let I =∫_(−∞) ^(+∞) ((t+1)/((t^2 −t+1)^2 ))dt find value of I .


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Question Number 53113 by maxmathsup by imad last updated on 17/Jan/19

$l e t I = \int_{- \infty}^{+ \infty} \frac{t + 1}{{(t^{2} - t + 1)}^{2}} d t$ $f i n d v a l u e o f I .$
Commented by maxmathsup by imad last updated on 18/Jan/19
$we have t^2 −t+1 =(t−(1/2))^2 +(3/4) so changement t−(1/2) =((√3)/2) x give I =∫_(−∞) ^(+∞) (((1/2)+((√3)/2)x +1)/(((3/4))^2 (x^2 +1)^2 )) ((√3)/2) dx =((16)/9) ((√3)/4) ∫_(−∞) ^(+∞) ((3+(√3)x)/((x^2 +1)^2 ))dx =((4(√3))/9) ∫_(−∞) ^(+∞) ((3+(√3)x)/((x^2 +1)^2 ))dx let ϕ(z) =((3 +(√3)z)/((z^2 +1)^2 )) we have ϕ(z) =((3+(√3)z)/((z−i)^2 (z+i)^2 )) so the ples of ϕ are +^− i(doubles) and residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { ((3+(√3)z)/((z+i)^2 ))}^((1)) =lim_(z→i) (((√3)(z+i)^2 −2(z+i)(3+(√3)z))/((z+i)^4 )) =lim_(z→i) (((√3)(z+i)−2(3+(√3)z))/((z+i)^3 )) =((2(√3)i−2(3+(√3)i))/(−8i)) =((−6)/(−8i)) =(3/(4i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (3/(4i)) =((3π)/2) ⇒ I =((4(√3))/9) ((3π)/2) =((2π(√3))/3) .$
$w e h a v e t^{2} - t + 1 = {(t - \frac{1}{2})}^{2} + \frac{3}{4} s o c h a n g e m e n t t - \frac{1}{2} = \frac{\sqrt{3}}{2} x g i v e$ $I = \int_{- \infty}^{+ \infty} \frac{\frac{1}{2} + \frac{\sqrt{3}}{2} x + 1}{{(\frac{3}{4})}^{2} {(x^{2} + 1)}^{2}} \frac{\sqrt{3}}{2} d x = \frac{16}{9} \frac{\sqrt{3}}{4} \int_{- \infty}^{+ \infty} \frac{3 + \sqrt{3} x}{{(x^{2} + 1)}^{2}} d x$ $= \frac{4 \sqrt{3}}{9} \int_{- \infty}^{+ \infty} \frac{3 + \sqrt{3} x}{{(x^{2} + 1)}^{2}} d x l e t φ (z) = \frac{3 + \sqrt{3} z}{{(z^{2} + 1)}^{2}} w e h a v e$ $φ (z) = \frac{3 + \sqrt{3} z}{{(z - i)}^{2} {(z + i)}^{2}} s o t h e p l e s o f φ a r e \bar{+} i (d o u b l e s) a n d r e s i d u s t h e o r e m$ $g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)} = {l i m}_{z \to i} {\frac{3 + \sqrt{3} z}{{(z + i)}^{2}}}^{(1)}$ $= {l i m}_{z \to i} \frac{\sqrt{3} {(z + i)}^{2} - 2 (z + i) (3 + \sqrt{3} z)}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{\sqrt{3} (z + i) - 2 (3 + \sqrt{3} z)}{{(z + i)}^{3}} = \frac{2 \sqrt{3} i - 2 (3 + \sqrt{3} i)}{- 8 i} = \frac{- 6}{- 8 i} = \frac{3}{4 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{3}{4 i} = \frac{3 π}{2} \Rightarrow I = \frac{4 \sqrt{3}}{9} \frac{3 π}{2} = \frac{2 π \sqrt{3}}{3} .$
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