My singular integral operator is defined by \begin{align} Sf(x)=\int_{\infty}^{\infty}f(tx) \frac{dt}{2\sinh\frac{\pi}{2}t}, \end{align} that is, a convolution $\frac{1 }{2\sinh\frac{\pi}2x}\ast f.$ To prove that this is bounded on $L^p(\mathbb{R})$, $1<p<\infty$, I have tried to make use of the Marcinkiewicz interpolation theorem. I have been able to show that it is strongly bounded on $L^2(\mathbb{R})$ and it remains to show that it is weakly bounded on $L^1(\mathbb{R})$ so that I can use the Marcinkiewicz interpolation theorem to conclude for $1<p<2$, and then use duality to complete the problem. Now, how do I show that it is weakly bounded on $L^1(\mathbb{R})$? Been trying whole day and night and I cannot try any more, please help.

4$\begingroup$ This is essentially the Hilbert transform. Weak estimate in $L^1$ is Kolmogorov's theorem, boundedness in $L^p$ is due to Riesz. Look in Zigmund, or in Koosis or in any other book on the subject. $\endgroup$ – Alexandre Eremenko Mar 4 '15 at 4:22
We have with $\chi\in C^\infty_c(\mathbb R)$, equal to $1$ near 0, $$ \frac{1}{\sinh t}=\frac{\chi(t)}{\sinh t}+\frac{1\chi(t)}{\sinh t}. $$ The second function belongs to $L^1$ and Young's inequality implies boundedness on $L^p$ for all $p\in [1,+\infty]$. We write, $$ \frac{\chi(t)}{\sinh t}=\frac1t\frac{t\chi(t)}{\sinh t}= \frac{\psi(t)}{t} $$ where $\psi$ is smooth compactly supported, $\psi(0)=1$. The Fourier transform of that function is, up to some normalization constant, equal to $ \hat \psi\ast \text{sign}. $ The HörmanderMihlin multiplier theorem (see e.g. theorem 7.9.5 in Springer Grunlehren 256) applies.
N.B. In fact a simpler argument should work for the latter term since $a=\hat \psi\ast \text{sign}$ is a regularization of the sign function and $ \vert\partial_\tau^k a\vert\le C_k(1+\vert \tau\vert)^{k}. $