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


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Question Number 35055 by math khazana by abdo last updated on 14/May/18

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{{(1 + x + x^{2})}^{3}}$
Commented by math khazana by abdo last updated on 15/May/18
$let put I = ∫_(−∞) ^(+∞) (dx/((1+x+x^2 )^3 )) I =∫_(−∞) ^(+∞) (dx/({(x+(1/2))^2 +(3/4)}^3 ))? changement x+(1/2) =((√3)/2)tant gibe I = ∫_(−(π/3)) ^(π/2) (1/(((3/4))^3 (1+tan^2 t)^3 )) ((√3)/2)( 1+tan^2 t)dt =(4^3 /3^3 ) ((√3)/2) ∫_(−(π/2)) ^(π/2) (dt/((1+tan^2 t)^2 )) =(4^3 /3^3 ) (√3) ∫_0 ^(π/2) cos^4 tdt = (4^3 /3^3 ) (√3) ∫_0 ^(π/2) (((1+cos(2t))^2 )/4)dt = ((16)/(27)) (√3) ∫_0 ^(π/2) (1+2cos(2t) +((1+cos(4t))/2))dt = ((8π(√3))/(27)) +((4π(√3))/(27)) = ((12π(√3))/(3.9)) =((4π(√3))/9) I = ((4π(√3))/9) .$
$l e t p u t I = \int_{- \infty}^{+ \infty} \frac{d x}{{(1 + x + x^{2})}^{3}}$ $I = \int_{- \infty}^{+ \infty} \frac{d x}{{{(x + \frac{1}{2})}^{2} + \frac{3}{4}}^{3}} ? c h a n g e m e n t$ $x + \frac{1}{2} = \frac{\sqrt{3}}{2} t a n t g i b e$ $I = \int_{- \frac{π}{3}}^{\frac{π}{2}} \frac{1}{{(\frac{3}{4})}^{3} {(1 + {t a n}^{2} t)}^{3}} \frac{\sqrt{3}}{2} (1 + {t a n}^{2} t) d t$ $= \frac{4^{3}}{3^{3}} \frac{\sqrt{3}}{2} \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{d t}{{(1 + {t a n}^{2} t)}^{2}}$ $= \frac{4^{3}}{3^{3}} \sqrt{3} \int_{0}^{\frac{π}{2}} {c o s}^{4} t d t$ $= \frac{4^{3}}{3^{3}} \sqrt{3} \int_{0}^{\frac{π}{2}} \frac{{(1 + c o s (2 t))}^{2}}{4} d t$ $= \frac{16}{27} \sqrt{3} \int_{0}^{\frac{π}{2}} (1 + 2 c o s (2 t) + \frac{1 + c o s (4 t)}{2}) d t$ $= \frac{8 π \sqrt{3}}{27} + \frac{4 π \sqrt{3}}{27} = \frac{12 π \sqrt{3}}{3.9} = \frac{4 π \sqrt{3}}{9}$ $I = \frac{4 π \sqrt{3}}{9} .$
Commented by abdo imad last updated on 17/May/18
$let introduce the complex function ϕ(z)= (1/((1+z+z^2 )^3 )) the[poles of ϕ are j=e^(i((2π)/3)) and j^− =e^(−i((2π)/3)) (triple poles ϕ(z)= (1/((z−j)^3 (z−j^− )^3 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,j) but Res(ϕ,j) =lim_(z→j) (1/((3−1)!)){ (z−j)^3 ϕ(z)}^((2)) =lim_(z→j) (1/2) {(z−j^− )^(−3) }^((2)) but we have {(z−j^− )^(−3) }^((1)) =−3(z−j^− )^(−4) {(z−j^− )}^((2)) = 12(z−j^− )^(−5) Res(ϕ,j) =lim_(z→j) 6(z−j^− )^(−5) =6(j−j^− )^(−5) =6(2i ((√3)/2))^(−5) =(6/((i(√3))^5 )) = (6/(i 9(√3))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (6/(i9(√3))) = ((2π .3.2)/(3.3(√3))) = ((4π)/(3(√3))) ⇒ ★∫_(−∞) ^(+∞) (dx/((1+x+x^2 )^3 )) = ((4π(√3))/9) ★$
$l e t i n t r o d u c e t h e c o m p l e x f u n c t i o n φ (z) = \frac{1}{{(1 + z + z^{2})}^{3}}$ $t h e [p o l e s o f φ a r e j = e^{i \frac{2 π}{3}} a n d \bar{j} = e^{- i \frac{2 π}{3}} (t r i p l e p o l e s$ $φ (z) = \frac{1}{{(z - j)}^{3} {(z - \bar{j})}^{3}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, j) b u t$ $R e s (φ, j) = {l i m}_{z \to j} \frac{1}{(3 - 1)!} {{(z - j)}^{3} φ (z)}^{(2)}$ $= {l i m}_{z \to j} \frac{1}{2} {{(z - \bar{j})}^{- 3}}^{(2)} b u t w e h a v e$ ${{(z - \bar{j})}^{- 3}}^{(1)} = - 3 {(z - \bar{j})}^{- 4}$ ${(z - \bar{j})}^{(2)} = 12 {(z - \bar{j})}^{- 5}$ $R e s (φ, j) = {l i m}_{z \to j} 6 {(z - \bar{j})}^{- 5} = 6 {(j - \bar{j})}^{- 5}$ $= 6 {(2 i \frac{\sqrt{3}}{2})}^{- 5} = \frac{6}{{(i \sqrt{3})}^{5}} = \frac{6}{i 9 \sqrt{3}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{6}{i 9 \sqrt{3}} = \frac{2 π . 3.2}{3.3 \sqrt{3}} = \frac{4 π}{3 \sqrt{3}} \Rightarrow$ $★ \int_{- \infty}^{+ \infty} \frac{d x}{{(1 + x + x^{2})}^{3}} = \frac{4 π \sqrt{3}}{9} ★$
	Answered by MJS last updated on 15/May/18

