find the value of I =∫_(−∞) ^(+∞) ((cos(αt))/((x^2 +x +1)^2 ))dx α from R. 2)calculate ∫


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Question Number 44309 by abdo.msup.com last updated on 26/Sep/18

$f i n d t h e v a l u e o f$ $I = \int_{- \infty}^{+ \infty} \frac{c o s (α t)}{{(x^{2} + x + 1)}^{2}} d x$ $α f r o m R .$ $2) c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + x + 1)}^{2}}$
Commented by maxmathsup by imad last updated on 28/Sep/18
$1) we have I =∫_(−∞) ^(+∞) ((cos(αx))/({(x+(1/2))^2 +(3/4)}^2 ))dx =_(x+(1/2)=((√3)/2) t) ∫_(−∞) ^(+∞) ((cos(α(((√3)/2)t−(1/2))))/(((3/4))^2 {t^2 +1}^2 ))((√3)/2) dt =((16)/9) ((√3)/2) ∫_(−∞) ^(+∞) ((cos{α(((t(√3)−1)/2))})/((t^2 +1)^2 ))dt =((8(√3))/9) Re( ∫_(−∞) ^(+∞) (e^(iα(((t(√3)−1)/2))) /((t^2 +1)^2 ))dt) let ϕ(z) = (e^(iα(((z(√3)−1)/2))) /((z^2 +1)^2 )) the poles f ϕ are i and −i(doubles) so ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ ,i) =lim_(z→i) (1/((2−1)!)){ (z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { (e^(iα(((z(√3)−1)/2))) /((z+i)^2 ))}^((1)) =lim_(z→i) e^(−((iα)/2)) { (e^(i((√3)/2)α z) /((z+i)^2 ))}^((1)) =lim_(z→i) e^(−((iα)/2)) ((i((√3)/2)α e^(i((√3)/2)αz) (z+i)^2 −2(z+i) e^(i((√3)/2)αz) )/((z+i)^4 )) =e^(−((iα)/2)) lim_(z→i) ((i((√3)/2)(z+i)e^(i((√3)/2)αz) −2 e^(i((√3)/2)αz) )/((z+i)^3 )) =e^(−((iα)/2)) ((−(√3)e^(−((α(√3))/2)) − 2 e^(−((α(√3))/2)) )/(−8i)) =((e^(−((iα)/2)) (2+(√3))e^(−((α(√3))/2)) )/(8i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2i π (2+(√3)) e^(−((iα)/2)) (e^(−((α(√3))/2)) /(8i)) =(π/4)(2+(√3)) e^(−((iα)/2)) e^(−((α(√3))/2)) = (π/4) (2+(√3)) e^(−((α(√3))/2)) (cos((α/2))−i sin((α/2))) ⇒ I =((8(√3))/9) (π/4)(2+(√3)) e^(−((α(√3))/2)) cos((α/2)) ⇒I =((2π (√3))/9)(2+(√3)) e^(−((α(√3))/2)) cos((α/2)). π$
$1) w e h a v e I = \int_{- \infty}^{+ \infty} \frac{c o s (α x)}{{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}^{2}} d x =_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t} \int_{- \infty}^{+ \infty} \frac{c o s (α (\frac{\sqrt{3}}{2} t - \frac{1}{2}))}{{(\frac{3}{4})}^{2} {t^{2} + 1}^{2}} \frac{\sqrt{3}}{2} d t$ $= \frac{16}{9} \frac{\sqrt{3}}{2} \int_{- \infty}^{+ \infty} \frac{c o s {α (\frac{t \sqrt{3} - 1}{2})}}{{(t^{2} + 1)}^{2}} d t = \frac{8 \sqrt{3}}{9} R e (\int_{- \infty}^{+ \infty} \frac{e^{i α (\frac{t \sqrt{3} - 1}{2})}}{{(t^{2} + 1)}^{2}} d t)$ $l e t φ (z) = \frac{e^{i α (\frac{z \sqrt{3} - 1}{2})}}{{(z^{2} + 1)}^{2}} t h e p o l e s f φ a r e i a n d - i (d o u b l e s) s o$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{e^{i α (\frac{z \sqrt{3} - 1}{2})}}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} e^{- \frac{i α}{2}} {\frac{e^{i \frac{\sqrt{3}}{2} α z}}{{(z + i)}^{2}}}^{(1)}$ $= {l i m}_{z \to i} e^{- \frac{i α}{2}} \frac{i \frac{\sqrt{3}}{2} α e^{i \frac{\sqrt{3}}{2} α z} {(z + i)}^{2} - 2 (z + i) e^{i \frac{\sqrt{3}}{2} α z}}{{(z + i)}^{4}}$ $= e^{- \frac{i α}{2}} {l i m}_{z \to i} \frac{i \frac{\sqrt{3}}{2} (z + i) e^{i \frac{\sqrt{3}}{2} α z} - 2 e^{i \frac{\sqrt{3}}{2} α z}}{{(z + i)}^{3}}$ $= e^{- \frac{i α}{2}} \frac{- \sqrt{3} e^{- \frac{α \sqrt{3}}{2}} - 2 e^{- \frac{α \sqrt{3}}{2}}}{- 8 i} = \frac{e^{- \frac{i α}{2}} (2 + \sqrt{3}) e^{- \frac{α \sqrt{3}}{2}}}{8 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (2 + \sqrt{3}) e^{- \frac{i α}{2}} \frac{e^{- \frac{α \sqrt{3}}{2}}}{8 i} = \frac{π}{4} (2 + \sqrt{3}) e^{- \frac{i α}{2}} e^{- \frac{α \sqrt{3}}{2}}$ $= \frac{π}{4} (2 + \sqrt{3}) e^{- \frac{α \sqrt{3}}{2}} (c o s (\frac{α}{2}) - i s i n (\frac{α}{2})) \Rightarrow$ $I = \frac{8 \sqrt{3}}{9} \frac{π}{4} (2 + \sqrt{3}) e^{- \frac{α \sqrt{3}}{2}} c o s (\frac{α}{2}) \Rightarrow I = \frac{2 π \sqrt{3}}{9} (2 + \sqrt{3}) e^{- \frac{α \sqrt{3}}{2}} c o s (\frac{α}{2}) .$ $π$
Commented by maxmathsup by imad last updated on 28/Sep/18

$2) l e t t a k e α = 0 \Rightarrow \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + x + 1)}^{2}} = \frac{2 π \sqrt{3}}{9} (2 + \sqrt{3}) .$
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