Fractional part of a log n is not equidistributed

complex-analysisequidistributionfourier analysis

In Exercise 9 in Fourier Analysis, Stein, I want to prove that fractional part of $a \log n$ is not equidistributed for any $a$.

Using Weyl's criterion, my attempt is to show that $$\frac1N\sum_{n=1}^N e^{2 \pi ib \log n}$$ does not converge to 0, for any nonzero $b$.

May I get any hints to prove above statement?

Best Answer

Apply the Euler–Maclaurin summation formula. For $f\in C^1\big([1,N]\big)$, it reads $$\sum_{n=1}^N f(n)=\int_1^N f(x)\,dx+\frac{f(1)+f(N)}{2}+\int_1^N\left(\{x\}-\frac12\right)f'(x)\,dx,$$ where $\{x\}=x-\lfloor x\rfloor$ is the fractional part of $x$.

For $f(x)=e^{2\pi ib\log x}$, we have $f'(x)=2\pi ib f(x)/x$ and $$W_N:=\frac1N\sum_{n=1}^N e^{2\pi ib\log N}=I_N+F_N+R_N, \\I_N=\frac1N\left.\frac{e^{(2\pi ib+1)\log x}}{2\pi ib+1}\right|_{x=1}^{x=N}=\frac{e^{2\pi ib\log N}-1/N}{2\pi ib+1}, \\F_N=\frac{1+e^{2\pi ib\log N}}{2N},\quad|R_N|\leqslant\frac{2\pi|b|}{N}\int_1^N\frac12\frac{dx}{x}=\pi|b|\frac{\log N}{N}.$$

As $N\to\infty$, clearly $F_N\to 0$ and $R_N\to 0$ but $I_N\not\to 0$. Hence, $W_N\not\to 0$.

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