calculate ∫_0 ^∞ xe^(−x^2 ) arctan(x−(1/x))dx


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Question Number 73182 by mathmax by abdo last updated on 07/Nov/19

$c a l c u l a t e \int_{0}^{\infty} {x e}^{- x^{2}} a r c t a n (x - \frac{1}{x}) d x$
Commented by mathmax by abdo last updated on 08/Nov/19
$let A =∫_0 ^∞ x e^(−x^2 ) arctan(x−(1/x))dx by parts u^′ =xe^(−x^2 ) and v =arctan(x−(1/x)) ⇒ A =[−(1/2)e^(−x^2 ) arctan(x−(1/x))]_0 ^(+∞) −∫_0 ^∞ (−(1/2)e^(−x^2 ) )×((1+(1/x^2 ))/(1+(x−(1/x))^2 ))dx =−(π/4) +(1/2)∫_0 ^∞ (((x^2 +1)e^(−x^2 ) )/(x^2 {1+(((x^2 −1)^2 )/x^2 )}))dx =−(π/4) +(1/2)∫_0 ^∞ (((x^2 +1)e^(−x^2 ) )/(x^2 +(x^2 −1)^2 )) =−(π/4) +(1/2)∫_0 ^∞ (((x^2 +1)e^(−x^2 ) )/(+x^4 −x^2 +1))dx we have ∫_0 ^∞ (((x^2 +1)e^(−x^2 ) )/(x^4 −x^2 +1))dx =(1/2)∫_(−∞) ^(+∞) (((x^2 +)e^(−x^2 ) )/(x^4 −x^2 +1)) let ϕ(z)=(((z^2 +1)e^(−z^2 ) )/(z^4 −z^2 +1)) poles of ϕ? z^4 −z^2 +1 =0 ⇒t^2 −t +1=0 (t=z^2 ) Δ=1−4=−3 ⇒t_1 =((1+i(√3))/2) =e^((iπ)/3) and t_2 =((1−i(√3))/2) =e^(−((iπ)/3)) z^4 −z^2 +1=(z^2 −e^((iπ)/3) )(z^2 −e^(−((iπ)/3)) ) ⇒ ϕ(z) =(((z^2 +1)e^(−z^2 ) )/((z^2 −e^((iπ)/3) )(z^2 −e^(−((iπ)/3)) ))) =(((z^2 +1)e^(−z^2 ) )/((z−e^((iπ)/6) )(z+e^((iπ)/6) )(z−e^(−((iπ)/6)) )(z+e^(−((iπ)/6)) ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,e^((iπ)/6) )+Res(ϕ,−e^(−((iπ)/6)) )}$
$l e t A = \int_{0}^{\infty} x e^{- x^{2}} a r c t a n (x - \frac{1}{x}) d x b y p a r t s u^{^{'}} = {x e}^{- x^{2}} a n d$ $v = a r c t a n (x - \frac{1}{x}) \Rightarrow$ $A = {[- \frac{1}{2} e^{- x^{2}} a r c t a n (x - \frac{1}{x})]}_{0}^{+ \infty} - \int_{0}^{\infty} (- \frac{1}{2} e^{- x^{2}}) \times \frac{1 + \frac{1}{x^{2}}}{1 + {(x - \frac{1}{x})}^{2}} d x$ $= - \frac{π}{4} + \frac{1}{2} \int_{0}^{\infty} \frac{(x^{2} + 1) e^{- x^{2}}}{x^{2} {1 + \frac{{(x^{2} - 1)}^{2}}{x^{2}}}} d x$ $= - \frac{π}{4} + \frac{1}{2} \int_{0}^{\infty} \frac{(x^{2} + 1) e^{- x^{2}}}{x^{2} + {(x^{2} - 1)}^{2}} = - \frac{π}{4} + \frac{1}{2} \int_{0}^{\infty} \frac{(x^{2} + 1) e^{- x^{2}}}{+ x^{4} - x^{2} + 1} d x$ $w e h a v e \int_{0}^{\infty} \frac{(x^{2} + 1) e^{- x^{2}}}{x^{4} - x^{2} + 1} d x = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{(x^{2} +) e^{- x^{2}}}{x^{4} - x^{2} + 1}$ $l e t φ (z) = \frac{(z^{2} + 1) e^{- z^{2}}}{z^{4} - z^{2} + 1} p o l e s o f φ ?$ $z^{4} - z^{2} + 1 = 0 \Rightarrow t^{2} - t + 1 = 0 (t = z^{2})$ $Δ = 1 - 4 = - 3 \Rightarrow t_{1} = \frac{1 + i \sqrt{3}}{2} = e^{\frac{i π}{3}} a n d t_{2} = \frac{1 - i \sqrt{3}}{2} = e^{- \frac{i π}{3}}$ $z^{4} - z^{2} + 1 = (z^{2} - e^{\frac{i π}{3}}) (z^{2} - e^{- \frac{i π}{3}}) \Rightarrow$ $φ (z) = \frac{(z^{2} + 1) e^{- z^{2}}}{(z^{2} - e^{\frac{i π}{3}}) (z^{2} - e^{- \frac{i π}{3}})}$ $= \frac{(z^{2} + 1) e^{- z^{2}}}{(z - e^{\frac{i π}{6}}) (z + e^{\frac{i π}{6}}) (z - e^{- \frac{i π}{6}}) (z + e^{- \frac{i π}{6}})}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{6}}) + R e s (φ, - e^{- \frac{i π}{6}})}$
Commented by mind is power last updated on 08/Nov/19

