let u_n =∫_0 ^∞ ((sin(nx^2 ))/(x^2 +6))dx 1) calculate u_n and lim u_n (n→+∞) 2) find nature of Σ u_n and calaculate it. 3) find nature of Σ u


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Question Number 54777 by maxmathsup by imad last updated on 10/Feb/19

$l e t u_{n} = \int_{0}^{\infty} \frac{s i n ({n x}^{2})}{x^{2} + 6} d x$ $1) c a l c u l a t e u_{n} a n d l i m u_{n} (n \to + \infty)$ $2) f i n d n a t u r e o f Σ u_{n} a n d c a l a c u l a t e i t .$ $3) f i n d n a t u r e o f Σ u_{n}^{2}$
Commented by Abdo msup. last updated on 11/Feb/19
$1) we have u_n =_(x=(√6)t) ∫_0 ^∞ ((sin(6nt^2 ))/(6(1+t^2 ))) (√6)dt =(1/(√6))∫_0 ^∞ ((sin(6nt^2 ))/(1+t^2 )) dt ⇒2u_n =(1/(√6)) ∫_(−∞) ^(+∞) ((sin(6nt^2 ))/(1+t^2 ))dt ⇒2(√6)u_n =Im( ∫_(−∞) ^(+∞) (e^(i6nt^2 ) /(1+t^2 ))dt) let ϕ(z) =((e^(6inz^2 ) )/(z^2 +1)) the poles of ϕ are i and −i residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Rez(ϕ,i) Rez(ϕ,i) =lim_(z→i) (z−i)ϕ(z) =(e^(−6ni) /(2i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(−6ni) /(2i)) =π{cos(6n)−i sin(6n)}⇒ 2(√6)u_n =−π sin(6n) ⇒u_n =−(π/(2(√6))) sin(6n) 2) the sequence (u_n ) diverges ⇒Σ u_n diverges also u_n ^2 =(π^2 /(24))sin^2 (6n) diverges ⇒Σ u_n ^2 diverges$
$1) w e h a v e u_{n} =_{x = \sqrt{6} t} \int_{0}^{\infty} \frac{s i n (6 {n t}^{2})}{6 (1 + t^{2})} \sqrt{6} d t$ $= \frac{1}{\sqrt{6}} \int_{0}^{\infty} \frac{s i n (6 {n t}^{2})}{1 + t^{2}} d t \Rightarrow 2 u_{n} = \frac{1}{\sqrt{6}} \int_{- \infty}^{+ \infty} \frac{s i n (6 {n t}^{2})}{1 + t^{2}} d t$ $\Rightarrow 2 \sqrt{6} u_{n} = I m (\int_{- \infty}^{+ \infty} \frac{e^{i 6 {n t}^{2}}}{1 + t^{2}} d t) l e t$ $φ (z) = \frac{e^{6 {i n z}^{2}}}{z^{2} + 1} t h e p o l e s o f φ a r e i a n d - i$ $r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e z (φ, i)$ $R e z (φ, i) = {l i m}_{z \to i} (z - i) φ (z) = \frac{e^{- 6 n i}}{2 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{e^{- 6 n i}}{2 i} = π {c o s (6 n) - i s i n (6 n)} \Rightarrow$ $2 \sqrt{6} u_{n} = - π s i n (6 n) \Rightarrow u_{n} = - \frac{π}{2 \sqrt{6}} s i n (6 n)$ $2) t h e s e q u e n c e (u_{n}) d i v e r g e s \Rightarrow Σ u_{n} d i v e r g e s$ $a l s o u_{n}^{2} = \frac{π^{2}}{24} {s i n}^{2} (6 n) d i v e r g e s \Rightarrow Σ u_{n}^{2} d i v e r g e s$
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