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Question Number 109373 by mathdave last updated on 23/Aug/20

Commented by mnjuly1970 last updated on 23/Aug/20

$m u r r a y s p i e g e l . a d v a n c e d$ $c a l c u l u s .$
Commented by PRITHWISH SEN 2 last updated on 23/Aug/20

$you want it ?$
Commented by bobhans last updated on 23/Aug/20
do you have the pdf file?
Commented by bobhans last updated on 23/Aug/20

$y e s$
Commented by PRITHWISH SEN 2 last updated on 23/Aug/20

$give your email address$
Commented by bobhans last updated on 23/Aug/20

$b o y h o n s o m e 70 @ g m a i l . c o m$
Commented by PRITHWISH SEN 2 last updated on 23/Aug/20

$please check your mail and reply if you get it .$
Commented by bobhans last updated on 23/Aug/20

$y e s . t h a n k y o u v e r y m u c h$
Commented by mathdave last updated on 23/Aug/20

$p l s i r e a l l y n e e d t h a d a d v a n c e c a l c u l u s p d f h e l p$ $m e s e n d i t t h r o u g h t h i s m y e m a i l$
Commented by mathdave last updated on 23/Aug/20

$m o n d a y m a t h e w 54 a t g m a i l . c o m$
Commented by PRITHWISH SEN 2 last updated on 26/Aug/20

