calculate ∫_0 ^∞ (dx/((x^2 +3)^2 (x^2 +4)^2 ))


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Question Number 146835 by mathmax by abdo last updated on 16/Jul/21

$calculate \int_{0}^{\infty} \frac{dx}{{(x^{2} + 3)}^{2} {(x^{2} + 4)}^{2}}$
	Answered by mathmax by abdo last updated on 16/Jul/21
	$Ψ=(1/2)∫_(−∞) ^(+∞) (dx/((x^2 +3)^2 (x^2 +4)^2 )) let ϕ(z)=(1/((z^2 +3)^2 (z^2 +4)^2 )) ⇒ϕ(z)=(1/((z−i(√3))^2 (z+i(√3))^2 (z−2i)^2 (z+2i)^2 )) the poles of ϕ are +^− i(√3) and +^− 2i(doubles) residus ⇒∫_(−∞) ^(+∞) ϕ(z)dz=2iπ{Res(ϕ,i(√3))+Res(ϕ,2i)} Res(ϕ,i(√3))=lim_(z→i(√3)) (1/((2−1)!)){(z−i(√3))^2 ϕ(z)}^((1)) =lim_(z→i(√3)) {(1/((z+i(√3))^2 (z^2 +4)^2 ))} =−lim_(z→i(√3)) ((2(z+i(√3))(z^2 +4)^2 +4z(z^2 +4)(z+i(√3))^2 )/((z+i(√3))^4 (z^2 +4)^4 )) =−lim_(z→i(√3)) ((2(z^2 +4)+4z(z+i(√3)))/((z+i(√3))^3 (z^2 +4)^3 )) =−((2(−3+4)+4i(√3)(2i(√3)))/((2i(√3))^3 (−3+4)^3 ))=−((2−24)/(−8i(3(√3))))=−((22)/(24(√3)i))=−((11)/(12(√3)i)) Res(ϕ,2i)=lim_(z→2i) (1/((2−1)!)){(z−2i)^2 ϕ(z)}^((1)) =lim_(z→2i) {(1/((z^2 +3)^2 (z+2i)^2 ))}^((1)) =−lim_(z→2i) ((4z(z^2 +3)(z+2i)^2 +2(z+2i)(z^2 +3)^2 )/((z^2 +3)^4 (z+2i)^4 )) =−lim_(z→2i) ((4z(z+2i)+2(z^2 +3))/((z^2 +3)^3 (z+2i)^3 )) =−(((8i)(4i)+2(−4+3))/((−4+3)^3 (4i)^3 ))=−((−32−2)/((−1)(−48i)))=−((−34)/(48i))=((34)/(48i)) ⇒ ⇒∫_(−∞) ^(+∞) ϕ(z)dz=2iπ{−((11)/(12(√3)i))+((17)/(24i))} =−((11π)/(6(√3))) +((17π)/(12)) =(((17)/(12))−((11)/(6(√3))))π ⇒Ψ=(π/2)(((17)/(12))−((11)/(6(√3))))$
	$Ψ = \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} + 3)}^{2} {(x^{2} + 4)}^{2}} let φ (z) = \frac{1}{{(z^{2} + 3)}^{2} {(z^{2} + 4)}^{2}}$ $\Rightarrow φ (z) = \frac{1}{{(z - i \sqrt{3})}^{2} {(z + i \sqrt{3})}^{2} {(z - 2 i)}^{2} {(z + 2 i)}^{2}}$ $the poles of φ are \bar{+} i \sqrt{3} and \bar{+} 2 i (doubles)$ $residus \Rightarrow \int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {Res (φ, i \sqrt{3}) + Res (φ, 2 i)}$ $Res (φ, i \sqrt{3}) = \lim_{z \to i \sqrt{3}} \frac{1}{(2 - 1)!