calculate by complex method ∫_0 ^∞ (dx/(x^2 −x+1))


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Question Number 86375 by mathmax by abdo last updated on 28/Mar/20

$c a l c u l a t e b y c o m p l e x m e t h o d \int_{0}^{\infty} \frac{d x}{x^{2} - x + 1}$
Commented by mathmax by abdo last updated on 30/Mar/20
$first let find the value of this integral we have ∫ (dx/(x^2 −x +1)) =∫ (dx/((x−(1/2))^2 +(3/4))) =_(x−(1/2)=((√3)/2)u) (4/3)∫ (1/(u^2 +1))×((√3)/2)du =(2/(√3)) arctan(u) +c =(2/(√3))arctan(((2x−1)/(√3)))+C ⇒ ∫_0 ^∞ (dx/(x^2 −x+1)) =[(2/(√3))arctan(((2x−1)/(√3)))]_0 ^(+∞) =(2/(√3)){ (π/2) +arctan((1/(√3)))} =(2/(√3)){(π/2) +(π/6)} =(2/(√3)){((4π)/6)} =(2/(√3))×((2π)/3) =((4π)/(3(√3))) now let use complex method the problem come from 0 ! x^2 −x+1 =0 →Δ =−3 ⇒x_1 =((1+i(√3))/2)=e^((iπ)/3) and x_2 =((1−i(√3))/2) =e^(−((iπ)/3)) ⇒∫ (dx/(x^2 −x+1)) =∫ (dx/((x−e^((iπ)/3) )(x−e^(−((iπ)/3)) )))=(1/(2isin((π/3))))∫ ((1/(x−e^((iπ)/3) ))−(1/(x−e^(−((iπ)/3)) )))dx =(1/(i(√3)))ln(((x−e^((iπ)/3) )/(x−e^(−((iπ)/3)) ))) +C ⇒ ∫_0 ^∞ (dx/(x^2 −x+1)) =(1/(i(√3)))[ln(((x−e^((iπ)/3) )/(x−e^(−((iπ)/3)) )))]_0 ^(+∞) =(1/(i(√3))){−ln( e^((2iπ)/3) )} =−(1/(i(√3)))(((2iπ)/3)) =((2π)/(3(√3))) not correct the problem comes from0 let change 0 by 1 ∫_1 ^(+∞) (dx/(x^2 −x+1)) =(2/(√3))[arctan(((2x−1)/(√3)))]_1 ^(+∞) =(2/(√3)){ (π/2) −arctan((1/(√3)))} =(2/(√3)){(π/2)−(π/6)} =(2/(√3)){(π/3)} =((2π)/(3(√3))) complex method→∫_1 ^(+∞) (dx/(x^2 −x+1)) =(1/(i(√3)))[ln(((x−e^((iπ)/3) )/(x−e^(−((iπ)/3)) )))]_1 ^(+∞) =(1/(i(√3)))(−ln(((1−e^((iπ)/3) )/(1−e^(−((iπ)/3)) )))) we have ((1−e^((iπ)/3) )/(1−e^(−((iπ)/3)) )) =((1−(1/2)−i((√3)/2))/(1−(1/2)+i((√3)/2))) =(((1/2)−i((√3)/2))/((1/2)+((i(√3))/2))) =(e^(−((iπ)/3)) /e^((iπ)/3) ) =e^(−((2iπ)/3)) ⇒ ln(...) =−((2iπ)/3) ⇒∫_1 ^(+∞) (dx/(x^2 −x+1)) =(1/(i(√3)))(((−2iπ)/3)) =((−2π)/(3(√3))) (probleme of −) i have 2 remark here dont use complex method if you have 0 in the integral the function x→(1/(x^2 −x+1)) is ≥0 ⇒∫_1 ^(+∞) (dx/(x^2 −x)) ≥0 so you must eradicate − in the value....$
$f i r s t l e t f i n d t h e v a l u e o f t h i s i n t e g r a l$ $w e h a v e \int \frac{d x}{x^{2} - x + 1} = \int \frac{d x}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}} =_{x - \frac{1}{2} = \frac{\sqrt{3}}{2} u} \frac{4}{3} \int \frac{1}{u^{2} + 1} \times \frac{\sqrt{3}}{2} d u$ $= \frac{2}{\sqrt{3}} a r c t a n (u) + c = \frac{2}{\sqrt{3}} a r c t a n (\frac{2 x - 1}{\sqrt{3}}) + C \Rightarrow$ $\int_{0}^{\infty} \frac{d x}{x^{2} - x + 1} = {[\frac{2}{\sqrt{3}} a r c t a n (\frac{2 x - 1}{\sqrt{3}})]}_{0}^{+ \infty} = \frac{2}{\sqrt{3}} {\frac{π}{2} + a r c t a n (\frac{1}{\sqrt{3}})}$ $= \frac{2}{\sqrt{3}} {\frac{π}{2} + \frac{π}{6}} = \frac{2}{\sqrt{3}} {\frac{4 π}{6}} = \frac{2}{\sqrt{3}} \times \frac{2 π}{3} = \frac{4 π}{3 \sqrt{3}}$ $n o w l e t u s e c o m p l e x m e t h o d t h e p r o b l e m c o m e f r o m 0!$ $x^{2} - x + 1 = 0 \to Δ = - 3 \Rightarrow x_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}} a n d x_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{3}}$ $\Rightarrow \int \frac{d x}{x^{2} - x + 1} = \int \frac{d x}{(x - e^{\frac{i π}{3}}) (x - e^{- \frac{i π}{3}})} = \frac{1}{2 i s i n (\frac{π}{3})} \int (\frac{1}{x - e^{\frac{i π}{3}}} - \frac{1}{x - e^{- \frac{i π}{3}}}) d x$ $= \frac{1}{i \sqrt{3}} l n (\frac{x - e^{\frac{i π}{3}}}{x - e^{- \frac{i π}{3}}}) + C \Rightarrow$ $\int_{0}^{\infty} \frac{d x}{x^{2} - x + 1} = \frac{1}{i \sqrt{3}} {[l n (\frac{x - e^{\frac{i π}{3}}}{x - e^{- \frac{i π}{3}}})]}_{0}^{+ \infty} = \frac{1}{i \sqrt{3}} {- l n (e^{\frac{2 i π}{3}})}$ $= - \frac{1}{i \sqrt{3}} (\frac{2 i π}{3}) = \frac{2 π}{3 \sqrt{3}} n o t c o r r e c t t h e p r o b l e m c o m e s f r o m 0$ $l e t c h a n g e 0 b y 1$ $\int_{1}^{+ \infty} \frac{d x}{x^{2} - x + 1} = \frac{2}{\sqrt{3}} {[a r c t a n (\frac{2 x - 1}{\sqrt{3}})]}_{1}^{+ \infty}$ $= \frac{2}{\sqrt{3}} {\frac{π}{2} - a r c t a n (\frac{1}{\sqrt{3}})} = \frac{2}{\sqrt{3}} {\frac{π}{2} - \frac{π}{6}} = \frac{2}{\sqrt{3}} {\frac{π}{3}} = \frac{2 π}{3 \sqrt{3}}$ $c o m p l e x m e t h o d \to \int_{1}^{+ \infty} \frac{d x}{x^{2} - x + 1} = \frac{1}{i \sqrt{3}} {[l n (\frac{x - e^{\frac{i π}{3}}}{x - e^{- \frac{i π}{3}}})]}_{1}^{+ \infty}$ $= \frac{1}{i \sqrt{3}} (- l n (\frac{1 - e^{\frac{i π}{3}}}{1 - e^{- \frac{i π}{3}}})) w e h a v e$ $\frac{1 - e^{\frac{i π}{3}}}{1 - e^{- \frac{i π}{3}}} = \frac{1 - \frac{1}{2} - i \frac{\sqrt{3}}{2}}{1 - \frac{1}{2} + i \frac{\sqrt{3}}{2}} = \frac{\frac{1}{2} - i \frac{\sqrt{3}}{2}}{\frac{1}{2} + \frac{i \sqrt{3}}{2}} = \frac{e^{- \frac{i π}{3}}}{e^{\frac{i π}{3}}} = e^{- \frac{2 i π}{3}} \Rightarrow$ $l n (. . .) = - \frac{2 i π}{3} \Rightarrow \int_{1}^{+ \infty} \frac{d x}{x^{2} - x + 1} = \frac{1}{i \sqrt{3}} (\frac{- 2 i π}{3}) = \frac{- 2 π}{3 \sqrt{3}} (p r o b l e m e o f -)$ $i h a v e 2 r e m a r k h e r e$ $d o n t u s e c o m p l e x m e t h o d i f y o u h a v e 0 i n t h e i n t e g r a l$ $t h e f u n c t i o n x \to \frac{1}{x^{2} - x + 1} i s ⩾ 0 \Rightarrow \int_{1}^{+ \infty} \frac{d x}{x^{2} - x} ⩾ 0 s o$ $y o u m u s t e r a d i c a t e - i n t h e v a l u e . . . .$
Commented by redmiiuser last updated on 31/Mar/20

