Oraux X Ens Analyse 4 24.djvu -

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

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

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 ). Thus [ I_n = -\frac\cos nn + \frac\sin nn^2