let n from N and A_n = ∫_(−∞) ^(+∞) ((cos(ax))/((x^2 +x+1)^n ))dx and B_n = ∫_(−∞) ^(+∞) ((sin(ax))/((x^2 +x+1)^n ))dx find the value of A_(n ) and B


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Question Number 38727 by maxmathsup by imad last updated on 28/Jun/18

$l e t n f r o m N a n d A_{n} = \int_{- \infty}^{+ \infty} \frac{c o s (a x)}{{(x^{2} + x + 1)}^{n}} d x a n d$ $B_{n} = \int_{- \infty}^{+ \infty} \frac{s i n (a x)}{{(x^{2} + x + 1)}^{n}} d x$ $f i n d t h e v a l u e o f A_{n} a n d B_{n} .$
Commented by math khazana by abdo last updated on 01/Jul/18
$we have A_n +iB_n = ∫_(−∞) ^(+∞) (e^(iax) /((x^2 +x+1)^n ))dx =W_n W_n = ∫_(−∞) ^(+∞) (e^(iax) /(((x+(1/2))^2 +(3/4))^n ))dx changement x+(1/2)=((√3)/2) t give W_n = ((4/3))^n ∫_(−∞) ^(+∞) (e^(ia(((√3)/2)t−(1/2))) /((t^2 +1)^n )) ((√3)/2)dt =((√3)/2)((4/3))^n e^(−((ia)/2)) ∫_(−∞) ^(+∞) (e^(i((√3)/2)at) /((t^2 +1)^n ))dt let λ =((√3)/2)a and find ∫_(−∞) ^(+∞) (e^(iλt) /((t^2 +1)^n )) =I_λ let ϕ(z)= (e^(iλz) /((z^2 +1)^n )) ϕ(z) = (e^(iλz) /((z−i)^n (z+i)^n )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((n−1)!)) { (z−i)^n ϕ(z)}^((n−1)) lim_(z→i) (1/((n−1)!)) { e^(iλz) (z+i)^(−n) }^((n−1)) but {e^(iλz) (z+i)^((−n)) }^((p)) =Σ_(k=0) ^p C_p ^k {(z+i)^((−n)) }^((k)) (e^(iλz) )^((p−k)) ...be continued...$
$w e h a v e A_{n} + {i B}_{n} = \int_{- \infty}^{+ \infty} \frac{e^{i a x}}{{(x^{2} + x + 1)}^{n}} d x = W_{n}$ $W_{n} = \int_{- \infty}^{+ \infty} \frac{e^{i a x}}{{({(x + \frac{1}{2})}^{2} + \frac{3}{4})}^{n}} d x c h a n g e m e n t$ $x + \frac{1}{2} = \frac{\sqrt{3}}{2} t g i v e$ $W_{n} = {(\frac{4}{3})}^{n} \int_{- \infty}^{+ \infty} \frac{e^{i a (\frac{\sqrt{3}}{2} t - \frac{1}{2})}}{{(t^{2} + 1)}^{n}} \frac{\sqrt{3}}{2} d t$ $= \frac{\sqrt{3}}{2} {(\frac{4}{3})}^{n} e^{- \frac{i a}{2}} \int_{- \infty}^{+ \infty} \frac{e^{i \frac{\sqrt{3}}{2} a t}}{{(t^{2} + 1)}^{n}} d t l e t λ = \frac{\sqrt{3}}{2} a$ $a n d f i n d \int_{- \infty}^{+ \infty} \frac{e^{i λ t}}{{(t^{2} + 1)}^{n}} = I_{λ}$ $l e t φ (z) = \frac{e^{i λ z}}{{(z^{2} + 1)}^{n}}$ $φ (z) = \frac{e^{i λ z}}{{(z - i)}^{n} {(z + i)}^{n}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(n - 1)!} {{(z - i)}^{n} φ (z)}^{(n - 1)}$ ${l i m}_{z \to i} \frac{1}{(n - 1)!} {e^{i λ z} {(z + i)}^{- n}}^{(n - 1)} b u t$ ${e^{i λ z} {(z + i)}^{(- n)}}^{(p)} = \sum_{k = 0}^{p} C_{p}^{k} {{(z + i)}^{(- n)}}^{(k)} {(e^{i λ z})}^{(p - k)}$ $. . . b e c o n t i n u e d . . .$
Commented by math khazana by abdo last updated on 02/Jul/18
$let find ((z +i)^(−n) )^((k)) (z+i)^(−n) }^((1)) =−n(z+i)^(−n−1) {(z+i)^(−n) }^((2)) =(−1)^2 n(n+1) (z+i)^(−n−2) ⇒ {(z+i)^(−n) }^((k)) =(−1)^k n(n+1)...(n+k−1)(z+i)^(−n−k) also we have (e^(iλz) )^((1)) =iλ e^(iλz) (e^(iλz) )^((2)) =(iλ)^2 e^(iλz) ⇒(e^(iλz) )^((p−k)) =(iλ)^(p−k) e^(iλz) ⇒ {e^(iλz) (z+i)^(−n) }^((p)) =Σ_(k=0) ^p C_p ^k (−1)^k n(n+1)...(n+k−1)(z+i)^(−n−k) (iλ)^(p−k) e^(iλz) p=n−1 ⇒ {e^(iλz) (z+i)^(−n) }^((n−1)) =Σ_(k=0) ^(n−1) C_(n−1) ^k (−1)^k n(n+1)...(2n−1)(z+i)^(−2n+1) (iλ)^(n−1−k) e^(iλz) Res(ϕ,i) = (1/((n−1)!)) Σ_(k=0) ^(n−1) C_(n−1) ^k (−1)^k n(n+1)...(2n−1)(2i)^(−2n+1) (iλ)^(n−1−k) e^(−λ)$
$l e t f i n d {({(z + i)}^{- n})}^{(k)}$ ${(z + i)}^{- n}}^{(1)} = - n {(z + i)}^{- n - 1}$ ${{(z + i)}^{- n}}^{(2)} = {(- 1)}^{2} n (n + 1) {(z + i)}^{- n - 2} \Rightarrow$ ${{(z + i)}^{- n}}^{(k)} = {(- 1)}^{k} n (n + 1) . . . (n + k - 1) {(z + i)}^{- n - k}$ $a l s o w e h a v e {(e^{i λ z})}^{(1)} = i λ e^{i λ z}$ ${(e^{i λ z})}^{(2)} = {(i λ)}^{2} e^{i λ z} \Rightarrow {(e^{i λ z})}^{(p - k)} = {(i λ)}^{p - k} e^{i λ z} \Rightarrow$ ${e^{i λ z} {(z + i)}^{- n}}^{(p)} = \sum_{k = 0}^{p} C_{p}^{k} {(- 1)}^{k} n (n + 1) . . . (n + k - 1) {(z + i)}^{- n - k} {(i λ)}^{p - k} e^{i λ z}$ $p = n - 1 \Rightarrow$ ${e^{i λ z} {(z + i)}^{- n}}^{(n - 1)} = \sum_{k = 0}^{n - 1} C_{n - 1}^{k} {(- 1)}^{k} n (n + 1) . . . (2 n - 1) {(z + i)}^{- 2 n + 1} {(i λ)}^{n - 1 - k} e^{i λ z}$ $R e s (φ, i) = \frac{1}{(n - 1)!} \sum_{k = 0}^{n - 1} C_{n - 1}^{k} {(- 1)}^{k} n (n + 1) . . . (2 n - 1) {(2 i)}^{- 2 n + 1} {(i λ)}^{n - 1 - k} e^{- λ}$
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