How to how that $x \operatorname{cosech}(x) $ is Schwartz function?

Question

We know $f (x)$ is in Schwartz space if for given $ m, k $ non negative integers the supremum of $ \left|x^{m } f^{(k)}(x) \right| $ over the real numbers is finite. How to show that $x \operatorname{cosech}(x) $ is in the Schwartz space? I proved this supremum is finite for initial particular values of $m ,k$ but don't know how to proceed in general case. Please help

Answer 1

Is complex analysis allowed? If so, you may exploit the representation: $$ \frac{x}{\sinh x} = 1+\sum_{n\geq 1}\frac{2x^2 (-1)^n}{x^2+n^2 \pi^2} $$ that comes from a Weierstrass product. As an alternative, you may prove that $$ \mathcal{L}\left(\frac{x}{\sinh x}\right) = \frac{1}{2}\psi'\left(\frac{s+1}{2}\right)$$ and both the given identities easily lead to $\frac{x}{\sinh x}\in\mathcal{S}(\mathbb{R})$. Also, the Fourier transform of $\frac{x}{\sinh x}$ behaves like $\frac{1}{\cosh(x)^2}$, and it is probably easier to check that $\frac{1}{\cosh(x)^2}=\frac{d^2}{dx^2}\log\cosh(x)\in\mathcal{S}(\mathbb{R})$.