I am solving an exercise in my textbook. After some steps, I need to compare $e^{i \alpha n^\beta}$ and $\int_n^{n+1}e^{i \alpha x^\beta}\, dx$, where $\alpha\neq 0,0<\beta<1$, in order to investigate the convergence property for some series. I hope I can get something like $$|e^{i \alpha n^\beta}-\int_{n}^{n+1}e^{i \alpha x^\beta}\, dx|\leq\quad \text{something related to }n^{\gamma},\gamma<0$$
But unlike real-valued function, I can't use monotonicity of the function to tell their relation. I also try to evaluate the integral by $y=x^\beta$, and by integration by part, we have $$\begin{align} \int_{n}^{n+1}e^{i \alpha x^\beta}\, dx&=\int_{n^\beta}^{(n+1)^\beta}e^{i \alpha y}\,\frac{1}{\beta}y^{\frac{1}{\beta}-1} dy\\&=e^{i \alpha y}\,y^{\frac{1}{\beta}}\bigg|_{n^\beta}^{(n+1)^\beta}-i \alpha\int_{n^\beta}^{(n+1)^\beta}e^{i \alpha y}\,y^{\frac{1}{\beta}} dy\\&=e^{i \alpha (n+1)^\beta}(n+1)-e^{i \alpha n^\beta}n-i \alpha\int_{n^\beta}^{(n+1)^\beta}e^{i \alpha y}\,y^{\frac{1}{\beta}} dy \end{align}$$
But I don't know how to continue. Maybe I should compare with $\int_{n-1}^{n}e^{i \alpha x^\beta}\, dx$?
Assuming $\alpha \in \mathbb{R}$, we can estimate
\begin{align} \left\lvert e^{i\alpha n^\beta} - \int_n^{n+1} e^{i\alpha x^\beta}\,dx\right\rvert &\leqslant \int_n^{n+1} \left\lvert e^{i\alpha n^\beta} - e^{i\alpha x^\beta}\right\rvert\,dx \\ &= \int_n^{n+1} 2 \left\lvert \sin \frac{\alpha n^\beta - \alpha x^\beta}{2}\right\rvert\,dx\\ &\leqslant \lvert\alpha\rvert \int_n^{n+1} \lvert x^\beta - n^\beta\rvert\,dx\\ &= \lvert\alpha\rvert \int_n^{n+1} \beta\cdot (n+\xi(x))^{\beta-1}\cdot (x-n)\,dx\\ &\leqslant \frac{\lvert\alpha\rvert\beta}{n^{1-\beta}} \int_n^{n+1} (x-n)\,dx\\ &= \frac{\lvert\alpha\rvert\beta}{2n^{1-\beta}} \end{align}
using $\lvert e^{ix}-e^{iy}\rvert = 2\left\lvert\sin \frac{x-y}{2}\right\rvert$, $\lvert \sin x\rvert \leqslant \lvert x\rvert$ for real $x$, and the mean value theorem.