calculate ∫_(−∞) ^(+∞) (dx/(1+x^2 +x^4 ))


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Question Number 42503 by maxmathsup by imad last updated on 26/Aug/18

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{d x}{1 + x^{2} + x^{4}}$
Commented by maxmathsup by imad last updated on 27/Aug/18
$method of residus let ϕ(z) =(1/(z^4 +z^2 +1)) .roots of z^4 +z^2 +1 =0 z^2 =t ⇒t^2 +t +1 =0 Δ =1−4=−3=(i(√3))^2 ⇒t_1 =((−1+i(√3))/2) =e^(i((2π)/3)) (=j) t_2 =((−1−i(√3))/2) =e^(−((i2π)/2)) ⇒ z^2 =e^((i2π)/3) ⇒ z =+^− e^((iπ)/3) and z_2 =e^(−((i2π)/3)) ⇒z =+^− e^(−((iπ)/3)) ⇒ϕ(z) = (1/((z−e^((iπ)/3) )(z+e^((iπ)/3) )(z−e^(−((iπ)/3)) )(z +e^(−((iπ)/3)) ))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,e^((iπ)/3) ) +Res(ϕ,−e^(−((iπ)/3)) )} but Res(ϕ, e^((iπ)/3) ) =lim_(z→e^((iπ)/3) ) (z−e^((iπ)/3) )ϕ(z) = (1/(2 e^((iπ)/3) (2i sin((π/3)))(2cos((π/3))))) = (e^(−((iπ)/3)) /(8i ((√3)/(2 )) (1/2))) = (e^(−((iπ)/3)) /(4i(√3))) Res(ϕ, −e^(−((iπ)/3)) ) =lim_(z→−e^(−((iπ)/3)) ) (z+e^(−((iπ)/3)) )ϕ(z) = (1/((−2isin((π/3))(2cos((π/3)))(−2e^(−((iπ)/3)) ))) = (e^((iπ)/3) /(4i(√3))) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ(1/(4i(√3))){ e^(−((iπ)/3)) +e^((iπ)/3) } = (π/(2(√3))) 2 cos((π/3)) =(π/(2(√3)))(1) ⇒ ∫_(−∞) ^(+∞) (dx/(1+x^2 +x^4 )) = (π/(2(√3))) .$
$m e t h o d o f r e s i d u s l e t φ (z) = \frac{1}{z^{4} + z^{2} + 1} . r o o t s o f z^{4} + z^{2} + 1 = 0$ $z^{2} = t \Rightarrow t^{2} + t + 1 = 0 Δ = 1 - 4 = - 3 = {(i \sqrt{3})}^{2} \Rightarrow t_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{i \frac{2 π}{3}} (= j)$ $t_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{2}} \Rightarrow z^{2} = e^{\frac{i 2 π}{3}} \Rightarrow z = \bar{+} e^{\frac{i π}{3}} a n d z_{2} = e^{- \frac{i 2 π}{3}} \Rightarrow z = \bar{+} e^{- \frac{i π}{3}}$ $\Rightarrow φ (z) = \frac{1}{(z - e^{\frac{i π}{3}}) (z + e^{\frac{i π}{3}}) (z - e^{- \frac{i π}{3}}) (z + e^{- \frac{i π}{3}})} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{3}}) + R e s (φ, - e^{- \frac{i π}{3}})} b u t$ $R e s (φ, e^{\frac{i π}{3}}) = {l i m}_{z \to e^{\frac{i π}{3}}} (z - e^{\frac{i π}{3}}) φ (z) = \frac{1}{2 e^{\frac{i π}{3}} (2 i s i n (\frac{π}{3})) (2 c o s (\frac{π}{3}))}$ $= \frac{e^{- \frac{i π}{3}}}{8 i \frac{\sqrt{3}}{2} \frac{1}{2}} = \frac{e^{- \frac{i