Integrate by parts twice: First: ( I_n = \frac1n \int_0^1 f'(t)\cos(nt) dt ) (boundary term vanishes because ( f(0)=f(1)=0 )). Second: Let ( K_n = \int_0^1 f'(t)\cos(nt) dt ). Integrate by parts: ( u = f'(t) ), ( dv = \cos(nt) dt ), ( du = f''(t) dt ), ( v = \sin(nt)/n ). Then [ K_n = \left[ f'(t) \frac\sin(nt)n \right]_0^1 - \frac1n \int_0^1 f''(t) \sin(nt) dt. ] Boundary term: at ( t=1 ), ( f'(1)\sin n /n = O(1/n) ); at ( t=0 ), ( f'(0)\sin 0 / n = 0 ). So ( K_n = O(1/n) ). Then [ I_n = \frac1n \cdot O\left(\frac1n\right) = O\left(\frac1n^2\right). ] With ( f'' ) integrable, the remaining integral ( \int f''(t)\sin(nt) dt \to 0 ) by Riemann–Lebesgue, giving ( o(1/n^2) ).
The goal of the oral exam is not to find the trick immediately. The jury wants to see organization. When you open a problem: Oraux X Ens Analyse 4 24.djvu