calculate ∫_(−∞) ^(+∞) ((cos(3x))/((x^2 +x+1)^2 ))dx


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Question Number 162016 by mathmax by abdo last updated on 25/Dec/21

$calculate \int_{- \infty}^{+ \infty} \frac{\cos (3 x)}{{(x^{2} + x + 1)}^{2}} dx$
Commented by MJS_new last updated on 25/Dec/21

$I can solve the indefinite integral but {it}^{'} s a$ $long hard way . . .$
Commented by mathmax by abdo last updated on 25/Dec/21

$use residus theorem sir$
	Answered by Ar Brandon last updated on 24/Mar/22
	$Υ=∫_(−∞) ^(+∞) ((cos3x)/((x^2 +x+1)^2 ))dx=Re∫_(−∞) ^(+∞) (e^(3ix) /((x^2 +x+1)^2 ))dx Let f(x)=(e^(3ix) /((x^2 +x+1)^2 )). Poles of f(x): e^((2/3)iπ) and e^(−(2/3)iπ) , order-2 Υ=Re∫_(−∞) ^(+∞) f(x)dx=Re(2iπRes(f, e^((2/3)iπ) )) Res (f, e^((2/3)iπ) )=lim_(x→e^((2/3)iπ) ) {(x−e^((2/3)iπ) )^2 f(x)}^((1)) =lim_(x→e^((2/3)iπ) ) {(e^(3ix) /((x−e^(−(2/3)iπ) )^2 ))}^((1)) =lim_(x→e^((2/3)iπ) ) {((3ie^(3ix) (x−e^(−(2/3)iπ) )^2 −2e^(3ix) (x−e^(−(2/3)iπ) ))/((x−e^(−(2/3)iπ) )^4 ))} =−(1/9)(9ie^(3i(−(1/2)+i((√3)/2))) +2(√3)ie^(3i(−(1/2)+i((√3)/2))) )=−(1/9)(9+2(√3))e^((−((3(√3))/2)−i((3−π)/2))) Υ=−((2π)/9)(9+2(√3))e^(−((3(√3))/2)) sin(((3−π)/2))=((2π)/9)(9+2(√3))e^(−((3(√3))/2)) cos((3/2))$
	$Υ = \int_{- \infty}^{+ \infty} \frac{\cos 3 x}{{(x^{2} + x + 1)}^{2}} d x = R e \int_{- \infty}^{+ \infty} \frac{e^{3 i x}}{{(x^{2} + x + 1)}^{2}} d x$ $Let f (x) = \frac{e^{3 i x}}{{(x^{2} + x + 1)}^{2}} . Poles of f (x) : e^{\frac{2}{3} i π} and e^{- \frac{2}{3} i π}, order - 2$ $Υ = R e \int_{- \infty}^{+ \infty} f (x) d x = R e (2 i π R e s (f, e^{\frac{2}{3} i π}))$ $R e s (f, e^{\frac{2}{3} i π}) = \lim_{x \to e^{\frac{2}{3} i π}} {{(x - e^{\frac{2}{3} i π})}^{2} f (x)}^{(1)} = \lim_{x \to e^{\frac{2}{3} i π}} {\frac{e^{3 i x}}{{(x - e^{- \frac{2}{3} i π})}^{2}}}^{(1)}$ $= \lim_{x \to e^{\frac{2}{3} i π}} {\frac{3 {i e}^{3 i x} {(x - e^{- \frac{2}{3} i π})}^{2} - 2 e^{3 i x} (x - e^{- \frac{2}{3} i π})}{{(x - e^{- \frac{2}{3} i π})}^{4}}}$ $= - \frac{1}{9} (9 {i e}^{3 i (- \frac{1}{2} + i \frac{\sqrt{3}}{2})} + 2 \sqrt{3} {i e}^{3 i (- \frac{1}{2} + i \frac{\sqrt{3}}{2})}) = - \frac{1}{9} (9 + 2 \sqrt{3}) e^{(- \frac{3 \sqrt{3}}{2} - i \frac{3 - π}{2})}$ $Υ = - \frac{2 π}{9} (9 + 2 \sqrt{3}) e^{- \frac{3 \sqrt{3}}{2}} \sin (\frac{3 - π}{2}) = \frac{2 π}{9} (9 + 2 \sqrt{3}) e^{- \frac{3 \sqrt{3}}{2}} \cos (\frac{3}{2})$
	Commented by Mathspace last updated on 26/Dec/21

	$e r r o r o f c a l c u l u s . .$
	Commented by Ar Brandon last updated on 26/Dec/21

