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\begin{equation} F_r(t)=\sum_{k=1}^\infty\cos kt\, \frac1{\Gamma(r)}\int_0^\infty dx\,x^{r-1}e^{-k x} \\ =\frac1{\Gamma(r)}\int_0^\infty dx\,x^{r-1}\sum_{k=1}^\infty\cos kt\, e^{-k x} \\ =\frac1{\Gamma(r)}\int_0^\infty dx\,x^{r-1}f(t,x), \end{equation} where $$f(t,x):=\frac{-\cos t-\sinh x+\cosh x}{2 (\cos t-\cosh x)}.$$ The partial derivative of $f(t,x)$ in $t$ is $$-\frac{\sin t \sinh x}{2 (\cos t-\cosh x)^2},$$ whose sign for $x>0$ is opposite to the sign of $\sin t$. So, the desired monotonicity follows.