Let $ t > 0 $ and $ k \in \{ 0,1,2,\ldots \} $. Does the following inequality hold?
$$ \int_{k + 1/2}^{k + 3/2} \frac{x \sin(2 \pi x)}{1 + 2 e^{2 \pi t} \cos(2 \pi x) + e^{4 \pi t}} \mathrm{d}{x} \leq \frac{1}{2 \pi} \cdot \frac{1}{(1 - e^{2 \pi t})^{2}}. $$
Such an inequality appears in the study of Selberg $ \zeta $-functions.
The inequality is true, and follows upon integrating by parts. The integral is $$ \int_{k+1/2}^{k+3/2} x d\Big( -\frac{\log (1+2 e^{2\pi t} \cos(2\pi x) +e^{4\pi t}}{4\pi e^{2\pi t}} \Big) $$ and integration by parts gives $$ = \frac{1}{4\pi e^{2\pi t}} \int_{k+1/2}^{k+3/2} \log \frac{1+2e^{2\pi t} \cos (2\pi x) + e^{4\pi t}}{1-2e^{2\pi t} +e^{4\pi t}} dx. $$ Using $\log (1+y) \le y$, the above is $$ \le \frac{1}{4\pi e^{2\pi t}} \int_{k+1/2}^{k+3/2} \frac{(2+2\cos(2\pi x))e^{2\pi t}}{(1-e^{2\pi t})^2} dx = \frac{1}{2\pi} \frac{1}{(1-e^{2\pi t})^2}. $$
I can estimate this as $c/(1-e^{4\pi t})$, where $c$ is a numerical constant which can be computed. It is possible that $c$ is less than $1/(2\pi)$, which will give what you need when $t<0$.
To do this set $r=e^{2\pi t}$, $\theta=2\pi(x-k-1)$. Than the integral becomes
$$\frac{u(-r)}{1-r^2},$$ where $u$ is the Poisson integral of the function $(\theta\sin\theta)/(2\pi), \;-\pi<\theta<\pi$. Trivial estimate (using Maple) shows that $u\leq 1.82/(2\pi)$, but certainly one can do better than that with more computation.
