In a textbook on analytic numbertheory we want to prove that \begin{equation} \Big\vert \{a\}-\frac{1}{2}-\sum\limits_{\substack{\vert m\vert\leq M \\ m\not=0}}\frac{\exp(-2\pi im\alpha)}{2\pi im}\Big\vert\leq\frac{1}{2\pi M\Vert \alpha\Vert} \end{equation}
In the prove we rewrite some stuff and end up with the following integral we want to estimate:
$$\int\limits_\alpha^{1/2}\frac{\sin((2M+1)\pi t)}{\sin \pi t}\mathrm{d}t$$
with $0<\alpha\leq 1/2$. The author claims that the second mean-value theorem gives us
$$\frac{1}{\sin(\pi \alpha)}\int\limits_\alpha^\xi \sin((2M+1)\pi t) \mathrm{d}t$$ with $\xi\in(\alpha,1/2)$ and that this is trivally bounded by $$\frac{1}{2\pi M\Vert \alpha\Vert}$$ First, I am not sure how we end up with this above integral using the second mean-value theorem. I found two versions of it here here, but in the first version we would have $\xi\in(\alpha,1/2]$, and in the second one I do not see how the second term vanishes.
Also, I do not see how this would be trivally bounded by the aboves term. Here $\Vert\alpha\Vert$ is supposed to be the distance from $\alpha$ to the nearest integer, but I don't see how this appears.
I would really appreciate any hints on this!