1)calculate A_n =∫_0 ^∞ ((sin(x^(2n) ))/((x^2 +1)^2 ))dx with n integr natural 2) study the convergene of Σ A


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Question Number 64392 by mathmax by abdo last updated on 17/Jul/19

$1) c a l c u l a t e A_{n} = \int_{0}^{\infty} \frac{s i n (x^{2 n})}{{(x^{2} + 1)}^{2}} d x w i t h n i n t e g r n a t u r a l$ $2) s t u d y t h e c o n v e r g e n e o f Σ A_{n}$
Commented by mathmax by abdo last updated on 18/Jul/19
$1) we have 2A_n =∫_(−∞) ^(+∞) ((sin(x^(2n) ))/((x^2 +1)^2 ))dx =Im(∫_(−∞) ^(+∞) (e^(ix^(2n) ) /((x^2 +1)^2 ))) let W(z) =(e^(iz^(2n) ) /((z^2 +1)^2 )) ⇒W(z)=(e^(iz^(2n) ) /((z−i)^2 (z+i)^2 )) so the poles of W are +^− i (doubles) residus theorem give ∫_(−∞) ^(+∞) W(z)dz =2iπRes(W,i) Res(W,i)=lim_(z→i) (1/((2−1)!)){(z−i)^2 W(z)}^((1)) =lim_(z→i) {(e^(iz^(2n) ) /((z+i)^2 ))}^((1)) =lim_(z→i) ((2niz^(2n−1) e^(iz^(2n) ) (z+i)^2 −2(z+i)e^(iz^(2n) ) )/((z+i)^4 )) =lim_(z→i) ((2niz^(2n−1) e^(iz^(2n) ) (z+i)−2e^(iz^(2n) ) )/((z+i)^3 )) lim_(z→i) (({2niz^(2n−1) (z+i)−2}e^(iz^(2n) ) )/((z+i)^3 )) =(({(2ni)(2i)(−1)^n i^(−1) −2})/((2i)^3 ))e^(i(−1)^n ) =(({4ni(−1)^n −2)e^(i(−1)^n ) )/(−8i)) =(((1−2ni(−1)^n )e^(i(−1)^n ) )/(4i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =((2iπ)/(4i)){1−2ni(−1)^n )e^(i(−1)^n ) =(π/2)(1−2ni(−1)^n )e^(i(−1)^n ) ) =(π/2)(1−2n(−1)^n i)(cos(−1)^n +isin(−1)^n ) =(π/2)(cos(−1)^n +isin(−1)^n −2n(−1)^n cos(−1)^n i +2n(−1)^n sin(−1)^n } =(π/2){ cos(−1)^n +2n(−1)^n sin(−1)^n +i(sin(−1)^n −2ncos(−1)^n )} 2A =Im(∫_(−∞) ^(+∞) W(z)dz) =(π/2){sin(−1)^n −2n cos(−1)^n } ⇒ A =(π/4){ sin(−1)^n −2ncos(−1)^n } .$
$1) w e h a v e 2 A_{n} = \int_{- \infty}^{+ \infty} \frac{s i n (x^{2 n})}{{(x^{2} + 1)}^{2}} d x = I m (\int_{- \infty}^{+ \infty} \frac{e^{{i x}^{2 n}}}{{(x^{2} + 1)}^{2}})$ $l e t W (z) = \frac{e^{{i z}^{2 n}}}{{(z^{2} + 1)}^{2}} \Rightarrow W (z) = \frac{e^{{i z}^{2 n}}}{{(z - i)}^{2} {(z + i)}^{2}} s o t h e p o l e s o f W a r e$ $\bar{+} i (d o u b l e s) r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, i)$ $R e s (W, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} W (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{e^{{i z}^{2 n}}}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} \frac{2 {n i z}^{2 n - 1} e^{{i z}^{2 n}} {(z + i)}^{2} - 2 (z + i) e^{{i z}^{2 n}}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{2 {n i z}^{2 n - 1} e^{{i z}^{2 n}} (z + i) - 2 e^{{i z}^{2 n}}}{{(z + i)}^{3}}$ ${l i m}_{z \to i} \frac{{2 {n i z}^{2 n - 1} (z + i) - 2} e^{{i z}^{2 n}}}{{(z + i)}^{3}}$ $= \frac{{(2 n i) (2 i) {(- 1)}^{n} i^{- 1} - 2}}{{(2 i)}^{3}} e^{i {(- 1)}^{n}}$ $= \frac{{4 n i {(- 1)}^{n} - 2) e^{i {(- 1)}^{n}}}{- 8 i} = \frac{(1 - 2 n i {(- 1)}^{n}) e^{i {(- 1)}^{n}}}{4 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = \frac{2 i π}{4 i} {1 - 2 n i {(- 1)}^{n}) e^{i {(- 1)}^{n}} = \frac{π}{2} (1 - 2 n i {(- 1)}^{n}) e^{i {(- 1)}^{n}})$ $= \frac{π}{2} (1 - 2 n {(- 1)}^{n} i) (c o s {(- 1)}^{n} + i s i n {(- 1)}^{n})$ $= \frac{π}{2} (c o s {(- 1)}^{n} + i s i n {(- 1)}^{n} - 2 n {(- 1)}^{n} c o s {(- 1)}^{n} i + 2 n {(- 1)}^{n} s i n {(- 1)}^{n}}$ $= \frac{π}{2} {c o s {(- 1)}^{n} + 2 n {(- 1)}^{n} s i n {(- 1)}^{n} + i (s i n {(- 1)}^{n} - 2 n c o s {(- 1)}^{n})}$ $2 A = I m (\int_{- \infty}^{+ \infty} W (z) d z) = \frac{π}{2} {s i n {(- 1)}^{n} - 2 n c o s {(- 1)}^{n}} \Rightarrow$ $A = \frac{π}{4} {s i n {(- 1)}^{n} - 2 n c o s {(- 1)}^{n}} .$
Commented by mathmax by abdo last updated on 18/Jul/19

$2) w e h a v e A_{n} = \frac{π}{4} s i n {(- 1)}^{n} - \frac{n π}{2} c o s {(- 1)}^{n} \Rightarrow$ $Σ A_{n} = \frac{π}{4} Σ s i n {(- 1)}^{n} - \frac{π}{2} Σ n c o s {(- 1)}^{n} b u t$ ${l i m}_{n \to + \infty} s i n {(- 1)}^{n} \neq 0 a l s o {l i m}_{n \to + \infty} n c o s {(- 1)}^{n} \neq 0 \Rightarrow$ $Σ A_{n} d i v e r g e s .$
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