π}{3}}}{4 i \sqrt{3}}$ $R e s (φ, - e^{- \frac{i π}{3}}) = {l i m}_{z \to - e^{- \frac{i π}{3}}} (z + e^{- \frac{i π}{3}}) φ (z) = \frac{1}{(- 2 i s i n (\frac{π}{3}) (2 c o s (\frac{π}{3})) (- 2 e^{- \frac{i π}{3}})}$ $= \frac{e^{\frac{i π}{3}}}{4 i \sqrt{3}} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{1}{4 i \sqrt{3}} {e^{- \frac{i π}{3}} + e^{\frac{i π}{3}}}$ $= \frac{π}{2 \sqrt{3}} 2 c o s (\frac{π}{3}) = \frac{π}{2 \sqrt{3}} (1) \Rightarrow \int_{- \infty}^{+ \infty} \frac{d x}{1 + x^{2} + x^{4}} = \frac{π}{2 \sqrt{3}} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/Aug/18
	$∫_(−∞) ^∞ ((1/x^2 )/(x^2 +(1/x^2 )+1))dx (1/2)∫_(−∞) ^∞ ((1+(1/x^2 )−(1−(1/x^2 )))/(x^2 +(1/x^2 )+1))dx (1/2)[∫_(−∞) ^∞ ((d(x−(1/x)))/((x−(1/x))^2 +3))−∫_(−∞) ^∞ ((d(x+(1/x)))/((x+(1/x))^2 −1))] (1/2)[∣(1/(√3))tan^(−1) (((x−(1/x))/(√3)))∣_(−∞) ^∞ −(1/(2×1))∣ln(((x+(1/x)−1)/(x+(1/x)+1)))∣_(−∞) ^∞ ] =(1/2)[(1/(√3))((Π/2)+(Π/2))−(1/(2 )){ln(((∞−1)/(∞+1)))−ln∣(((−∞−1)/(−∞+2)))∣] (1/2)[(Π/(√3))−(1/2){ln(((1−0)/(1+0)))−ln∣(((∞+1)/(2−∞)))∣}] (1/2)[(Π/(√3))−(1/2){0−ln∣((1+(1/∞))/((2/∞)−1))∣}] (1/2)×(Π/(√3))−(1/2)(0−0) (Π/(2(√3)))$
	$\int_{- \infty}^{\infty} \frac{\frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} + 1} d x$ $\frac{1}{2} \int_{- \infty}^{\infty} \frac{1 + \frac{1}{x^{2}} - (1 - \frac{1}{x^{2}})}{x^{2} + \frac{1}{x^{2}} + 1} d x$ $\frac{1}{2} [\int_{- \infty}^{\infty} \frac{d (x - \frac{1}{x})}{{(x - \frac{1}{x})}^{2} + 3} - \int_{- \infty}^{\infty} \frac{d (x + \frac{1}{x})}{{(x + \frac{1}{x})}^{2} - 1}]$ $\frac{1}{2} [∣ \frac{1}{\sqrt{3}} {t a n}^{- 1} (\frac{x - \frac{1}{x}}{\sqrt{3}}) ∣_{- \infty}^{\infty} - \frac{1}{2 \times 1} ∣ l n (\frac{x + \frac{1}{x} - 1}{x + \frac{1}{x} + 1}) ∣_{- \infty}^{\infty}]$ $= \frac{1}{2} [\frac{1}{\sqrt{3}} (\frac{Π}{2} + \frac{Π}{2}) - \frac{1}{2} {l n (\frac{\infty - 1}{\infty + 1}) - l n ∣ (\frac{- \infty - 1}{- \infty + 2}) ∣]$ $\frac{1}{2} [\frac{Π}{\sqrt{3}} - \frac{1}{2} {l n (\frac{1 - 0}{1 + 0}) - l n ∣ (\frac{\infty + 1}{2 - \infty}) ∣}]$ $\frac{1}{2} [\frac{Π}{\sqrt{3}} - \frac{1}{2} {0 - l n ∣ \frac{1 + \frac{1}{\infty}}{\frac{2}{\infty} - 1} ∣}]$ $\frac{1}{2} \times \frac{Π}{\sqrt{3}} - \frac{1}{2} (0 - 0)$ $\frac{Π}{2 \sqrt{3}}$
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