calculate ∫_0 ^(+∞) (dx/((x^4 +1)^3 ))


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Question Number 121492 by mathmax by abdo last updated on 08/Nov/20

$calculate \int_{0}^{+ \infty} \frac{dx}{{(x^{4} + 1)}^{3}}$
	Answered by mnjuly1970 last updated on 09/Nov/20

	$s o l u t i o n :$ $Θ \overset{x^{4} = y}{=} \frac{1}{4} \int_{0}^{\infty} \frac{y^{(\frac{- 3}{4} + 1) - 1} d y}{{(y + 1)}^{3}} = \frac{1}{4} β (\frac{1}{4}, 3 - \frac{1}{4})$ $= \frac{1}{4} \frac{Γ (\frac{1}{4}) Γ (3 - \frac{1}{4})}{Γ (3)} = \frac{1}{4} * \frac{\frac{7}{4} Γ (\frac{7}{4}) Γ (\frac{1}{4})}{2}$ $= \frac{1}{4} * \frac{\frac{7}{4} * \frac{3}{4} Γ (\frac{3}{4}) Γ (\frac{1}{4})}{2}$ $= \frac{21}{128} * \frac{π}{s i n (\frac{π}{4})} = \frac{42 π}{128 \sqrt{2}}$ $= \frac{21 π}{64 \sqrt{2}}$
	Answered by MJS_new last updated on 09/Nov/20

	$\int \frac{d x}{{(x^{4} + 1)}^{3}} =$ $[Ostrogradski]$ $= \frac{x (7 x^{4} + 11)}{32 {(x^{4} + 1)}^{2}} + \frac{21}{32} \int \frac{d x}{x^{4} + 1} =$ $= \frac{x (7 x^{4} + 11)}{32 {(x^{4} + 1)}^{2}} + \frac{21 \sqrt{2}}{256} \ln \frac{x^{2} + \sqrt{2} x + 1}{x^{2} - \sqrt{2} x + 1} + \frac{21 \sqrt{2}}{128} (\tan^{- 1} (\sqrt{2} x - 1) + \tan^{- 1} (\sqrt{2} x + 1)) + C$ $\Rightarrow answer is \frac{21 \sqrt{2}}{128} π$
	Answered by mathmax by abdo last updated on 09/Nov/20
	$A =∫_0 ^∞ (dx/((x^4 +1)^3 )) ⇒2A =∫_(−∞) ^(+∞) (dx/((x^4 +1)^3 )) =∫_(−∞) ^(+∞) (dx/((x^2 −i)^2 (x^2 +i)^2 )) let ϕ(z) =(1/((z^2 −i)^2 (z^2 +i)^2 )) ⇒ϕ(z)=(1/((z−(√i))^2 (z+(√i))^2 (z−(√(−i)))^2 (z+(√(−i)))^2 )) =(1/((z−e^((iπ)/4) )^2 (z+e^((iπ)/4) )^2 (z−e^(−((iπ)/4)) )^2 (z+e^(−((iπ)/4)) )^2 )) the poles of ϕ are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) (doubles) so ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,e^((iπ)/4) ) +Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^((iπ)/4) ) =lim_(z→e^((iπ)/4) ) (1/((2−1)!)){(z−e^((iπ)/4) )^2 ϕ(z)}^((1)) =lim_(z→e^((iπ)/4) ) {(1/((z+e^((iπ)/4) )^2 (z^2 +i)^2 ))}^((1)) =−lim_(z→e^((iπ)/4) ) ((2(z+e^((iπ)/4) )(z^2 +i)^2 +4z(z^2 +i)(z+e^((iπ)/4) )^2 )/((z+e^((iπ)/4) )^4 (z^2 +i)^4 )) =−lim_(z→e^((iπ)/4) ) ((2(z^2 +i)+4z(z+e^((iπ)/4) ))/((z+e^((iπ)/4) )^3 (z^2 +i)^3 )) =−((2(2i)+4e^((iπ)/4) ×2e^((iπ)/4) )/((2e^((iπ)/4) )^3 (2i)^3 )) =−((4i+8i)/(2^3 (−8i)))e^(−((i3π)/4)) =((12)/(64)) e^(−((i3π)/4)) rest to find Res(ϕ,−e^(−((iπ)/4)) )....be continued....$
	$A = \int_{0}^{\infty} \frac{dx}{{(x^{4} + 1)}^{3}} \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{4} + 1)}^{3}} = \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} - i)}^{2} {(x^{2} + i)}^{2}}$ $let φ (z) = \frac{1}{{(z^{2} - i)}^{2} {(z^{2} + i)}^{2}} \Rightarrow φ (z) = \frac{1}{{(z - \sqrt{i})}^{2} {(z + \sqrt{i})}^{2} {(z - \sqrt{- i})}^{2} {(z + \sqrt{- i})}^{2}}$ $= \frac{1}{{(z - e^{\frac{i π}{4}})}^{2} {(z + e^{\frac{i π}{4}})}^{2} {(z - e^{- \frac{i π}{4}})}^{2} {(z + e^{- \frac{i π}{4}})}^{2}}$ $the poles of φ are \bar{+} e^{\frac{i π}{4}} and \bar{+} e^{- \frac{i π}{4}} (doubles) so$ $\int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, e^{\frac{i π}{4}}) + Res (φ, - e^{- \frac{i π}{4}})}$ $Res (φ, e^{\frac{i π}{4}}) = \lim_{z \to e^{\frac{i π}{4}}} \frac{1}{(2 - 1)!} {{(z - e^{\frac{i π}{4}})}^{2} φ (z)}^{(1)}$ $= \lim_{z \to e^{\frac{i π}{4}}} {\frac{1}{{(z + e^{\frac{i π}{4}})}^{2} {(z^{2} + i)}^{2}}}^{(1)}$ $= - \lim_{z \to e^{\frac{i π}{4}}} \frac{2 (z + e^{\frac{i π}{4}}) {(z^{2} + i)}^{2} + 4 z (z^{2} + i) {(z + e^{\frac{i π}{4}})}^{2}}{{(z + e^{\frac{i π}{4}})}^{4} {(z^{2} + i)}^{4}}$ $= - \lim_{z \to e^{\frac{i π}{4}}} \frac{2 (z^{2} + i) + 4 z (z + e^{\frac{i π}{4}})}{{(z + e^{\frac{i π}{4}})}^{3} {(z^{2} + i)}^{3}}$ $= - \frac{2 (2 i) + {4 e}^{\frac{i π}{4}} \times {2 e}^{\frac{i π}{4}}}{{({2 e}^{\frac{i π}{4}})}^{3} {(2 i)}^{3}} = - \frac{4 i + 8 i}{2^{3} (- 8 i)} e^{- \frac{i 3 π}{4}} = \frac{12}{64} e^{- \frac{i 3 π}{4}}$ $rest to find Res (φ, - e^{- \frac{i π}{4}}) . . . . be continued . . . .$
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