find the value of ∫_(−∞) ^(+∞) (dx/(x^4 −x^2 +3))


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Question Number 56827 by maxmathsup by imad last updated on 24/Mar/19

$f i n d t h e v a l u e o f \int_{- \infty}^{+ \infty} \frac{d x}{x^{4} - x^{2} + 3}$
Commented by maxmathsup by imad last updated on 25/Mar/19
$residus method let ϕ(z)=(1/(z^4 −z^2 +3)) poles of ϕ z^4 −z^2 +3 =0 ⇒t^2 −t+3=0 with t=z^2 Δ =1−12 =−11 ⇒t_1 =((1+i(√(11)))/2) and t_2 =((1−i(√(11)))/2) ⇒ϕ(z)=(1/((z^2 −t_1 )(z^2 −t_2 ))) = (1/((z−(√t_1 ))(z+(√t_1 ))(z−(√t_2 ))(z+(√t_2 )))) we have ∣t_1 ∣ =(√((1/4)+((11)/4)))=(√3) ⇒t_1 =(√3)((1/(2(√3))) +((i(√(11)))/(2(√3))))=r e^(iθ) ⇒cosθ =(1/(2(√3))) and sinθ=((√(11))/(2(√3))) ⇒tanθ =(√(11)) ⇒θ=arctan((√(11))) ⇒t_1 =(√3)e^(iarctan((√(11)))) ⇒ϕ(z) =(1/((z−(√(√3)) e^((i/2)arctan((√(11)))) )(z+(√(√3))e^((i/2)arctan((√(11)))) )(z−(√(√3))e^(−(i/2)arctan((√(11)))) )(z+(√(√3))e^(−(i/2)arctan((√(11)))) ))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,(√(√3))e^((i/2)arctan((√(11)))) +Res(ϕ,−(√(√3))e^(−(i/2)arctan((√(11)))) )} Res(ϕ,(√(√3))e^((i/2)arctan((√(11)))) )= (1/(2(√(√3))e^((i/2)arctan((√(11)))) 2i(√3) sin(((arctan((√(11))))/2))2cos(((arctan((√(11))))/2)))) = (e^(−(i/2)arctan((√(11)))) /(4i(√3)(√(√3))sin(arctan((√(11)))))) Res(ϕ,−(√(√3))e^(−(i/2)arctan((√(11)))) ) =(1/(−2(√(√3))e^(−(i/2)arctan((√(11)))) (−(√(√3)))2cos(((arctan((√(11))))/2))2i(√(√3))sin(((arctan((√(11))))/2)))) = (e^((i/2)arctan((√(11)))) /(4i(√3)(√(√3))sin(arctan((√(11)))))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =((2iπ)/(4i(√3)(√(√3))sin(arctan((√(11)))))) { e^((i/2)arctan((√(11)))) +e^(−(i/2)arctan((√(11)))) } =(π/(2(√3)(√(√3))sin(arctan((√(11))))) .2cos(((arctan((√(11))))/2)) ⇒ I =(π/((√3)(√(√3)))) ((cos(((arctan((√(11))))/2)))/(2sin(((arctan((√(11))))/2))cos(((arctan((√(11))))/2)))) I =(π/(2(√3)(√(√3))sin(((arctan((√(11))))/2)))) .$
$r e s i d u s m e t h o d l e t φ (z) = \frac{1}{z^{4} - z^{2} + 3} p o l e s o f φ$ $z^{4} - z^{2} + 3 = 0 \Rightarrow t^{2} - t + 3 = 0 w i t h t = z^{2}$ $Δ = 1 - 12 = - 11 \Rightarrow t_{1} = \frac{1 + i \sqrt{11}}{2} a n d t_{2} = \frac{1 - i \sqrt{11}}{2} \Rightarrow φ (z) = \frac{1}{(z^{2} - t_{1}) (z^{2} - t_{2})}$ $= \frac{1}{(z - \sqrt{t_{1}}) (z + \sqrt{t_{1}}) (z - \sqrt{t_{2}}) (z + \sqrt{t_{2}})} w e h a v e$ $∣ t_{1} ∣ = \sqrt{\frac{1}{4} + \frac{11}{4}} = \sqrt{3} \Rightarrow t_{1} = \sqrt{3} (\frac{1}{2 \sqrt{3}} + \frac{i \sqrt{11}}{2 \sqrt{3}}) = r e^{i θ} \Rightarrow c