Oraux X Ens Analyse 4 24.djvu [ No Login ]

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 \frac\cos nn ), 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) ).

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[ 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 = -\frac\cos nn + \frac\sin nn^2

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: Indeed ( |I_n| \sim \frac\cos nn ), not

Thus [ I_n = \frac1n J_n - \fracf(1)\cos nn = \frac1n \left( O(1/n) \right) - \fracf(1)\cos nn = -\fracf(1)\cos nn + O\left(\frac1n^2\right). ] So ( I_n = O(1/n) ), not yet ( o(1/n^2) ). Hmm — but the problem statement says: if ( f'(0)=0 ) and ( f \in C^2 ), prove ( I_n = o(1/n^2) ). That suggests extra cancellation in the boundary term? Let's check carefully.