$s i r c a n y o u p l s c o m m e n t$ $o n m y p r o c e s s .$
	Answered by redmiiuser last updated on 30/Mar/20

	$l e t x^{2} - x = z$ ${(z + 1)}^{- 1}$ $= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . z^{n}$ $= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . {(x^{2} - x)}^{n}$ $= \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} . \underset{k = 0}{\sum^{n}} \underset{k}{\overset{n}{C}} . {(- 1)}^{n - k} . {(x^{2})}^{k} . {(x)}^{n - k}$ $= \underset{n = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{n}} {(- 1)}^{2 n - k} . \underset{k}{\overset{n}{C}} . x^{2 k + n - k}$ ${\int_{0}}^{\infty} \underset{n = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{n}} i^{2 n - k} . \underset{k}{\overset{n}{C}} . x^{k + n} . d x$ $= \frac{\underset{n = 0}{\sum^{\infty}} \underset{k = 0}{\sum^{n}} i^{2 n - k} . \underset{k}{\overset{n}{C}} . {(\infty)}^{k + n + 1}}{k + n + 1}$
	Commented by redmiiuser last updated on 30/Mar/20

	$s i r c a n y o u p l s c h e c k$ $t h e a n s w e r .$
	Commented by mathmax by abdo last updated on 01/Apr/20

	$n o t c o r r e c t s i r b e c s u s e \frac{1}{1 + z} = \sum_{n = 0}^{\infty} {(- 1)}^{n} z^{n} i f w e h a v e ∣ z ∣< 1$ $y o u c a n t h i s i f y o u h a v e ∣ x^{2} - x ∣< 1 . . .$
	Commented by redmiiuser last updated on 01/Apr/20

	$s i r c a n y o u p l s c h e c k$ $\frac{1}{z + 1} e x p a n s i o n b y$ ${T a l o y r}^{'} s f o r m u l a .$
	Commented by redmiiuser last updated on 03/Apr/20

	$f o r z < 1 l i k e z = - l$ $∴ {(1 + z)}^{- 1} = {(1 - l)}^{- 1}$ $= \underset{n = 0}{\sum^{\infty}} l^{n}$
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