o s θ = \frac{1}{2 \sqrt{3}} a n d s i n θ = \frac{\sqrt{11}}{2 \sqrt{3}}$ $\Rightarrow t a n θ = \sqrt{11} \Rightarrow θ = a r c t a n (\sqrt{11}) \Rightarrow t_{1} = \sqrt{3} e^{i a r c t a n (\sqrt{11})}$ $\Rightarrow φ (z) = \frac{1}{(z - \sqrt{\sqrt{3}} e^{\frac{i}{2} a r c t a n (\sqrt{11})}) (z + \sqrt{\sqrt{3}} e^{\frac{i}{2} a r c t a n (\sqrt{11})}) (z - \sqrt{\sqrt{3}} e^{- \frac{i}{2} a r c t a n (\sqrt{11})}) (z + \sqrt{\sqrt{3}} e^{- \frac{i}{2} a r c t a n (\sqrt{11})})}$ $r e s i d u s t h e o r e m g i v e \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, \sqrt{\sqrt{3}} e^{\frac{i}{2} a r c t a n (\sqrt{11})} + R e s (φ, - \sqrt{\sqrt{3}} e^{- \frac{i}{2} a r c t a n (\sqrt{11})})}$ $R e s (φ, \sqrt{\sqrt{3}} e^{\frac{i}{2} a r c t a n (\sqrt{11})}) = \frac{1}{2 \sqrt{\sqrt{3}} e^{\frac{i}{2} a r c t a n (\sqrt{11})} 2 i \sqrt{3} s i n (\frac{a r c t a n (\sqrt{11})}{2}) 2 c o s (\frac{a r c t a n (\sqrt{11})}{2})}$ $= \frac{e^{- \frac{i}{2} a r c t a n (\sqrt{11})}}{4 i \sqrt{3} \sqrt{\sqrt{3}} s i n (a r c t a n (\sqrt{11}))}$ $R e s (φ, - \sqrt{\sqrt{3}} e^{- \frac{i}{2} a r c t a n (\sqrt{11})}) = \frac{1}{- 2 \sqrt{\sqrt{3}} e^{- \frac{i}{2} a r c t a n (\sqrt{11})} (- \sqrt{\sqrt{3}}) 2 c o s (\frac{a r c t a n (\sqrt{11})}{2}) 2 i \sqrt{\sqrt{3}} s i n (\frac{a r c t a n (\sqrt{11})}{2})}$ $= \frac{e^{\frac{i}{2} a r c t a n (\sqrt{11})}}{4 i \sqrt{3} \sqrt{\sqrt{3}} s i n (a r c t a n (\sqrt{11}))} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 i π}{4 i \sqrt{3} \sqrt{\sqrt{3}} s i n (a r c t a n (\sqrt{11}))} {e^{\frac{i}{2} a r c t a n (\sqrt{11})} + e^{- \frac{i}{2} a r c t a n (\sqrt{11)}}}$ $= \frac{π}{2 \sqrt{3} \sqrt{\sqrt{3}} s i n (a r c t a n (\sqrt{11})} . 2 c o s (\frac{a r c t a n (\sqrt{11})}{2}) \Rightarrow$ $I = \frac{π}{\sqrt{3} \sqrt{\sqrt{3}}} \frac{c o s (\frac{a r c t a n (\sqrt{11})}{2})}{2 s i n (\frac{a r c t a n (\sqrt{11})}{2}) c o s (\frac{a r c t a n (\sqrt{11})}{2})}$ $I = \frac{π}{2 \sqrt{3} \sqrt{\sqrt{3}} s i n (\frac{a r c t a n (\sqrt{11})}{2})} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 25/Mar/19
