X Ens Analyse 4 24.djvu - Oraux

Better: By Riemann–Lebesgue lemma, for any ( g \in L^1 ), ( \int g(t) \cos(nt) dt \to 0 ). Here ( g = f' \in L^1 ). Therefore [ \int_0^1 f'(t) \cos(nt) , dt \to 0. ] Hence [ I_n = \frac1n \cdot o(1) = o\left(\frac1n\right). ] Example with ( I_n \sim C/n ) Take ( f(t) = t ). Then ( f(0)=0 ), ( f \in C^1 ).

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.

Actually, known result: If ( f ) is ( C^1 ) and ( f(0)=0 ), ( I_n = o(1/n) ). If ( f ) is ( C^2 ) and ( f(0)=f(1)=0 ), then ( I_n = O(1/n^2) ). But here they only give ( f'(0)=0 ), not ( f(1)=0 ). Possibly a misprint? Let's assume they intended ( f(0)=f(1)=0 ) for (3). Then: Oraux X Ens Analyse 4 24.djvu

Compute: [ I_n = \int_0^1 t \sin(nt) dt. ] Integration by parts: ( u = t ), ( dv = \sin(nt)dt ), ( du = dt ), ( v = -\cos(nt)/n ): [ I_n = \left[ -t \frac\cos(nt)n \right]_0^1 + \frac1n \int_0^1 \cos(nt) dt. ] First term: ( -\frac\cos nn ). Second: ( \frac1n \left[ \frac\sin(nt)n \right]_0^1 = \frac\sin nn^2 ).

We made a mistake: The boundary term at ( t=0 ) in the second integration by parts: ( f'(0) \sin(0)/n = 0 ) indeed, but the first integration by parts gave the term ( -f(1)\cos n / n ). That term is ( O(1/n) ), not smaller. So we cannot get ( o(1/n^2) ) unless ( f(1)=0 ). But the problem didn't assume ( f(1)=0 ). Possibly the intended condition is ( f(0)=f(1)=0 ) and ( f'(0)=0 )? Or perhaps the statement in (3) is: prove ( I_n = o(1/n) ) (already done) but with ( C^2 ) and ( f'(0)=0 ) we can improve? Wait, let's recompute properly with a view to ( o(1/n^2) ). Better: By Riemann–Lebesgue lemma, for any ( g

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

Thus [ I_n = -\frac\cos nn + \frac\sin nn^2. ] As ( n \to \infty ), ( I_n = -\frac\cos nn + o\left(\frac1n\right) ). The amplitude of ( I_n ) is ( \sim \frac1n ) up to a bounded oscillatory factor. Indeed ( |I_n| \sim \fracn ), not ( C/n ) with constant sign, but in the sense of equivalence modulo ( o(1/n) ), it's ( O(1/n) ) and not ( o(1/n) ). ] Hence [ I_n = \frac1n \cdot o(1) = o\left(\frac1n\right)

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) ).