let f∈L^1 (R) let u_n = ∫_a ^b f(t)sin(nt)dt , v_n =∫_a ^b ((f(t))/t)sin(nt) 1)Prove that lim_(n→∞) u_n =0 2)Deduce in term of a,b,f(0) the value of lim_(n→∞) v


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Question Number 80334 by ~blr237~ last updated on 02/Feb/20

$l e t f \in L^{1} (R)$ $l e t u_{n} = \int_{a}^{b} f (t) s i n (n t) d t, v_{n} = \int_{a}^{b} \frac{f (t)}{t} s i n (n t)$ $1) P r o v e t h a t \lim_{n \to \infty} u_{n} = 0$ $2) D e d u c e i n t e r m o f a, b, f (0) t h e v a l u e o f \lim_{n \to \infty} v_{n}$
Commented by abdomathmax last updated on 02/Feb/20
$1) by parts u=f and v^′ =sin(nt) ⇒ u_n =[−(1/n)cos(nt)f(t)]_a ^b +(1/n)∫_a ^b f^′ (t)cos(nt)dt =(1/n){f(a)cos(na)−f(b)cos(nb)} +(1/n) ∫_a ^b f^′ (t)cos(nt)dt ⇒ ∣u_n ∣≤(1/n){f(a)+f(b)} +((m(b−a))/n)→0(n→+∞) with m=sup∣f^′ (t)∣_(t∈[a,b])$
$1) b y p a r t s u = f a n d v^{^{'}} = s i n (n t) \Rightarrow$ $u_{n} = {[- \frac{1}{n} c o s (n t) f (t)]}_{a}^{b} + \frac{1}{n} \int_{a}^{b} f^{^{'}} (t) c o s (n t) d t$ $= \frac{1}{n} {f (a) c o s (n a) - f (b) c o s (n b)}$ $+ \frac{1}{n} \int_{a}^{b} f^{^{'}} (t) c o s (n t) d t \Rightarrow$ $∣ u_{n} ∣⩽ \frac{1}{n} {f (a) + f (b)} + \frac{m (b - a)}{n} \to 0 (n \to + \infty) w i t h$ $m = s u p ∣ f^{^{'}} (t) ∣_{t \in [a, b]}$
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