Fourier series analysis \begin{equation}
F_{r}(\theta)=\sum_{k=1}^{\infty} \frac{\cos (k \theta)}{k^{r}} \qquad 0\leq \theta\leq 2\pi 
\end{equation}
\begin{equation}
F_{2}(\theta)=\frac{1}{6} \pi^{2}-\frac{1}{2} \pi \theta+\frac{1}{4} \theta^{2} \qquad 0\leq \theta\leq 2\pi 
\end{equation}
When $r>1$, how to prove that $F_{r}$ is monotonically decreasing on $[0,\pi]$ and is monotonically increasing on $[\pi,2\pi]$?
\begin{equation}
F_{1}(\theta)=-\log|2\sin(\frac{\theta}{2})| \qquad 0\leq \theta\leq 2\pi 
\end{equation}
When $0<r<1$, how to prove that $[0,\pi]$ is monotonically decreasing and $F(0+)=+\infty$ is on $[\pi,2\pi]$, and  it is monotonically increasing and $F(2\pi-)=+\infty$?
 A: Take any $t\in(0,2\pi)$. Then
\begin{equation*}
\begin{aligned}
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{aligned}
\tag{1}
\end{equation*}
where
$$f(t,x):=\sum_{k=1}^\infty\cos kt\, e^{-k x}=\frac{\cos t-e^{-x}}{2 (\cosh x-\cos t)}.$$
(The convergence of the series defining $F_r(t)$, as well as
interchange of the summation and the integration, are justified at the end of this answer.)
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.

Also,
$$f(t,x)=\frac{\sinh x}{2 (\cosh x-\cos t)}-\frac12.$$
So, by monotone convergence, for $r\in(0,1)$ we have
$$\int_0^1 dx\,x^{r-1}f(t,x)\to\int_0^1 dx\,x^{r-1}\Big(\frac{\sinh x}{2 (\cosh x-1)}-\frac12\Big)=\infty$$
as $t\to2n\pi$, for any integer $n$,
since $\sinh x\sim x$ and $\cosh x-1\sim x^2/2$ as $x\to0$.
On the other hand, for real $x>1$,
$$|f(t,x)|\le\sum_{k=1}^\infty e^{-k x}\le 2e^{-x}$$
and hence for $r\in(0,1)$ we have
$$\int_1^\infty dx\,x^{r-1}|f(t,x)|\le2\int_1^\infty dx\,e^{-x}<1.$$
Thus,
$$\Gamma(r)F_r(t)\ge\int_0^1 dx\,x^{r-1}f(t,x)-\int_1^\infty dx\,x^{r-1}|f(t,x)|\to\infty$$
as $t\to2n\pi$, for any integer $n$. So, the claims about the limits follow as well.

(Justification of the convergence of the series defining $F_r(t)$, as well as of the interchange of the summation and the integration in (1).)
Take any real $r\in(0,2)$. Take any $t\in(0,2\pi)$.
Let us show that the limit
\begin{equation*}
F_r(t)=\lim_{n\to\infty}F_{r,n}(t)  
\end{equation*}
exists, where
\begin{equation*}
    F_{r,n}(t):=\sum_{k=1}^n\frac{\cos kt}{k^r}. 
\end{equation*}
Indeed, for all natural $m$ and $n>m$,
\begin{equation*}
\begin{aligned}
F_{r,n}(t)-F_{r,m}(t)&=\sum_{k=m+1}^n\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=m+1}^n\cos kt\, e^{-k x}. 
\end{aligned} 
\end{equation*}
Next, for $x>0$,
\begin{align*}
    \Big|\sum_{k=m+1}^n\cos kt\, e^{-k x}\Big|
&   =\Big|\Re\sum_{k=m+1}^n e^{-k(it+x)}\Big| \\ 
&   \le\Big|\sum_{k=m+1}^n e^{-k(it+x)}\Big| \\ 
&=e^{-(m+1)x}\,\frac{|1-e^{-(n-m)(it+x)}|}{|1-e^{-it-x}|} \\ 
&\le e^{-(m+1)x}\,\frac2{2\sin(t/2)\,e^{-x/2}} \\ 
&\le \,\frac{e^{-mx}}{\sin(t/2)}. 
\end{align*}
So, $F_{r,n}(t)$ is Cauchy-convergent in $n$ and hence convergent in $n$. Also, by dominated convergence, we can interchange the summation and the integration in (1).
