Oraux X Ens Analyse 4 24.djvu Apr 2026

[ I_n = \left[ -f(t) \frac\cos(nt)n \right]_0^1 + \frac1n \int_0^1 f'(t) \cos(nt) , dt. ] Boundary term: at ( t=1 ): ( -f(1) \frac\cos nn ). At ( t=0 ): ( + f(0) \frac1n = 0 ). So boundary term is ( O(1/n) ).

Thus ( I_n = o(1/n^2) ).

Let ( u = f'(t) ), ( dv = \cos(nt)dt ), ( du = f''(t) dt ), ( v = \frac\sin(nt)n ). Oraux X Ens Analyse 4 24.djvu

[ J_n = \left[ f'(t) \frac\sin(nt)n \right]_0^1 - \frac1n \int_0^1 f''(t) \sin(nt) dt. ] Boundary: at ( t=1 ): ( f'(1) \sin n / n ); at ( t=0 ): ( f'(0) \cdot 0 / n = 0 ). So ( J_n = O(1/n) ).

I cannot directly access external files such as Oraux X Ens Analyse 4 24.djvu . However, if you provide the text or a specific exercise from that document (e.g., by copying the statement or describing the problem), I can certainly help produce a detailed solution, commentary, or a synthetic correction typical of an oral examination at ENS/X level in analysis. [ I_n = \left[ -f(t) \frac\cos(nt)n \right]_0^1 +

If you want a strictly positive constant ( C ), take ( f(t) = t ) and look at subsequence ( n = 2k\pi ) not possible, but better: ( f(t)=1 ) fails ( f(0)=0 ). Try ( f(t)=t ): Then ( \limsup n|I_n| = 1 ), so not ( o(1/n) ). If ( f \in C^2 ) and ( f'(0)=0 ) Integrate by parts twice. First as before: [ I_n = \frac1n \int_0^1 f'(t) \cos(nt) dt - \fracf(1)\cos nn. ] Now integrate by parts again on ( J_n := \int_0^1 f'(t) \cos(nt) dt ).

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) ). So boundary term is ( O(1/n) )

The integral term: ( \left| \int_0^1 f'(t) \cos(nt) , dt \right| \leq \int_0^1 |f'(t)| dt < \infty ), hence it is bounded. Thus the whole integral term is ( O(1/n) ). Wait — but we need ( o(1/n) ), not just ( O(1/n) ).