Let $a>0$. How to prove the following inequality $$\exists c>0,\exists A>0,\forall t>0:\quad\int^\infty\_0 \frac{\sin(rt)}{rt}\frac{r^4}{\sinh^2(r)} e^{-ar\coth(r)}dr\leq c \big(e^{-At}\big)$$
for some constants $A>0,c>0$



The inequality $\int^\infty_0 \frac{\sin(rt)}{rt}\frac{r^4}{\sinh^2(r)} e^{-ar\coth(r)}dr\leq c \big(e^{-At}\big)$
$\newcommand{\R}{\mathbb R}\renewcommand{\S}{\mathcal S}$This follows from [Theorem IX.14](https://www.google.com/books/edition/II_Fourier_Analysis_Self_Adjointness/Kz7s7bgVe8gC?hl=en), which states the following:



> 
> Let $T\in\S'(\R^n)$. Suppose that the Fourier transform $\hat T$ of $T$ can be continued analytically to the set $\{z\colon|\Im z|<A\}$ for some $A>0$. Suppose also that for each $B<A$ we have $\sup\{\|\hat T(\cdot+iy)\|\_1\colon y\in(-B,B)\}<\infty$. Then $T$ is a bounded continuous function and for each $B<A$ there is a real number $C\_B$ such that
> \begin{equation\*}
>  |T(x)|\le C\_B e^{-B|x|}
> \end{equation\*}
> for $x\in\R^n$.
> 
> 
> 


Indeed, for the integral in question,
\begin{equation\*}
 I(t):=\int^\infty\_0 \frac{\sin rt}{rt}\frac{r^4}{\sinh^2r} e^{-ar\coth(r)}\,dr, 
\end{equation\*}
and
\begin{equation\*}
 T(t):=t I(t), \tag{1}\label{1}
\end{equation\*}
we have $T=\frac{\sqrt{2\pi}}{2i}\,\check f$, where
\begin{equation\*}
 f(r):=\frac{r^3}{\sinh^2r} e^{-ar\coth(r)}
\end{equation\*}
and $\check f$ is the inverse Fourier transform of $f$ (defined by the formula $\check f(t):=\frac1{\sqrt{2\pi}}\,\int\_\R e^{irt}f(r)\,dr$), so that
\begin{equation\*}
 \hat T=\frac{\sqrt{2\pi}}{2i}\,f. 
\end{equation\*}


It is not hard to see that the function $T$ defined by \eqref{1} satisfies all the conditions of [Theorem IX.14](https://www.google.com/books/edition/II_Fourier_Analysis_Self_Adjointness/Kz7s7bgVe8gC?hl=en), cited above, with $A=\pi$. So, for each $B\in(0,\pi)$ there is a real number $C\_B$ such that $|T(t)|\le C\_B e^{-Bt}$ for all $t\ge0$ and hence
\begin{equation\*}
 |I(t)|\le C\_B e^{-Bt} 
\end{equation\*}
for $t\ge1$. On the other hand, for all $t\in[0,1]$,
\begin{equation\*}
 |I(t)|\le c\_a:=\int^\infty\_0\frac{r^4}{\sinh^2r} e^{-ar\coth(r)}\,dr
 \le c\_{a,B}\, e^{-Bt},
\end{equation\*}
where $c\_{a,B}:=c\_a e^B\in(0,\infty)$.


Thus, for each $B\in(0,\pi)$ and all $t\ge0$
\begin{equation\*}
 |I(t)|\le\max(C\_B,c\_{a,B}) e^{-Bt},
\end{equation\*}
as desired.


