find A_λ =∫_0 ^∞ ((cos(λx))/((x^2 −x+1)^2 ))dx with λ>0 2)find the value of ∫


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Question Number 88930 by mathmax by abdo last updated on 13/Apr/20

$f i n d A_{λ} = \int_{0}^{\infty} \frac{c o s (λ x)}{{(x^{2} - x + 1)}^{2}} d x w i t h λ > 0$ $2) f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{c o s (3 x)}{{(x^{2} - x + 1)}^{2}} d x$
Commented bymathmax by abdo last updated on 14/Apr/20
$A_λ =∫_0 ^∞ ((cos(λx))/((x^2 −x+1)^2 ))dx ⇒2A_λ =∫_(−∞) ^(+∞) ((cos(λx))/((x^2 −x+1)^2 ))dx =Re(∫_(−∞) ^(+∞) (e^(iλx) /((x^2 −x+1)^2 ))dx) let ϕ(z) =(e^(iλz) /((z^2 −z +1)^2 )) poles of ϕ? z^2 −z +1 =→Δ =1−4 =−3 ⇒z_1 =((1+i(√3))/2) =e^((iπ)/3) z_2 =((1−i(√3))/2) =e^(−((iπ)/3)) ⇒ϕ(z) =(e^(iλz) /((z−e^((iπ)/3) )^2 (z−e^(−((iπ)/3)) )^2 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,e^((iπ)/3) ) Res(ϕ,e^((iπ)/3) ) =lim_(z→e^((iπ)/3) ) (1/((2−1)!)){(z−e^((iπ)/3) )^2 ϕ(z)}^((1)) =lim_(z→e^((iπ)/3) ) {(z−e^(−((iπ)/3)) )^(−2) e^(iλz) }^((1)) =lim_(z→e^((iπ)/3) ) {−2(z−e^(−((iπ)/3)) )^(−3) e^(iλz) +iλ(z−e^(−((iπ)/3)) )^(−2) e^(iλz) } ={−2 (2isin((π/3))^(−3) +iλ (2i sin((π/3)))^(−2) } e^(iλe^((iπ)/3) ) ={((−2)/((i(√3))^3 )) +((iλ)/((i(√3))^2 ))} e^(iλ((1/2)+((i(√3))/2))) ={ (2/(3i(√3))) −((iλ)/3)} e^(−((√3)/2)λ) {cos((λ/2))+isin((λ/2))} ={((−2i)/(3(√3)))−((iλ)/3)}{ cos((λ/2))+isin((λ/2))}e^(−((√3)/2)λ) =((2/(3(√3)))+(λ/3))(−icos((λ/2))+sin((λ/2))e^(−((√3)/2)λ) ⇒ 2A_λ =((2/(3(√3)))+(λ/3))sin((λ/2))e^(−((√3)/2)λ) ⇒ A_λ =((1/(3(√3))) +(λ/6))sin((λ/2))e^(−((√3)/2)λ)$
$A_{λ} = \int_{0}^{\infty} \frac{c o s (λ x)}{{(x^{2} - x + 1)}^{2}} d x \Rightarrow 2 A_{λ} = \int_{- \infty}^{+ \infty} \frac{c o s (λ x)}{{(x^{2} - x + 1)}^{2}} d x$ $= R e (\int_{- \infty}^{+ \infty} \frac{e^{i λ x}}{{(x^{2} - x + 1)}^{2}} d x) l e t φ (z) = \frac{e^{i λ z}}{{(z^{2} - z + 1)}^{2}} p o l e s o f φ ?$ $z^{2} - z + 1 =\to Δ = 1 - 4 = - 3 \Rightarrow z_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}}$ $z_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{3}} \Rightarrow φ (z) = \frac{e^{i λ z}}{{(z - e^{\frac{i π}{3}})}^{2} {(z - e^{- \frac{i π}{3}})}^{2}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, e^{\frac{i π}{3}})$ $R e s (φ, e^{\frac{i π}{3}}) = {l i m}_{z \to e^{\frac{i π}{3}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i π}{3}})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to e^{\frac{i π}{3}}} {{(z - e^{- \frac{i π}{3}})}^{- 2} e^{i λ z}}^{(1)}$ $= {l i m}_{z \to e^{\frac{i π}{3}}} {- 2 {(z - e^{- \frac{i π}{3}})}^{- 3} e^{i λ z} + i λ {(z - e^{- \frac{i π}{3}})}^{- 2} e^{i λ z}}$ $= {- 2 (2 i s i n {(\frac{π}{3})}^{- 3} + i λ {(2 i s i n (\frac{π}{3}))}^{- 2}} e^{i λ e^{\frac{i π}{3}}}$ $= {\frac{- 2}{{(i \sqrt{3})}^{3}} + \frac{i λ}{{(i \sqrt{3})}^{2}}} e^{i λ (\frac{1}{2} + \frac{i \sqrt{3}}{2})}$ $= {\frac{2}{3 i \sqrt{3}} - \frac{i λ}{3}} e^{- \frac{\sqrt{3}}{2} λ} {c o s (\frac{λ}{2}) + i s i n (\frac{λ}{2})}$ $= {\frac{- 2 i}{3 \sqrt{3}} - \frac{i λ}{3}} {c o s (\frac{λ}{2}) + i s i n (\frac{λ}{2})} e^{- \frac{\sqrt{3}}{2} λ}$ $= (\frac{2}{3 \sqrt{3}} + \frac{λ}{3}) (- i c o s (\frac{λ}{2}) + s i n (\frac{λ}{2}) e^{- \frac{\sqrt{3}}{2} λ} \Rightarrow$ $2 A_{λ} = (\frac{2}{3 \sqrt{3}} + \frac{λ}{3}) s i n (\frac{λ}{2}) e^{- \frac{\sqrt{3}}{2} λ} \Rightarrow$ $A_{λ} = (\frac{1}{3 \sqrt{3}} + \frac{λ}{6}) s i n (\frac{λ}{2}) e^{- \frac{\sqrt{3}}{2} λ}$
Commented bymathmax by abdo last updated on 14/Apr/20

$\int_{0}^{\infty} \frac{c o s (3 x)}{{(x^{2} - x + 1)}^{2}} d x = A_{3} = (\frac{1}{3 \sqrt{3}} + \frac{1}{2}) s i n (\frac{3}{2}) e^{- \frac{3 \sqrt{3}}{2}}$
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