prove for (0<a<2) ∫_0 ^( ∞) ((x^(a−1) dx)/(1+x+x^2 )) = ((2π)/(√3))cos (((2πa+π)/6))cosec πa .


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Question Number 88547 by ajfour last updated on 11/Apr/20

$p r o v e f o r (0 < a < 2)$ $\int_{0}^{\infty} \frac{x^{a - 1} d x}{1 + x + x^{2}} = \frac{2 π}{\sqrt{3}} \cos (\frac{2 π a + π}{6}) cosec π a .$
	Answered by mind is power last updated on 12/Apr/20
	$∫_(−∞) ^(+∞) (x^(a−1) /(1+x+x^2 ))dx=∫_0 ^∞ ((x^(a−1) dx)/(1+x+x^2 ))+∫_0 ^(+∞) (((−1)^(a−1) x^(a−1) )/(1−x+x^2 ))dx =∫_C_R (x^(a−1) /(1+x+x^2 ))dx≤∫_0 ^π ((R^a e^(i(a−1)π) )/(R^2 −R+1))→0 ⇒∫_0 ^(+∞) ((e^(iπ(a−1)) x^(a−1) )/(x^2 −x+1))dx+∫_0 ^∞ ((x^(a−1) dx)/(x^2 +x+1))=2iπRes((z^(a−1) /(z^2 +z+1)),e^(i((2π)/3)) ) =((2iπe^(i(a−1)((2π)/3)) )/((3e^((2iπ)/3) +1)))=−isin(πa)∫_0 ^∞ ((x^(a−1) dx)/(x^2 −x+1))+−cos(πa)∫_0 ^(+∞) ((x^(a−1) dx)/(x^2 −x+1)) +∫_0 ^∞ ((x^(a−1) dx)/(x^2 +x+1)) ((2iπe^(i(a−1)((2π)/3)) )/(2(−(1/2)+((i(√3))/2))+1))=((2iπe^(i(a−1)((2π)/3)) )/(i(√3)))=((2π)/(√3))(cos(((2π(a−1))/3))+isin(((2π(a−1))/3))) =−isin(πa)∫_0 ^∞ ((x^(a−1) dx)/(x^2 −x+1))−cos(πa)∫_0 ^∞ ((x^(a−1) dx)/(x^2 −x+1))+∫_0 ^(+∞) ((x^(a−1) dx)/(x^2 +x+1)) ⇒((2iπ)/(√3))sin(((2π(a−1))/3))=−isin(πa)∫_0 ^(+∞) ((x^(a−1) dx)/(x^2 −x+1)) ⇒∫_0 ^(+∞) ((x^(a−1) dx)/(x^2 −x+1))=((−2πsin(((2π(a−1))/3)))/(sin(πa)(√3))) ∫_0 ^(+∞) ((x^(a−1) dx)/(x^2 +x+1))=((2π)/(√3))cos(((2π(a−1))/3))+((−2πsin(((2π(a−1))/3))cos(πa))/(sin(πa)(√3))) =((2π)/(√3)).(1/(sin(πa))){cos(((2π(a−1))/3))sin(πa)−sin(((2π(a−1))/3))cos(πa)) =((2π)/(sin(πa)(√3))){sin(πa−((2π(a−1))/3))} =((2π)/((√3)sin(πa))){sin(((πa)/3)+((2π)/3))} =((2π)/(sin(πa)(√3)))cos((π/2)−((πa)/3)−((2π)/3))=((2π)/(sin(πa)(√3)))(cos(((−πa)/3)−(π/6))) =((2π)/(sin(πa)(√3)))cos(((2πa+π)/6))=((2π)/(√3))cos(((2πa+π)/6))cosec(πa)$
	$\int_{- \infty}^{+ \infty} \frac{x^{a - 1}}{1 + x + x^{2}} d x = \int_{0}^{\infty} \frac{x^{a - 1} d x}{1 + x + x^{2}} + \int_{0}^{+ \infty} \frac{{(- 1)}^{a - 1} x^{a - 1}}{1 - x + x^{2}} d x$ $= \int_{C_{R}} \frac{x^{a - 1}}{1 + x + x^{2}} d x ⩽ \int_{0}^{π} \frac{R^{a} e^{i (a - 1) π}}{R^{2} - R + 1} \to 0$ $\Rightarrow \int_{0}^{+ \infty} \frac{e^{i π (a - 1)} x^{a - 1}}{x^{2} - x + 1} d x + \int_{0}^{\infty} \frac{x^{a - 1} d x}{x^{2} + x + 1} = 2 i π R