$you miss e^{- x^{2}} in lign 4$
Commented by mathmax by abdo last updated on 08/Nov/19
$Res(ϕ,e^((iπ)/6) ) =(((1+e^((iπ)/3) )e^(−e^((iπ)/3) ) )/(2e^((iπ)/6) (e^((iπ)/3) −e^(−((iπ)/3)) ))) =(1/2)e^(−((iπ)/6)) ×(((1+e^((iπ)/3) )e^(−e^((iπ)/3) ) )/(2i((√3)/2))) =(1/(2i(√3)))( e^(−((iπ)/6)) +e^((iπ)/6) )e^(−e^((iπ)/3) ) =((2((√3)/2) e^(−e^((iπ)/3) ) )/(2i(√3))) =(e^(−e^((iπ)/3) ) /(2i)) Res(ϕ,−e^(−((iπ)/6)) ) =(((1+e^(−((iπ)/3)) ) e^(−e^(−((iπ)/3)) ) )/((−2e^(−((iπ)/6)) )(e^(−((iπ)/3)) −e^((iπ)/3) ))) =((e^((iπ)/6) (1+e^(−((iπ)/3)) )e^(−e^(−((iπ)/3)) ) )/(2(2i((√3)/2)))) =((2((√3)/2)e^(−e^(−((iπ)/3)) ) )/(2i(√3))) =(e^(−e^(−((iπ)/3)) ) /(2i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{(e^(−e^((iπ)/3) ) /(2i)) +(e^(−e^(−((iπ)/3)) ) /(2i))} =π{ e^(−cos((π/3))−isin((π/3))) +e^(−cos((π/3))+isin((π/3))) } =π{ e^(−(1/2)−i((√3)/2)) +e^(−(1/2)+i((√3)/2)) } =π{e^(−(1/2)) ( e^((i(√3))/2) +e^(−((i(√3))/2)) )} =(π/(√e)) (2 Re(e^((i(√3))/2) )) e^((i(√3))/2) =cos(((√3)/2))+isin(((√3)/2)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =((2π)/(√e)) cos(((√3)/2)) ⇒ A =−(π/4)+(1/4)×((2π)/(√e)) cos(((√3)/2)) ⇒ A =(π/(2(√e))) cos(((√3)/2))−(π/4)$
$R e s (φ, e^{\frac{i π}{6}}) = \frac{(1 + e^{\frac{i π}{3}}) e^{- e^{\frac{i π}{3}}}}{2 e^{\frac{i π}{6}} (e^{\frac{i π}{3}} - e^{- \frac{i π}{3}})} = \frac{1}{2} e^{- \frac{i π}{6}} \times \frac{(1 + e^{\frac{i π}{3}}) e^{- e^{\frac{i π}{3}}}}{2 i \frac{\sqrt{3}}{2}}$ $= \frac{1}{2 i \sqrt{3}} (e^{- \frac{i π}{6}} + e^{\frac{i π}{6}}) e^{- e^{\frac{i π}{3}}} = \frac{2 \frac{\sqrt{3}}{2} e^{- e^{\frac{i π}{3}}}}{2 i \sqrt{3}} = \frac{e^{- e^{\frac{i π}{3}}}}{2 i}$ $R e s (φ, - e^{- \frac{i π}{6}}) = \frac{(1 + e^{- \frac{i π}{3}}) e^{- e^{- \frac{i π}{3}}}}{(- 2 e^{- \frac{i π}{6}}) (e^{- \frac{i π}{3}} - e^{\frac{i π}{3}})}$ $= \frac{e^{\frac{i π}{6}} (1 + e^{- \frac{i π}{3}}) e^{- e^{- \frac{i π}{3}}}}{2 (2 i \frac{\sqrt{3}}{2})} = \frac{2 \frac{\sqrt{3}}{2} e^{- e^{- \frac{i π}{3}}}}{2 i \sqrt{3}} = \frac{e^{- e^{- \frac{i π}{3}}}}{2 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{e^{- e^{\frac{i π}{3}}}}{2 i} + \frac{e^{- e^{- \frac{i π}{3}}}}{2 i}}$ $= π {e^{- c o s (\frac{π}{3}) - i s i n (\frac{π}{3})} + e^{- c o s (\frac{π}{3}) + i s i n (\frac{π}{3})}}$ $= π {e^{- \frac{1}{2} - i \frac{\sqrt{3}}{2}} + e^{- \frac{1}{2} + i \frac{\sqrt{3}}{2}}}$ $= π {e^{- \frac{1}{2}} (e^{\frac{i \sqrt{3}}{2}} + e^{- \frac{i \sqrt{3}}{2}})} = \frac{π}{\sqrt{e}} (2 R e (e^{\frac{i \sqrt{3}}{2}}))$ $e^{\frac{i \sqrt{3}}{2}} = c o s (\frac{\sqrt{3}}{2}) + i s i n (\frac{\sqrt{3}}{2}) \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = \frac{2 π}{\sqrt{e}} c o s (\frac{\sqrt{3}}{2}) \Rightarrow$ $A = - \frac{π}{4} + \frac{1}{4} \times \frac{2 π}{\sqrt{e}} c o s (\frac{\sqrt{3}}{2}) \Rightarrow A = \frac{π}{2 \sqrt{e}} c o s (\frac{\sqrt{3}}{2}) - \frac{π}{4}$
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