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 following result from Marcinkiewicz interpolation theorem:
For $1\le p\le\infty$ there exists a constant $C_p$ such that if $f\in L^p$ and $g\in L^1$, then $f\ast g\in L^p$ and \begin{align*} ||f\ast g||_p\le C_p||g||_1||f||_p. \end{align*}
This result means it suffices for me to show that $g=-\frac{1 }{2\sinh\frac{\pi}2x}$ is integrable. I have tried to do this as follows:
Let $f\in L^p(\mathbb{R})$ and $g=-1/(2\sinh\frac{\pi}2x)$, then \begin{align*} ||g||_{L^1(\mathbb{R})}=\int_{-\infty}^{\infty} \frac{dx}{2\sinh\frac{\pi}{2}x} &=\left[\frac1{\pi}\ln\left(\tanh\frac{\pi}4x\right)\right]_{-\infty}^{\infty}\\ &=\frac1{\pi}\left[\ln\left(\frac{e^{\frac{\pi}2x}-1}{e^{\frac{\pi}2x}+1}\right)\right]_{-\infty}^{\infty}\\ &=-\frac1{\pi}\ln(-1)\\ &=\text{ diverges} \end{align*} that is $g\notin L^1(\mathbb{R})$, and my conclusion is that $S$ is not bounded on $L^p(\mathbb{R})$. However, my professor insists that it is bounded. Is there another way of showing that $S$ is indeed bounded on $L^p(\mathbb{R})$, $1<p<\infty$.
$\bf\underline{\text{SO FAR}}.$ 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 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.