	$Thank you for checking, Sir . Rectified!$
	Answered by Mathspace last updated on 26/Dec/21
	$Ψ=∫_(−∞) ^(+∞ ) ((cos(3x))/((x^2 +x+1)^2 ))dx ⇒ Ψ=Re(∫_(−∞) ^(+∞) (e^(3ix) /((x^2 +x+1)^2 ))dx) let ϕ(z)=(e^(3iz) /((z^2 +z+1)^2 )) poles of ϕ? z^2 +z+1=0→Δ=1−4=−3 ⇒ z_1 =((−1+i(√3))/2)=e^((i2π)/3) z_2 =((−1−i(√3))/2)=e^(−((i2π)/3)) the poles are z_i (with ordre=2) ϕ(z)=(e^(3iz) /((z−z_1 )^2 (z−z_2 )^2 )) ∫_(−∞) ^(+∞) ϕ(z)dz=2iπRes(ϕ,z_1 ) Res(ϕ,z_1 )=lim_(z→z_1 ) (1/((2−1)!)){(z−z_1 )^2 ϕ(z)}^((1)) =lim_(z→z_1 ) {(e^(3iz) /((z−z_2 )^2 ))}^((1)) =lim_(z→z1) ((3ie^(3iz) (z−z_2 )^2 −2(z−z_2 )e^(3iz) )/((z−z_2 )^4 )) =lim_(z→z_1 ) (({3i(z−z_2 )−2}e^(3iz) )/((z−z_2 )^3 )) =(({3i(z_1 −z_2 )−2}e^(3iz_1 ) )/((z_1 −z_2 )^3 )) =(({3i(i(√3))−2}e^(3i(((−1+i(√3))/2))) )/((i(√3))^3 )) =(({−3(√3)−2}e^(−((3(√3))/2)) {cos((3/2))−isin((3/2))})/(−3i(√3))) =(({3(√3)+2}e^(−((3(√3))/2)) {cos((3/2))−isin((3/2))})/(3i(√3))) ∫_(−∞) ^(+∞) ϕ(z)dz=((2iπ)/(3i(√3))){3(√3)+2}(...) ⇒Ψ=((2π)/(3(√3)))(3(√3)+2)e^(−((3(√3))/2)) cos((3/2))$
	$Ψ = \int_{- \infty}^{+ \infty} \frac{c o s (3 x)}{{(x^{2} + x + 1)}^{2}} d x \Rightarrow$ $Ψ = R e (\int_{- \infty}^{+ \infty} \frac{e^{3 i x}}{{(x^{2} + x + 1)}^{2}} d x)$ $l e t φ (z) = \frac{e^{3 i z}}{{(z^{2} + z + 1)}^{2}} p o l e s o f φ ?$ $z^{2} + z + 1 = 0 \to Δ = 1 - 4 = - 3 \Rightarrow$ $z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}}$ $z_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}}$ $t h e p o l e s a r e z_{i} (w i t h o r d r e = 2)$ $φ (z) = \frac{e^{3 i z}}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{1})$ $R e s (φ, z_{1}) = {l i m}_{z \to z_{1}} \frac{1}{(2 - 1)!} {{(z - z_{1})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to z_{1}} {\frac{e^{3 i z}}{{(z - z_{2})}^{2}}}^{(1)}$ $= {l i m}_{z \to z 1} \frac{3 {i e}^{3 i z} {(z - z_{2})}^{2} - 2 (z - z_{2}) e^{3 i z}}{{(z - z_{2})}^{4}}$ $= {l i m}_{z \to z_{1}} \frac{{3 i (z - z_{2}) - 2} e^{3 i z}}{{(z - z_{2})}^{3}}$ $= \frac{{3 i (z_{1} - z_{2}) - 2} e^{3 {i z}_{1}}}{{(z_{1} - z_{2})}^{3}}$ $= \frac{{3 i (i \sqrt{3}) - 2} e^{3 i (\frac{- 1 + i \sqrt{3}}{2})}}{{(i \sqrt{3})}^{3}}$ $= \frac{{- 3 \sqrt{3} - 2} e^{- \frac{3 \sqrt{3}}{2}} {c o s (\frac{3}{2}) - i s i n (\frac{3}{2})}}{- 3 i \sqrt{3}}$ $= \frac{{3 \sqrt{3} + 2} e^{- \frac{3 \sqrt{3}}{2}} {c o s (\frac{3}{2}) - i s i n (\frac{3}{2})}}{3 i \sqrt{3}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{3 i \sqrt{3}} {3 \sqrt{3} + 2} (. . .)$ $\Rightarrow Ψ = \frac{2 π}{3 \sqrt{3}} (3 \sqrt{3} + 2) e^{- \frac{3 \sqrt{3}}{2}} c o s (\frac{3}{2})$
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