e s (\frac{z^{a - 1}}{z^{2} + z + 1}, e^{i \frac{2 π}{3}})$ $= \frac{2 i π e^{i (a - 1) \frac{2 π}{3}}}{(3 e^{\frac{2 i π}{3}} + 1)} = - i s i n (π a) \int_{0}^{\infty} \frac{x^{a - 1} d x}{x^{2} - x + 1} + - c o s (π a) \int_{0}^{+ \infty} \frac{x^{a - 1} d x}{x^{2} - x + 1}$ $+ \int_{0}^{\infty} \frac{x^{a - 1} d x}{x^{2} + x + 1}$ $\frac{2 i π e^{i (a - 1) \frac{2 π}{3}}}{2 (- \frac{1}{2} + \frac{i \sqrt{3}}{2}) + 1} = \frac{2 i π e^{i (a - 1) \frac{2 π}{3}}}{i \sqrt{3}} = \frac{2 π}{\sqrt{3}} (c o s (\frac{2 π (a - 1)}{3}) + i s i n (\frac{2 π (a - 1)}{3}))$ $= - i s i n (π a) \int_{0}^{\infty} \frac{x^{a - 1} d x}{x^{2} - x + 1} - c o s (π a) \int_{0}^{\infty} \frac{x^{a - 1} d x}{x^{2} - x + 1} + \int_{0}^{+ \infty} \frac{x^{a - 1} d x}{x^{2} + x + 1}$ $\Rightarrow \frac{2 i π}{\sqrt{3}} s i n (\frac{2 π (a - 1)}{3}) = - i s i n (π a) \int_{0}^{+ \infty} \frac{x^{a - 1} d x}{x^{2} - x + 1}$ $\Rightarrow \int_{0}^{+ \infty} \frac{x^{a - 1} d x}{x^{2} - x + 1} = \frac{- 2 π s i n (\frac{2 π (a - 1)}{3})}{s i n (π a) \sqrt{3}}$ $\int_{0}^{+ \infty} \frac{x^{a - 1} d x}{x^{2} + x + 1} = \frac{2 π}{\sqrt{3}} c o s (\frac{2 π (a - 1)}{3}) + \frac{- 2 π s i n (\frac{2 π (a - 1)}{3}) c o s (π a)}{s i n (π a) \sqrt{3}}$ $= \frac{2 π}{\sqrt{3}} . \frac{1}{s i n (π a)} {c o s (\frac{2 π (a - 1)}{3}) s i n (π a) - s i n (\frac{2 π (a - 1)}{3}) c o s (π a))$ $= \frac{2 π}{s i n (π a) \sqrt{3}} {s i n (π a - \frac{2 π (a - 1)}{3})}$ $= \frac{2 π}{\sqrt{3} s i n (π a)} {s i n (\frac{π a}{3} + \frac{2 π}{3})}$ $= \frac{2 π}{s i n (π a) \sqrt{3}} c o s (\frac{π}{2} - \frac{π a}{3} - \frac{2 π}{3}) = \frac{2 π}{s i n (π a) \sqrt{3}} (c o s (\frac{- π a}{3} - \frac{π}{6}))$ $= \frac{2 π}{s i n (π a) \sqrt{3}} c o s (\frac{2 π a + π}{6}) = \frac{2 π}{\sqrt{3}} c o s (\frac{2 π a + π}{6}) c o s e c (π a)$
	Commented byajfour last updated on 12/Apr/20

	$T h a n k y o u s i r, h o p e y o u l i k e d$ $s o l v i n g i t .$
	Commented bymind is power last updated on 12/Apr/20

	$n i c e o n e S i r$ $s i n c e 4 o r 5 m o n t h a g o i l o s t m y m o t i v a t i o n$ $l o s t p l e a s u r o f s o l v i n g p r o b l e m e s i d o n t k n o w w h y!$
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