let u_n =∫_(−∞) ^∞ ((sin(nx^2 ))/(x^2 +x +n)) dx 1) calculate u_n 2) find lim_(n→+∞) u_n 3) study the serie Σ u


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Question Number 56189 by maxmathsup by imad last updated on 11/Mar/19

$l e t u_{n} = \int_{- \infty}^{\infty} \frac{s i n ({n x}^{2})}{x^{2} + x + n} d x$ $1) c a l c u l a t e u_{n}$ $2) f i n d {l i m}_{n \to + \infty} u_{n}$ $3) s t u d y t h e s e r i e Σ u_{n}$
Commented by maxmathsup by imad last updated on 12/Mar/19
$we have u_n =Im(∫_(−∞) ^(+∞) (e^(inx^2 ) /(x^2 +x +n))dx) let ϕ(z) =(e^(inz^2 ) /(z^2 +z+n)) poles of ϕ? z^2 +z +n =0 →Δ =1−4n =(i(√(4n−1)))^2 ⇒the roots are z_1 =((−1+i(√(4n−1)))/2) and z_2 =((−1−i(√(4n−1)))/2) ⇒ ϕ(z) =(e^(inz^2 ) /((z−z_1 )(z−z_2 ))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_1 ) Res(ϕ,z_1 ) =(e^(inz_1 ^2 ) /(z_1 −z_2 )) =(e^(inz_1 ^2 ) /(i(√(4n−1)))) but z_1 ^2 =(1/4){1−2i(√(4n−1))−4n+1} =(1/4){2−4n −2i(√(4n−1))} =(1/2){1−2n −i(√(4n−1))} ⇒ inz^2 =((in)/2)(1−2n −i(√(4n−1))} =(n/2)(√(4n−1)) +((n(1−2n))/2) i ⇒ e^(inz^2 ) =e^((n/2)(√(4n−1))) {cos(((n(1−2n))/2))+i sin(((n(1−2n))/2))} ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (1/(i(√(4n−1)))) e^((n/2)(√(4n−1))) { cos(((n(1−2n))/2))+i sin(((n(1−2n))/2))} ⇒ u_n =−((2π)/(√(4n−1))) e^((n/2)(√(4n−1))) sin(((n(2n−1))/2)) with n integr and n≥1 .$
$w e h a v e u_{n} = I m (\int_{- \infty}^{+ \infty} \frac{e^{{i n x}^{2}}}{x^{2} + x + n} d x) l e t φ (z) = \frac{e^{{i n z}^{2}}}{z^{2} + z + n} p o l e s o f φ ?$ $z^{2} + z + n = 0 \to Δ = 1 - 4 n = {(i \sqrt{4 n - 1})}^{2} \Rightarrow t h e r o o t s a r e$ $z_{1} = \frac{- 1 + i \sqrt{4 n - 1}}{2} a n d z_{2} = \frac{- 1 - i \sqrt{4 n - 1}}{2} \Rightarrow$ $φ (z) = \frac{e^{{i n z}^{2}}}{(z - z_{1}) (z - z_{2})} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{1})$ $R e s (φ, z_{1}) = \frac{e^{{i n z}_{1}^{2}}}{z_{1} - z_{2}} = \frac{e^{{i n z}_{1}^{2}}}{i \sqrt{4 n - 1}} b u t z_{1}^{2} = \frac{1}{4} {1 - 2 i \sqrt{4 n - 1} - 4 n + 1}$ $= \frac{1}{4} {2 - 4 n - 2 i \sqrt{4 n - 1}} = \frac{1}{2} {1 - 2 n - i \sqrt{4 n - 1}} \Rightarrow$ ${i n z}^{2} = \frac{i n}{2} (1 - 2 n - i \sqrt{4 n - 1}} = \frac{n}{2} \sqrt{4 n - 1} + \frac{n (1 - 2 n)}{2} i \Rightarrow$ $e^{{i n z}^{2}} = e^{\frac{n}{2} \sqrt{4 n - 1}} {c o s (\frac{n (1 - 2 n)}{2}) + i s i n (\frac{n (1 - 2 n)}{2})} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{1}{i \sqrt{4 n - 1}} e^{\frac{n}{2} \sqrt{4 n - 1}} {c o s (\frac{n (1 - 2 n)}{2}) + i s i n (\frac{n (1 - 2 n)}{2})} \Rightarrow$ $u_{n} = - \frac{2 π}{\sqrt{4 n - 1}} e^{\frac{n}{2} \sqrt{4 n - 1}} s i n (\frac{n (2 n - 1)}{2}) w i t h n i n t e g r a n d n ⩾ 1 .$
Commented by maxmathsup by imad last updated on 12/Mar/19

$2) n o l i m i t f o r (u_{n}) b u t ∣ u_{n} ∣ \to + \infty$ $3) Σ u_{n} i s a d i v e r g e n t s e r i e .$
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