Show that for any $a\ne0$, and $\sigma$ with $0<\sigma<1$, the sequence $\langle an^\sigma\rangle$ is equidistributed in $[0,1)$.
[Hint: Prove that $\sum_{n=1}^Ne^{2\pi ibn^\sigma}=O(N^\sigma)+O(N^{1-\sigma})$ if $b\ne0$.] In fact, note the following $$\sum_{n=1}^Ne^{2\pi ibn^\sigma}-\int_1^Ne^{2\pi ibn^\sigma}\mathrm dx=O\left(\sum_{n=1}^Nn^{-1+\sigma}\right)$$
I have proved $O(N^\sigma)$,but I can't prove that $\int=O(N^{1-\sigma})$
Put $$u_n=\exp(2\pi bn^{\sigma})-\int_n^{n+1}\exp(2\pi ix^{\sigma}dx=\int_n^{n+1}(\exp(2\pi bn^{\sigma})-\exp(2\pi ix^{\sigma}))dx$$
We have $$\exp(2i\pi bn^{\sigma})-\exp(2\pi ix^{\sigma})=\exp(i\pi (n^{\sigma}+x^{\sigma}))(2\sin(\pi b(n^{\sigma}-x^{\sigma}))$$
Hence $$|\exp(2i\pi bn^{\sigma})-\exp(2\pi ix^{\sigma})|\leq 2|\sin(\pi b(n^{\sigma}-x^{\sigma}))|\leq 2|b|\pi (x^{\sigma}-n^{\sigma})$$ and then: $$|u_n|\leq 2\pi |b|(\frac{(n+1)^{\sigma+1}-n^{\sigma+12}}{\sigma+1}-n^{\sigma})$$
Now there exist $c_1\in ]n,n+1[$ such that $\displaystyle (\frac{(n+1)^{\sigma+1}-n^{\sigma+12}}{\sigma+1}=c_1^{\sigma}\leq (n+1)^{\sigma}$, and in the same way $(n+1)^{\sigma}-n^{\sigma}=\sigma c_2^{\sigma-1}\leq n^{\sigma-1}$ (as $-1+\sigma<0$), and it is easy to finish.
