My singular integral operator is defined by
\begin{align}
Sf(x)=-\int\_{-\infty}^{\infty}f(t-x) \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.



Boundedness of a singular integral operator on $L^p(\mathbb{R})$, $1<p<\infty$
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 Hörmander-Mihlin 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}.
$


