Here $\alpha >1 $. The function is defined as
$$f(x) = \sum_{n\geq 1 } \sin n x /n^\alpha .$$
The domain is $(0, \pi)$.
We know that if $\alpha = 1$, $f(x) = (\pi -x )/2$.
If $\alpha >1$, the jump at $x =0 $ is softened into a boundary layer.
The function has the following picture. The question is, where is the maximum? How does it scale with $\alpha$? Is it power of $( \alpha -1 )$?
For the general case, we have $$f(x)=\frac{1}{2} i \left(\text{Li}_{\alpha }\left(e^{-i x}\right)-\text{Li}_{\alpha }\left(e^{i x}\right)\right)$$ So, for a given $\alpha$, we need to solve for $x$ $$f'(x)=\frac{1}{2} \left(\text{Li}_{\alpha-1}\left(e^{-i x}\right)+\text{Li}_{\alpha-1}\left(e^{i x}\right)\right)=0$$ which does not show analytical solutions.
Using numerical methods, some results $$\left( \begin{array}{cc} \alpha& x_{\text{max}} \\ 1.5 & 0.74377 \\ 2.0 & 1.04720 \\ 2.5 & 1.21912 \\ 3.0 & 1.32779 \\ 3.5 & 1.40038 \\ 4.0 & 1.45035 \\ 4.5 & 1.48534 \\ 5.0 & 1.51007 \\ 5.5 & 1.52763 \\ 6.0 & 1.54013 \\ 6.5 & 1.54902 \\ 7.0 & 1.55534 \\ 7.5 & 1.55984 \\ 8.0 & 1.56303 \\ 8.5 & 1.56530 \\ 9.0 & 1.56690 \\ 9.5 & 1.56804 \\ 10.0 & 1.56885 \end{array} \right)$$
I suppose that you already noticed where this is going.