calculate I=∫_1 ^(+∞) ((x^2 −1)/(x^4 −x^2 +1))dx


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Question Number 86258 by lémùst last updated on 27/Mar/20

$c a l c u l a t e I = \int_{1}^{+ \infty} \frac{x^{2} - 1}{x^{4} - x^{2} + 1} d x$
Commented by mathmax by abdo last updated on 27/Mar/20
$I =∫_1 ^(+∞) ((x^2 −1)/(x^4 −x^2 +1))dx (complex method) x^4 −x^2 +1 =0 →t^2 −t+1 =0 (t=x^2 ) Δ=1−4=−3 ⇒t_1 =((1+i(√3))/2)=e^((iπ)/6) and t_2 =((1−i(√3))/2)=e^(−((iπ)/6)) ⇒((x^2 −1)/(x^4 −x^2 +1)) =((x^2 −1)/((x^2 −e^((iπ)/6) )(x^2 −e^(−((iπ)/6)) ))) =((x^2 −1)/(2isin((π/6)))){(1/(x^2 −e^((iπ)/6) ))−(1/(x^2 −e^(−((iπ)/6)) ))} =−i(((x^2 −1)/(x^2 −e^((iπ)/6) )) −((x^2 −1)/(x^2 −e^(−((iπ)/6)) ))) =−i(((x^2 −e^((iπ)/6) +e^((iπ)/6) −1)/(x^2 −e^((iπ)/6) ))−((x^2 −e^(−((iπ)/6)) +e^(−((iπ)/6)) −1)/(x^2 −e^(−((iπ)/6)) ))) =−i{((e^((iπ)/6) −1)/(x^2 −e^((iπ)/6) ))−((e^(−((iπ)/6)) −1)/(x^2 −e^(−((iπ)/6)) ))} ⇒I=i(e^(−((iπ)/6)) −1)∫_1 ^(+∞) (dx/(x^2 −e^(−((iπ)/6)) )) −i(e^((iπ)/6) −1)∫_1 ^(+∞) (dx/(x^2 −e^((iπ)/6) )) ∫_1 ^(+∞) (dx/(x^2 −e^((iπ)/6) )) =∫_1 ^(+∞) (dx/((x−e^((iπ)/(12)) )(x+e^((iπ)/(12)) ))) =(1/2)e^(−((iπ)/(12))) ∫_1 ^(+∞) ((1/(x−e^((iπ)/(12)) ))−(1/(x+e^((iπ)/(12)) )))dx =(1/2)e^(−((iπ)/(12))) [ln(((x−e^((iπ)/(12)) )/(x+e^((iπ)/(12)) )))]_1 ^(+∞) =(1/2)e^(−((iπ)/(12))) {−ln(((1−e^((iπ)/(12)) )/(1+e^((iπ)/(12)) )))} ...becontinued...$
$I = \int_{1}^{+ \infty} \frac{x^{2} - 1}{x^{4} - x^{2} + 1} d x (c o m p l e x m e t h o d)$ $x^{4} - x^{2} + 1 = 0 \to t^{2} - t + 1 = 0 (t = x^{2})$ $Δ = 1 - 4 = - 3 \Rightarrow t_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{6}} a n d t_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{6}}$ $\Rightarrow \frac{x^{2} - 1}{x^{4} - x^{2} + 1} = \frac{x^{2} - 1}{(x^{2} - e^{\frac{i π}{6}}) (x^{2} - e^{- \frac{i π}{6}})} = \frac{x^{2} - 1}{2 i s i n (\frac{π}{6})} {\frac{1}{x^{2} - e^{\frac{i π}{6}}} - \frac{1}{x^{2} - e^{- \frac{i π}{6}}}}$ $= - i (\frac{x^{2} - 1}{x^{2} - e^{\frac{i π}{6}}} - \frac{x^{2} - 1}{x^{2} - e^{- \frac{i π}{6}}}) = - i (\frac{x^{2} - e^{\frac{i π}{6}} + e^{\frac{i π}{6}} - 1}{x^{2} - e^{\frac{i π}{6}}} - \frac{x^{2} - e^{- \frac{i π}{6}} + e^{- \frac{i π}{6}} - 1}{x^{2} - e^{- \frac{i π}{6}}})$ $= - i {\frac{e^{\frac{i π}{6}} - 1}{x^{2} - e^{\frac{i π}{6}}} - \frac{e^{- \frac{i π}{6}} - 1}{x^{2} - e^{- \frac{i π}{6}}}} \Rightarrow I = i (e^{- \frac{i π}{6}} - 1) \int_{1}^{+ \infty} \frac{d x}{x^{2} - e^{- \frac{i π}{6}}}$ $- i (e^{\frac{i π}{6}} - 1) \int_{1}^{+ \infty} \frac{d x}{x^{2} - e^{\frac{i π}{6}}}$ $\int_{1}^{+ \infty} \frac{d x}{x^{2} - e^{\frac{i π}{6}}} = \int_{1}^{+ \infty} \frac{d x}{(x - e^{\frac{i π}{12}}) (x + e^{\frac{i π}{12}})}$ $= \frac{1}{2} e^{- \frac{i π}{12}} \int_{1}^{+ \infty} (\frac{1}{x - e^{\frac{i π}{12}}} - \frac{1}{x + e^{\frac{i π}{12}}}) d x$ $= \frac{1}{2} e^{- \frac{i π}{12}} {[l n (\frac{x - e^{\frac{i π}{12}}}{x + e^{\frac{i π}{12}}})]}_{1}^{+ \infty} = \frac{1}{2} e^{- \frac{i π}{12}} {- l n (\frac{1 - e^{\frac{i π}{12}}}{1 + e^{\frac{i π}{12}}})}$ $. . . b e c o n t i n u e d . . .$
Commented by lémùst last updated on 28/Mar/20

$s i r h o w d o y o u c a l c u l a t e t h e l o g a r i t h o f a$ $c o m p l e x ?$
Commented by mathmax by abdo last updated on 31/Mar/20

$l n (x + i y) = l n (\sqrt{x^{2} + y^{2}} e^{i a r c t a n (\frac{y}{x})})$ $= \frac{1}{2} l n (x^{2} + y^{2}) + i a r c t a n (\frac{y}{x}) (p r i n c i p l e d e t e r m i n a t i o n o f l n)$
	Answered by TANMAY PANACEA. last updated on 28/Mar/20

	$\int \frac{1 - \frac{1}{x^{2}}}{x^{2} - 1 + \frac{1}{x^{2}}} d x$ $\int \frac{d (x + \frac{1}{x})}{{(x + \frac{1}{x})}^{2} - 3} = \frac{1}{2 \sqrt{3}} l n (\frac{x + \frac{1}{x} - \sqrt{3}}{x + \frac{1}{x} + \sqrt{3}})$ $u s e i n g f o r m u l a \int \frac{d p}{p^{2} - a^{2}}$
	Commented by lémùst last updated on 28/Mar/20

	$t h a n k s s i r$
	Commented by TANMAY PANACEA. last updated on 28/Mar/20

	$m o s t w e l c o m e . . .$
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