	$(1/(x^4 −x^2 +3)) =((1/x^2 )/(x^2 −1+(3/x^2 ))) =(1/(2(√3)))((((2(√3))/x^2 )/(x^2 −1+(3/x^2 )))) =(1/(2(√3)))(((1+((√3)/x^2 )−(1−((√3)/x^2 )))/(x^2 +(3/x^2 )−1))) =(1/(2(√3)))(((d(x−((√3)/x)))/((x−((√3)/x))^2 +2(√3) −1)))−(1/(2(√3)))(((d(x+((√3)/x)))/((x+((√3)/x))^2 −2(√3) −1))) (1/(2(√3)))∫_(−∞) ^(+∞) ((d(x−((√3)/x)))/((x−((√3)/x))^2 +((√(2(√3) −1)) )^2 ))←I_1 (1/(2(√3)))×(1/(√(2(√3) −1)))×∣tan^(−1) (((x−((√3)/x))/(√(2(√3) −1))))∣_(−∞) ^(+∞) =(1/(2(√3)))×(1/(√(2(√3) −1)))×{((π/2))−(((−π)/2))} =(π/(2(√3)))×(1/(√(2(√3) −1)))←value of I_1 (1/(2(√3)))∫_(−∞) ^(+∞) ((d(x+((√3)/x)))/((x+((√3)/x))^2 −(2(√3) +1))) ←I_2 (1/(2(√3)))×(1/(2((√(2(√3) +1)) )))∣ln(((x+((√3)/x)−(√(2(√3) +1)))/(x+((√3)/x)+(√(2(√3) +1)))))∣_(−∞) ^(+∞) now ∣ln(((x+(a/x)−b)/(x+(a/x)+b)))∣_(−∞) ^(+∞) =∣ln(((x^2 +a−bx)/(x^2 +a+bx)))∣_(−∞) ^(+∞) =∣ln(((1+(a/x^2 )−(b/x))/(1+(a/x^2 )+(b/x))))∣_(−∞) ^(+∞) =0 so I_2 =0 hence answer is (π/(2(√3)))×(1/(√(2(√3) −1))) this is the answer$
	$\frac{1}{x^{4} - x^{2} + 3}$ $= \frac{\frac{1}{x^{2}}}{x^{2} - 1 + \frac{3}{x^{2}}}$ $= \frac{1}{2 \sqrt{3}} (\frac{\frac{2 \sqrt{3}}{x^{2}}}{x^{2} - 1 + \frac{3}{x^{2}}})$ $= \frac{1}{2 \sqrt{3}} (\frac{1 + \frac{\sqrt{3}}{x^{2}} - (1 - \frac{\sqrt{3}}{x^{2}})}{x^{2} + \frac{3}{x^{2}} - 1})$ $= \frac{1}{2 \sqrt{3}} (\frac{d (x - \frac{\sqrt{3}}{x})}{{(x - \frac{\sqrt{3}}{x})}^{2} + 2 \sqrt{3} - 1}) - \frac{1}{2 \sqrt{3}} (\frac{d (x + \frac{\sqrt{3}}{x})}{{(x + \frac{\sqrt{3}}{x})}^{2} - 2 \sqrt{3} - 1})$ $\frac{1}{2 \sqrt{3}} \int_{- \infty}^{+ \infty} \frac{d (x - \frac{\sqrt{3}}{x})}{{(x - \frac{\sqrt{3}}{x})}^{2} + {(\sqrt{2 \sqrt{3} - 1})}^{2}} \leftarrow I_{1}$ $\frac{1}{2 \sqrt{3}} \times \frac{1}{\sqrt{2 \sqrt{3} - 1}} \times ∣ {t a n}^{- 1} (\frac{x - \frac{\sqrt{3}}{x}}{\sqrt{2 \sqrt{3} - 1}}) ∣_{- \infty}^{+ \infty}$ $= \frac{1}{2 \sqrt{3}} \times \frac{1}{\sqrt{2 \sqrt{3} - 1}} \times {(\frac{π}{2}) - (\frac{- π}{2})}$ $= \frac{π}{2 \sqrt{3}} \times \frac{1}{\sqrt{2 \sqrt{3} - 1}} \leftarrow v a l u e o f I_{1}$ $\frac{1}{2 \sqrt{3}} \int_{- \infty}^{+ \infty} \frac{d (x + \frac{\sqrt{3}}{x})}{{(x + \frac{\sqrt{3}}{x})}^{2} - (2 \sqrt{3} + 1)} \leftarrow I_{2}$ $\frac{1}{2 \sqrt{3}} \times \frac{1}{2 (\sqrt{2 \sqrt{3} + 1})} ∣ l n (\frac{x + \frac{\sqrt{3}}{x} - \sqrt{2 \sqrt{3} + 1}}{x + \frac{\sqrt{3}}{x} + \sqrt{2 \sqrt{3} + 1}}) ∣_{- \infty}^{+ \infty}$ $n o w ∣ l n (\frac{x + \frac{a}{x} - b}{x + \frac{a}{x} + b}) ∣_{- \infty}^{+ \infty}$ $=∣ l n (\frac{x^{2} + a - b x}{x^{2} + a + b x}) ∣_{- \infty}^{+ \infty}$ $=∣ l n (\frac{1 + \frac{a}{x^{2}} - \frac{b}{x}}{1 + \frac{a}{x^{2}} + \frac{b}{x}}) ∣_{- \infty}^{+ \infty} = 0$ $s o I_{2} = 0$ $h e n c e a n s w e r i s$ $\frac{π}{2 \sqrt{3}} \times \frac{1}{\sqrt{2 \sqrt{3} - 1}} t h i s i s t h e a n s w e r$
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