calculate ∫_1 ^∞ (dx/((x^2 +x+1)^9 ))


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Question Number 133537 by mathmax by abdo last updated on 22/Feb/21

$calculate \int_{1}^{\infty} \frac{dx}{{(x^{2} + x + 1)}^{9}}$
	Answered by Ñï= last updated on 23/Feb/21

	$\int_{1}^{\infty} \frac{d x}{{(1 + x + x^{2})}^{9}}$ $= \int_{1}^{\infty} \frac{d x}{{[{(x + \frac{1}{2})}^{2} + \frac{3}{4}]}^{9}}$ $= \sqrt{\frac{3}{4}} \int_{\frac{π}{3}}^{\frac{π}{2}} \frac{{s e c}^{2} θ d θ}{{(\frac{3}{4} {t a n}^{2} θ + \frac{3}{4})}^{9}}$ $= \frac{\sqrt{\frac{3}{4}}}{{(\frac{3}{4})}^{9}} \int_{π / 3}^{π / 2} {c o s}^{16} θ d θ$ $= {(\frac{3}{4})}^{- 17 / 2} R e (\int_{π / 3}^{π / 2} e^{16 i θ} d θ)$ $= {(\frac{3}{4})}^{- 17 / 2} R e (\frac{1}{16 i} (e^{8 i π} - e^{16 i π / 3})$ $= \frac{2^{12}}{3^{8}}$
	Answered by mathmax by abdo last updated on 24/Feb/21
	$Φ=∫_1 ^∞ (dx/((x^2 +x+1)^9 )) ⇒Φ =∫_1 ^(∞ ) (dx/(((x+(1/2))^2 +(3/4))^9 )) =_(x+(1/2)=((√3)/2)t) ∫_(√3) ^∞ ((4/3))^9 (1/((t^2 +1)^9 ))((√3)/2)dt =((√3)/2)((4/3))^9 ∫_(√3) ^∞ (dt/((t^2 +1)^9 )) wehave ∫_(√3) ^∞ (dt/((t^2 +1)^9 )) =∫_(√3) ^∞ (dt/((t−i)^9 (t+i)^9 ))=∫_(√3) ^∞ (dt/((((t−i)/(t+i)))^9 (t+i)^(18) )) changement ((t−i)/(t+i))=z give t−i=zt+iz ⇒(1−z)t=i(1+z) ⇒ t=i((1+iz)/(1−iz)) ⇒(dt/dz)=i((i(1−iz)−(1+iz)(−i))/((1−iz)^2 ))=i((i+z+i−z)/((1−iz)^2 ))=((−2)/((1−iz)^2 )) and t+i =((i−z)/(1−iz))+i =((i−z+i+z)/((1−iz)))=((2i)/((1−iz))) ⇒ Φ =∫_(((√3)−i)/( (√3)+i)) ^1 ((−2)/((1−iz)^2 z^9 (((2i)/(1−iz)))^(18) ))dz =(2/((2i)^(18) ))∫_1 ^(((√3)−i)/( (√3)+i)) (((1−iz)^(16) )/z^9 )dz =(2/((2i)^(18) ))∫_1 ^(((√3)−i)/( (√3)+i)) ((Σ_(k=0) ^(16) C_(16) ^k (iz)^k (−1)^(16−k) )/z^9 )dz =−(2/(2^(18) )) ∫_1 ^(((√3)−i)/( (√3) +i)) Σ_(k=0) ^(16) (−1)^k C_(16) ^k i^k z^(k−9) dz =−(1/2^(17) )Σ_(k=0 ,k≠8) ^(16) (−i)^k C_(16) ^k [(1/(k−8))z^(k−8) ]_1 ^(((√3)−i)/( (√(3 ))+i)) −(1/2^(17) )C_(16) ^8 [lnz]_1 ^(((√3)−i)/( (√3)+i)) =−(1/2^(17) )Σ_(k=0 and k≠8) ^(16) (((−1)^k C_(16) ^k )/(k−8)){((((√3)−i)/( (√3)+i)))^(k−8) −1} −(1/2^(17) )C_(16) ^8 ln((((√3)−i)/( (√3)+i)))$
	$Φ = \int_{1}^{\infty} \frac{dx}{{(x^{2} + x + 1)}^{9}} \Rightarrow Φ = \int_{1}^{\infty} \frac{dx}{{({(x + \frac{1}{2})}^{2} + \frac{3}{4})}^{9}}$ $=_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} t} \int_{\sqrt{3}}^{\infty} {(\frac{4}{3})}^{9} \frac{1}{{(t^{2} + 1)}^{9}} \frac{\sqrt{3}}{2} dt = \frac{\sqrt{3}}{2} {(\frac{4}{3})}^{9} \int_{\sqrt{3}}^{\infty} \frac{dt}{{(t^{2} + 1)}^{9}}$ $wehave \int_{\sqrt{3}}^{\infty} \frac{dt}{{(t^{2} + 1)}^{9}} = \int_{\sqrt{3}}^{\infty} \frac{dt}{{(t - i)}^{9} {(t + i)}^{9}} = \int_{\sqrt{3}}^{\infty} \frac{dt}{{(\frac{t - i}{t + i})}^{9} {(t + i)}^{18}}$ $changement \frac{t - i}{t + i} = z give t - i = zt + iz \Rightarrow (1 - z) t = i (1 + z) \Rightarrow$ $t = i \frac{1 + iz}{1 - iz} \Rightarrow \frac{dt}{dz} = i \frac{i (1 - iz) - (1 + iz) (- i)}{{(1 - iz)}^{2}} = i \frac{i + z + i - z}{{(1 - iz)}^{2}} = \frac{- 2}{{(1 - iz)}^{2}}$ $and t + i = \frac{i - z}{1 - iz} + i = \frac{i - z + i + z}{(1 - iz)} = \frac{2 i}{(1 - iz)} \Rightarrow$ $Φ = \int_{\frac{\sqrt{3} - i}{\sqrt{3} + i}}^{1} \frac{- 2}{{(1 - iz)}^{2} z^{9} {(\frac{2 i}{1 - iz})}^{18}} dz$ $= \frac{2}{{(2 i)}^{18}} \int_{1}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}} \frac{{(1 - iz)}^{16}}{z^{9}} dz = \frac{2}{{(2 i)}^{18}} \int_{1}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}} \frac{\sum_{k = 0}^{16} C_{16}^{k} {(iz)}^{k} {(- 1)}^{16 - k}}{z^{9}} dz$ $= - \frac{2}{2^{18}} \int_{1}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}} \sum_{k = 0}^{16} {(- 1)}^{k} C_{16}^{k} i^{k} z^{k - 9} dz$ $= - \frac{1}{2^{17}} \sum_{k = 0, k \neq 8}^{16} {(- i)}^{k} C_{16}^{k} {[\frac{1}{k - 8} z^{k - 8}]}_{1}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}}$ $- \frac{1}{2^{17}} C_{16}^{8} {[lnz]}_{1}^{\frac{\sqrt{3} - i}{\sqrt{3} + i}}$ $= - \frac{1}{2^{17}} \sum_{k = 0 and k \neq 8}^{16} \frac{{(- 1)}^{k} C_{16}^{k}}{k - 8} {{(\frac{\sqrt{3} - i}{\sqrt{3} + i})}^{k - 8} - 1}$ $- \frac{1}{2^{17}} C_{16}^{8} \ln (\frac{\sqrt{3} - i}{\sqrt{3} + i})$
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