$something is wrong . I {can}^{'} t send you . Is the$ $mail address given correct ?$
	Answered by 1549442205PVT last updated on 23/Aug/20
	$we find resider of the function f(z)=(z^2 /((z^2 −z+1)^3 )). For this,we need to solve ( z^2 −z+1)^3 =0⇔z^2 −z+1=0 Δ=1−4=3i^2 ⇒z=((1±3i)/2)=e^(±((πi)/3)) Thus,a_1 =e^((πi)/3) and a_2 =e^((−πi)/3) are the poles of degree 3 of f(z).Next,we calculate the residers of f(z) at this poles Res(f(z),e^((πi)/3) )=(1/(2!))lim_(z→e^((πi)/3) ) (d^2 /dz^2 ){(z−e^((πi)/3) ).(z^2 /((z−e^((πi)/3) )^3 (z−e^(−((πi)/3)) )^3 ))} =lim_(z→a_1 ) {(d^2 /dz^2 )[(z^2 /((z−e^((−πi)/3) )^3 ))]}= we have (d/dz)[(z^2 /((z−a)^3 ))]=((2z.(z−a)^3 −3z^2 (z−a)^2 )/z^6 ) ((2z(z−a)−3z^2 )/((z−a)^4 ))=((−z^2 −2az)/((z−a)^4 )).Hence, (d^2 /z^2 )[(z^2 /((z−e^((−πi)/3) )^3 ))]=(((−2z−2a)(z−a)^4 +4(z−a)^3 (z^2 +2az))/((z−a)^8 )) =((−2(z+a)(z−a)+4z^2 +8az)/((z−a)^5 ))=((2z^2 +2a^2 +8az)/((z−a)^5 )) =⇒Res(f(z),e^((πi)/3) )=(1/2).((2.e^((2πi)/3) +2.e^((−2πi)/3) +8e^((πi)/3) .e^((−πi)/3) )/((e^((πi)/3) −e^((−πi)/3) )^5 )) =(1/2).((−1+(−1)+8)/((i(√3))^5 ))=(3/(9i(√3)))=((−i)/(3(√3))) Similarly, Res(f(z),e^((−πi)/3) )=(1/2)lim_(z→e^((−πi)/3) ) {(d^2 /dz^2 )[(z−e^(−((πi)/3)) ).(z^2 /((z−e^((πi)/3) )^3 (z−e^(−((πi)/3)) )^3 ))]} =(1/2).((2.e^((−2πi)/3) +2.e^((2πi)/3) +8e^((−πi)/3) .e^((πi)/3) )/((e^((−πi)/3) −e^((πi)/3) )^5 ))= =(1/2).(((−1)+(−1)+8)/((−i(√3))^5 ))=(3/(9i(√3)))=((−i)/(3(√3))) Σ_a_k Res(f(z))=((−2i)/( (√3))).Therefore, ∫_(−∞) ^(+∞) f(x)dx=2πiΣRes(f(z),a_k ) =2πi×((−2i)/(3(√3)))=((4π)/(3(√3)))$
	$we find resider of the function$ $f (z) = \frac{z^{2}}{{(z^{2} - z + 1)}^{3}} . For this, we need to$ $solve {(z^{2} - z + 1)}^{3} = 0 \Leftrightarrow z^{2} - z + 1 = 0$ $Δ = 1 - 4 = {3 i}^{2} \Rightarrow z = \frac{1 \pm 3 i}{2} = e^{\pm \frac{π i}{3}}$ $Thus, a_{1} = e^{\frac{π i}{3}} and a_{2} = e^{\frac{- π i}{3}} are the poles$ $of degree 3 of f (z) . Next, we calculate the residers of f (z) at this poles$ $Res (f (z), e^{\frac{π i}{3}}) = \frac{1}{2!} \lim_{z \to e^{\frac{π i}{3}}} \frac{d^{2}}{{dz}^{2}} {(z - e^{\frac{π i}{3}}) . \frac{z^{2}}{{(z - e^{\frac{π i}{3}})}^{3} {(z - e^{- \frac{π i}{3}})}^{3}}}$ $= \lim_{z \to a_{1}} {\frac{d^{2}}{{dz}^{2}} [\frac{z^{2}}{{(z - e^{\frac{- π i}{3}})}^{3}}]} =$ $we have \frac{d}{dz} [\frac{z^{2}}{{(z - a)}^{3}}] = \frac{2 z . {(z - a)}^{3} - {3 z}^{2} {(z - a)}^{2}}{z^{6}}$ $\frac{2 z (z - a) - {3 z}^{2}}{{(z - a)}^{4}} = \frac{- z^{2} - 2 az}{{(z - a)}^{4}} . Hence,$ $\frac{d^{2}}{z^{2}} [\frac{z^{2}}{{(z - e^{\frac{- π i}{3}})}^{3}}] = \frac{(- 2 z - 2 a) {(z - a)}^{4} + 4 {(z - a)}^{3} (z^{2} + 2 az)}{{(z - a)}^{8}}$ $= \frac{- 2 (z + a) (z - a) + {4 z}^{2} + 8 az}{{(z - a)}^{5}} = \frac{{2 z}^{2} + {2 a}^{2} + 8 az}{{(z - a)}^{5}}$ $=\Rightarrow Res (f (z), e^{\frac{π i}{3}}) = \frac{1}{2} . \frac{2 . e^{\frac{2 π i}{3}} + 2 . e^{\frac{- 2 π i}{3}} + {8 e}^{\frac{π i}{3}} . e^{\frac{- π i}{3}}}{{(e^{\frac{π i}{3}} - e^{\frac{- π i}{3}})}^{5}}$ $= \frac{1}{2} . \frac{- 1 + (- 1) + 8}{{(i \sqrt{3})}^{5}} = \frac{3}{9 i \sqrt{3}} = \frac{- i}{3 \sqrt{3}}$ $Similarly,$ $Res (f (z), e^{\frac{- π i}{3}}) = \frac{1}{2} \lim_{z \to e^{\frac{- π i}{3}}} {\frac{d^{2}}{{dz}^{2}} [(z - e^{- \frac{π i}{3}}) . \frac{z^{2}}{{(z - e^{\frac{π i}{3}})}^{3} {(z - e^{- \frac{π i}{3}})}^{3}}]}$ $= \frac{1}{2} . \frac{2 . e^{\frac{- 2 π i}{3}} + 2 . e^{\frac{2 π i}{3}} + {8 e}^{\frac{- π i}{3}} . e^{\frac{π i}{3}}}{{(e^{\frac{- π i}{3}} - e^{\frac{π i}{3}})}^{5}} =$ $= \frac{1}{2} . \frac{(- 1) + (- 1) + 8}{{(- i \sqrt{3})}^{5}} = \frac{3}{9 i \sqrt{3}} = \frac{- i}{3 \sqrt{3}}$ $\sum_{a_{k}} Res (f (z)) = \frac{- 2 i}{\sqrt{3}} . Therefore,$ $\int_{- \infty}^{+ \infty} f (x) dx = 2 π i Σ Res (f (z), a_{k})$ $= 2 π i \times \frac{- 2 i}{3 \sqrt{3}} = \frac{4 π}{3 \sqrt{3}}$
	Commented by mathdave last updated on 23/Aug/20

	$r e c h e c k u r w i r k i n g w e l l$
	Commented by 1549442205PVT last updated on 24/Aug/20

