Inequality for an integral involving $ \exp $, $ \sin $ and $ \cos $ 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.
 A: 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 
$$ 
\le \frac{1}{4\pi e^{2\pi t}} \int_{k+1/2}^{k+3/2} \frac{(2+2\cos(2\pi x))e^{2\pi t}}{(1-e^{2\pi t})^2} dx = \frac{1}{2\pi} \frac{1}{(1-e^{2\pi t})^2}. 
$$
A: 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.
