The inequality $\int^\infty_0 (\sin(rt)r^3/\sinh^2(r)) dr\leq cte^{-At}$ How to prove the following inequality $$\forall t>0,\quad\int^\infty_0 \sin(rt)\frac{r^3}{\sinh^2(r)} dr\leq c \big(te^{-At}\big)$$
for some constants $A>0,c>0$
 A: $$\int^\infty_0 \frac{r^3\sin rt}{\sinh^2 r} \,dr=\tfrac{1}{8} \pi ^3 \frac{\pi  t \cosh \pi  t+2\pi t-3 \sinh \pi  t}{\sinh^4(\pi t/2)}, $$
which decays as $e^{-\pi t}$ for large $t$, while it vanishes $\propto t$ for small $t$, so choosing $A$ a bit smaller than $\pi$ will satisfy the inequality for all $t>0$ for a sufficiently large $c$.
I checked that $A=3$ and $c=100$ works.
The exponential decay $\propto e^{-\pi t}$ could be obtained more directly from the pole of the integrand at $r=i\pi$, since the residue decays as $e^{irt}$.
A: This is to complement the answer by Carlo Beenakker by showing that
$$I(t)\le\pi^4 te^{-\pi t}\tag{1}\label{1}$$
for real $t\ge0$, where $I(t)$ is the integral in question.
Indeed, according to Carlo Beenakker,
$$I(t)=\frac{\pi ^3}{8}  \frac{\pi  t \cosh \pi  t+2\pi t-3 \sinh \pi  t}{\sinh^4(\pi t/2)}.$$
This expression for $I(t)$ can be obtained from formula 3.986.4 of Gradshteyn and Ryzhik by using substitutions $\pi x=r$ and $\beta=\pi t/2$, to get
$$J(t):=\int_0^\infty\frac{1-\cos rt}{\sinh^2 r}\,dr
=\frac{\pi t}2\,\coth\frac{\pi t}2-1,$$
and then noting that $I(t)=-J'''(t)$.
So, using the substitution $t=\dfrac{\ln(1+x)}\pi$, one rewrites
inequality \eqref{1} as
$$d(x):=\left(8 x^3+19 x^2+18 x+6\right) \ln(x+1)-3 x (x+1)^2 (x+2)\le0\tag{2}\label{2}$$
for real $x\ge0$.
In turn, inequality \eqref{2} follows immediately because $d(0)=d'(0)=d''(0)=d'''(0)=0$ and
$$d''''(x)=-\frac{2 x \left(36 x^3+120 x^2+139 x+58\right)}{(x+1)^4}\le0$$
for real $x\ge0$. $\quad\Box$
One may note that $I(t)\sim\pi^4 te^{-\pi t}$ as $t\to\infty$, so that the upper bound $\pi^4 te^{-\pi t}$ on $I(t)$ in \eqref{1} is exact.
