let A_n =∫_0 ^∞ ((x sin(nx))/((x^2 +n^2 )^2 ))dx with n integr natural not 0 1) find the value of A_n 2) study the convergence of Σ A


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Question Number 53114 by maxmathsup by imad last updated on 17/Jan/19

$l e t A_{n} = \int_{0}^{\infty} \frac{x s i n (n x)}{{(x^{2} + n^{2})}^{2}} d x w i t h n i n t e g r n a t u r a l n o t 0$ $1) f i n d t h e v a l u e o f A_{n}$ $2) s t u d y t h e c o n v e r g e n c e o f Σ A_{n}$
Commented by maxmathsup by imad last updated on 18/Jan/19
$1) changement x=nt give A_n =∫_0 ^∞ ((nt sin(n^2 t))/(n^4 (t^2 +1)^2 )) ndt =(1/n^2 )∫_0 ^∞ ((tsin(n^2 t))/((t^2 +1)^2 ))dt ⇒ 2n^2 A_n =∫_(−∞) ^(+∞) ((tsin(n^2 t))/((t^2 +1)^2 ))dt =Im( ∫_(−∞) ^(+∞) ((t e^(in^2 t) )/((t^2 +1)^2 ))dt) let consider the complex function ϕ(z) =((z e^(in^2 z) )/((z^2 +1)^2 )) ⇒ϕ(z)=((z e^(in^2 z) )/((z−i)^2 (z+i)^2 )) so the poles of ϕ are i and −i (doubles) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπRes(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { ((z e^(in^2 z) )/((z+i)^2 ))}^((1)) =lim_(z→i) (((e^(in^2 z) +in^2 z e^(in^2 z) )(z+i)^2 −2(z+i)z e^(in^2 z) )/((z+i)^4 )) =lim_(z→i) ((((1+in^2 z)(z+i)−2z)e^(in^2 z) )/((z+i)^3 )) =((((1−n^2 )(2i)−2i)e^(−n^2 ) )/((2i)^3 )) =((−2in^2 e^(−n^2 ) )/(−8i)) =(n^2 /4) e^(−n^2 ) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (n^2 /4) e^(−n^2 ) =((iπn^2 )/2) e^(−n^2 ) 2n^2 A_n =((πn^2 )/2) e^(−n^2 ) ⇒ A_n =(π/4) e^(−n^2 ) . we have Σ_(n=0) ^∞ A_n =(π/4) Σ_(n=0) ^∞ e^(−n^2 ) but n^2 ≥n ⇒−n^2 ≤−n ⇒e^(−n^2 ) ≤e^(−n) ⇒ Σ_(n=0) ^∞ A_n ≤(π/4) Σ_(n=0) ^∞ ((1/e))^n =(π/4)(1/(1−(1/e))) =(π/4) (e/(e−1)) ⇒ 0< Σ_(n=0) ^∞ A_n ≤((πe)/(4(e−1))) .$
$1) c h a n g e m e n t x = n t g i v e A_{n} = \int_{0}^{\infty} \frac{n t s i n (n^{2} t)}{n^{4} {(t^{2} + 1)}^{2}} n d t = \frac{1}{n^{2}} \int_{0}^{\infty} \frac{t s i n (n^{2} t)}{{(t^{2} + 1)}^{2}} d t \Rightarrow$ $2 n^{2} A_{n} = \int_{- \infty}^{+ \infty} \frac{t s i n (n^{2} t)}{{(t^{2} + 1)}^{2}} d t = I m (\int_{- \infty}^{+ \infty} \frac{t e^{{i n}^{2} t}}{{(t^{2} + 1)}^{2}} d t) l e t c o n s i d e r t h e c o m p l e x$ $f u n c t i o n φ (z) = \frac{z e^{{i n}^{2} z}}{{(z^{2} + 1)}^{2}} \Rightarrow φ (z) = \frac{z e^{{i n}^{2} z}}{{(z - i)}^{2} {(z + i)}^{2}} s o t h e p o l e s o f φ a r e i a n d - i$ $(d o u b l e s) r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{z e^{{i n}^{2} z}}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} \frac{(e^{{i n}^{2} z} + {i n}^{2} z e^{{i n}^{2} z}) {(z + i)}^{2} - 2 (z + i) z e^{{i n}^{2} z}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{((1 + {i n}^{2} z) (z + i) - 2 z) e^{{i n}^{2} z}}{{(z + i)}^{3}} = \frac{((1 - n^{2}) (2 i) - 2 i) e^{- n^{2}}}{{(2 i)}^{3}}$ $= \frac{- 2 {i n}^{2} e^{- n^{2}}}{- 8 i} = \frac{n^{2}}{4} e^{- n^{2}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{n^{2}}{4} e^{- n^{2}} = \frac{i π n^{2}}{2} e^{- n^{2}}$ $2 n^{2} A_{n} = \frac{π n^{2}}{2} e^{- n^{2}} \Rightarrow A_{n} = \frac{π}{4} e^{- n^{2}} .$ $w e h a v e \sum_{n = 0}^{\infty} A_{n} = \frac{π}{4} \sum_{n = 0}^{\infty} e^{- n^{2}} b u t n^{2} ⩾ n \Rightarrow - n^{2} ⩽ - n \Rightarrow e^{- n^{2}} ⩽ e^{- n} \Rightarrow$ $\sum_{n = 0}^{\infty} A_{n} ⩽ \frac{π}{4} \sum_{n = 0}^{\infty} {(\frac{1}{e})}^{n} = \frac{π}{4} \frac{1}{1 - \frac{1}{e}} = \frac{π}{4} \frac{e}{e - 1} \Rightarrow$ $0 < \sum_{n = 0}^{\infty} A_{n} ⩽ \frac{π e}{4 (e - 1)} .$
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