	$\int \frac{d x}{{(x^{2} + x + 1)}^{3}} = \int \frac{d x}{{({(x + \frac{1}{2})}^{2} + \frac{3}{4})}^{3}} =$ $= 64 \int \frac{d x}{{({(2 x + 1)}^{2} + + 3)}^{3}} =$ $[u = 2 x + 1 \to d x = \frac{d u}{2}]$ $= 32 \int \frac{d u}{{(u^{2} + 3)}^{3}}$ $[\int \frac{d u}{{({a u}^{2} + b)}^{n}} = \frac{u}{2 b (n - 1) {({a u}^{2} + b)}^{n - 1}} + \frac{2 n - 3}{2 b (n - 1)} \int \frac{d u}{{({a u}^{2} + b)}^{n - 1}}]$ $a = 1; b = 3; n = 3$ $= 32 (\frac{u}{12 {(u^{2} + 3)}^{2}} + \frac{1}{4} \int \frac{d u}{{(u^{2} + 3)}^{2}}) =$ $a = 1; b = 3; n = 2$ $= 32 (\frac{u}{12 {(u^{2} + 3)}^{2}} + \frac{1}{4} (\frac{u}{6 (u^{2} + 3)} + \frac{1}{6} \int \frac{d u}{u^{2} + 3})) =$ $[v = \frac{u \sqrt{3}}{3} \to d u = \sqrt{3} d v]$ $= 32 (\frac{u}{12 {(u^{2} + 3)}^{2}} + \frac{1}{4} (\frac{u}{6 (u^{2} + 3)} + \frac{\sqrt{3}}{18} \int \frac{d v}{v^{2} + 1})) =$ $= 32 (\frac{u}{12 {(u^{2} + 3)}^{2}} + \frac{1}{4} (\frac{u}{6 (u^{2} + 3)} + \frac{\sqrt{3}}{18} \arctan (v))) =$ $= 32 (\frac{u}{12 {(u^{2} + 3)}^{2}} + \frac{1}{4} (\frac{u}{6 (u^{2} + 3)} + \frac{\sqrt{3}}{18} \arctan (\frac{u \sqrt{3}}{3}))) =$ $= \frac{8 u}{3 {(u^{2} + 3)}^{2}} + \frac{4 u}{3 (u^{2} + 3)} + \frac{4 \sqrt{3}}{9} \arctan (\frac{u \sqrt{3}}{3}) =$ $= \frac{4 u (u^{2} + 5)}{3 {(u^{2} + 3)}^{2}} + \frac{4 \sqrt{3}}{9} \arctan (\frac{u \sqrt{3}}{3}) =$ $= \frac{4 (2 x + 1) ({(2 x + 1)}^{2} + 5)}{3 {({(2 x + 1)}^{2} + 3)}^{2}} + \frac{4 \sqrt{3}}{9} \arctan (\frac{\sqrt{3}}{3} (2 x + 1)) =$ $= \frac{(2 x + 1) (2 x^{2} + 2 x + 3)}{6 {(x^{2} + x + 1)}^{2}} + \frac{4 \sqrt{3}}{9} \arctan (\frac{\sqrt{3}}{3} (2 x + 1)) + C$ $\underset{- \infty}{\int^{\infty}} \frac{d x}{{(x^{2} + x + 1)}^{3}} = \frac{4 \sqrt{3}}{9} π$
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