Is this a bounded sequence?
5
$\begingroup$
analysis
sequences-and-series
fourier-analysis
2 Answers
7
Yes, and it converges to $\int_1^\infty x^{-a}e^{ix}dx$. To see that this improper Riemann integral exists, you could break up the integrand into real and imaginary parts, break up the resulting real integrals over intervals of length $\pi$, and apply the alternating series test.
8
I just thought of another answer. Using integration by parts, we get $$\frac{1}{i}(N^{-a}e^{iN}-e^i)+\frac{a}{i}\int_1^N y^{-(1+a)}e^{iy} dy$$ The first term is bounded and the second term is dominated by the integral $\int_1^\infty y^{-(1+a)} dy$.