Question Number 80334 by ~blr237~ last updated on 02/Feb/20
$$\:{let}\:\:\:{f}\in{L}^{\mathrm{1}} \left(\mathbb{R}\right)\:\:\: \\ $$$${let}\:\:{u}_{{n}} =\:\int_{{a}} ^{{b}} {f}\left({t}\right){sin}\left({nt}\right){dt}\:,\:{v}_{{n}} =\int_{{a}} ^{{b}} \frac{{f}\left({t}\right)}{{t}}{sin}\left({nt}\right)\: \\ $$$$\left.\mathrm{1}\right){Prove}\:{that}\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{u}_{{n}} =\mathrm{0} \\ $$$$\left.\mathrm{2}\right){Deduce}\:\:{in}\:{term}\:{of}\:{a},{b},{f}\left(\mathrm{0}\right)\:{the}\:{value}\:{of}\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:{v}_{{n}} \:\: \\ $$
Commented by abdomathmax last updated on 02/Feb/20
![1) by parts u=f and v^′ =sin(nt) ⇒ u_n =[−(1/n)cos(nt)f(t)]_a ^b +(1/n)∫_a ^b f^′ (t)cos(nt)dt =(1/n){f(a)cos(na)−f(b)cos(nb)} +(1/n) ∫_a ^b f^′ (t)cos(nt)dt ⇒ ∣u_n ∣≤(1/n){f(a)+f(b)} +((m(b−a))/n)→0(n→+∞) with m=sup∣f^′ (t)∣_(t∈[a,b])](https://www.tinkutara.com/question/Q80350.png)
$$\left.\mathrm{1}\right)\:\:{by}\:{parts}\:\:{u}={f}\:{and}\:{v}^{'} ={sin}\left({nt}\right)\:\Rightarrow \\ $$$${u}_{{n}} =\left[−\frac{\mathrm{1}}{{n}}{cos}\left({nt}\right){f}\left({t}\right)\right]_{{a}} ^{{b}} +\frac{\mathrm{1}}{{n}}\int_{{a}} ^{{b}} \:{f}^{'} \left({t}\right){cos}\left({nt}\right){dt} \\ $$$$=\frac{\mathrm{1}}{{n}}\left\{{f}\left({a}\right){cos}\left({na}\right)−{f}\left({b}\right){cos}\left({nb}\right)\right\} \\ $$$$+\frac{\mathrm{1}}{{n}}\:\int_{{a}} ^{{b}} \:{f}^{'} \left({t}\right){cos}\left({nt}\right){dt}\:\Rightarrow \\ $$$$\mid{u}_{{n}} \mid\leqslant\frac{\mathrm{1}}{{n}}\left\{{f}\left({a}\right)+{f}\left({b}\right)\right\}\:+\frac{{m}\left({b}−{a}\right)}{{n}}\rightarrow\mathrm{0}\left({n}\rightarrow+\infty\right)\:{with}\: \\ $$$${m}={sup}\mid{f}^{'} \left({t}\right)\mid_{{t}\in\left[{a},{b}\right]} \\ $$