frac{1}{y} d y$ is not bounded, while $\|H \phi\|_1$ is bounded.
But the other side of this question I am not sure.