[ 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) ).
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) ). Oraux X Ens Analyse 4 24.djvu
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 ). [ J_n = \left[ f'(t) \frac\sin(nt)n \right]_0^1 -
Thus ( I_n = o(1/n^2) ).
Let ( u = f'(t) ), ( dv = \cos(nt)dt ), ( du = f''(t) dt ), ( v = \frac\sin(nt)n ). We made a mistake: The boundary term at
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 ).