	$Thsnk you, sir . You are welcome$
	Answered by mathmax by abdo last updated on 23/Aug/20
	$I =∫_(−∞) ^(+∞) (x^2 /((x^2 −x+1)^3 ))dx we hsve x^2 −x +1 =(x−(1/2))^2 +(3/4) we do the changement x−(1/2) =((√3)/2)t ⇒ I =∫_(−∞) ^(+∞) ((1(1+(√3)t)^2 )/(4((3/4))^2 (t^2 +1)^3 )) ×((√3)/2) dt =((2(√3))/9)∫_(−∞) ^(+∞) (((1+(√3)z)^2 )/((t^2 +1)^3 )) dt let ϕ(z) =(((1+(√3)z)^2 )/((z^2 +1)^3 )) ⇒ ϕ(z) =(((1+(√3)z)^2 )/((z−i)^3 (z+i)^3 )) residus theorem ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((3−1)!)){(z−i)^3 ϕ(z)}^((2)) =(1/2)lim_(z→i) {(((1+(√3)z)^2 )/((z+i)^3 ))}^((2)) =(1/2)lim_(z→i) { ((2(√3)(1+(√3)z)(z+i)^3 −3(z+i)^2 (1+(√3)z)^2 )/((z+i)^6 ))}^((1)) =(1/2)lim_(z→i) {((2(√3)(1+(√3)z)(z+i)−3(1+(√3)z)^2 )/((z+i)^4 ))}^((1)) =(1/2)lim_(z→i) (((6(z+i)+2(√3)(1+(√3)z)−6(√3)(1+(√3)z))(z+i)^4 −4(z+i)^3 (2(√3)(1+(√3)z)(z+i)−3(1+(√3)z)^2 )/((z+i)^8 )) rest to finish the calculus....$
	$I = \int_{- \infty}^{+ \infty} \frac{x^{2}}{{(x^{2} - x + 1)}^{3}} dx we hsve x^{2} - x + 1 = {(x - \frac{1}{2})}^{2} + \frac{3}{4}$ $we do the changement x - \frac{1}{2} = \frac{\sqrt{3}}{2} t \Rightarrow$ $I = \int_{- \infty}^{+ \infty} \frac{1 {(1 + \sqrt{3} t)}^{2}}{4 {(\frac{3}{4})}^{2} {(t^{2} + 1)}^{3}} \times \frac{\sqrt{3}}{2} dt$ $= \frac{2 \sqrt{3}}{9} \int_{- \infty}^{+ \infty} \frac{{(1 + \sqrt{3} z)}^{2}}{{(t^{2} + 1)}^{3}} dt let φ (z) = \frac{{(1 + \sqrt{3} z)}^{2}}{{(z^{2} + 1)}^{3}} \Rightarrow$ $φ (z) = \frac{{(1 + \sqrt{3} z)}^{2}}{{(z - i)}^{3} {(z + i)}^{3}} residus theorem \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π Res (φ, i)$ $Res (φ, i) = \lim_{z \to i} \frac{1}{(3 - 1)!} {{(z - i)}^{3} φ (z)}^{(2)}$ $= \frac{1}{2} \lim_{z \to i} {\frac{{(1 + \sqrt{3} z)}^{2}}{{(z + i)}^{3}}}^{(2)}$ $= \frac{1}{2} \lim_{z \to i} {\frac{2 \sqrt{3} (1 + \sqrt{3} z) {(z + i)}^{3} - 3 {(z + i)}^{2} {(1 + \sqrt{3} z)}^{2}}{{(z + i)}^{6}}}^{(1)}$ $= \frac{1}{2} \lim_{z \to i} {\frac{2 \sqrt{3} (1 + \sqrt{3} z) (z + i) - 3 {(1 + \sqrt{3} z)}^{2}}{{(z + i)}^{4}}}^{(1)}$ $= \frac{1}{2} \lim_{z \to i} \frac{(6 (z + i) + 2 \sqrt{3} (1 + \sqrt{3} z) - 6 \sqrt{3} (1 + \sqrt{3} z)) {(z + i)}^{4} - 4 {(z + i)}^{3} (2 \sqrt{3} (1 + \sqrt{3} z) (z + i) - 3 {(1 + \sqrt{3} z)}^{2}}{{(z + i)}^{8}}$ $rest to finish the calculus . . . .$
	Answered by Sarah85 last updated on 24/Aug/20

	$remember {Ostrogradski}^{'} s Method introduced$ $by former member MJS$ $\Rightarrow$ $\int \frac{x^{2}}{{(x^{2} - x + 1)}^{3}} d x = \frac{2 x^{3} - 3 x^{2} + 2 x - 2}{{(x^{2} - x + 1)}^{2}} + \frac{1}{3} \int \frac{d x}{x^{2} - x + 1} =$ $= \frac{2 x^{3} - 3 x^{2} + 2 x - 2}{{(x^{2} - x + 1)}^{2}} + \frac{2 \sqrt{3}}{9} \arctan \frac{(2 x - 1) \sqrt{3}}{3} + C$ $\Rightarrow$ $\underset{- \infty}{\int^{+ \infty}} \frac{x^{2}}{{(x^{2} - x + 1)}^{3}} d x = {[\frac{2 \sqrt{3}}{9} \arctan \frac{(2 x - 1) \sqrt{3}}{3}]}_{- \infty}^{+ \infty} =$ $= \frac{2 π \sqrt{3}}{9}$
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