} {{(z - i \sqrt{3})}^{2} φ (z)}^{(1)}$ $= \lim_{z \to i \sqrt{3}} {\frac{1}{{(z + i \sqrt{3})}^{2} {(z^{2} + 4)}^{2}}}$ $= - \lim_{z \to i \sqrt{3}} \frac{2 (z + i \sqrt{3}) {(z^{2} + 4)}^{2} + 4 z (z^{2} + 4) {(z + i \sqrt{3})}^{2}}{{(z + i \sqrt{3})}^{4} {(z^{2} + 4)}^{4}}$ $= - \lim_{z \to i \sqrt{3}} \frac{2 (z^{2} + 4) + 4 z (z + i \sqrt{3})}{{(z + i \sqrt{3})}^{3} {(z^{2} + 4)}^{3}}$ $= - \frac{2 (- 3 + 4) + 4 i \sqrt{3} (2 i \sqrt{3})}{{(2 i \sqrt{3})}^{3} {(- 3 + 4)}^{3}} = - \frac{2 - 24}{- 8 i (3 \sqrt{3})} = - \frac{22}{24 \sqrt{3} i} = - \frac{11}{12 \sqrt{3} i}$ $Res (φ, 2 i) = \lim_{z \to 2 i} \frac{1}{(2 - 1)!} {{(z - 2 i)}^{2} φ (z)}^{(1)}$ $= \lim_{z \to 2 i} {\frac{1}{{(z^{2} + 3)}^{2} {(z + 2 i)}^{2}}}^{(1)}$ $= - \lim_{z \to 2 i} \frac{4 z (z^{2} + 3) {(z + 2 i)}^{2} + 2 (z + 2 i) {(z^{2} + 3)}^{2}}{{(z^{2} + 3)}^{4} {(z + 2 i)}^{4}}$ $= - \lim_{z \to 2 i} \frac{4 z (z + 2 i) + 2 (z^{2} + 3)}{{(z^{2} + 3)}^{3} {(z + 2 i)}^{3}}$ $= - \frac{(8 i) (4 i) + 2 (- 4 + 3)}{{(- 4 + 3)}^{3} {(4 i)}^{3}} = - \frac{- 32 - 2}{(- 1) (- 48 i)} = - \frac{- 34}{48 i} = \frac{34}{48 i} \Rightarrow$ $\Rightarrow \int_{- \infty}^{+ \infty} φ (z) dz = 2 i π {- \frac{11}{12 \sqrt{3} i} + \frac{17}{24 i}}$ $= - \frac{11 π}{6 \sqrt{3}} + \frac{17 π}{12} = (\frac{17}{12} - \frac{11}{6 \sqrt{3}}) π \Rightarrow Ψ = \frac{π}{2} (\frac{17}{12} - \frac{11}{6 \sqrt{3}})$
	Answered by Olaf_Thorendsen last updated on 16/Jul/21

	$f (a, b) = \int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2}) (x^{2} + b^{2})} (1)$ $f (a, b) = \frac{1}{b^{2} - a^{2}} \int_{0}^{\infty} (\frac{1}{x^{2} + a^{2}} - \frac{1}{x^{2} + b^{2}}) d x$ $f (a, b) = \frac{1}{b^{2} - a^{2}} {[\frac{1}{a} \arctan \frac{x}{a} - \frac{1}{b} \arctan \frac{x}{b}]}_{0}^{\infty}$ $f (a, b) = \frac{1}{b^{2} - a^{2}} . \frac{π}{2} (\frac{1}{a} - \frac{1}{b})$ $f (a, b) = \frac{π}{2 a b (a + b)} (2)$ $(1) : \frac{\partial f (a, b)}{\partial a} = - 2 a \int_{0}^{\infty} \frac{d x}{{(x^{2} + a^{2})}^{2} (x^{2} + b^{2})}$ $\frac{\partial^{2} f (a, b)}{\partial a \partial b} = 4 a b \int_{0}^{\infty} \frac{d x}{{(x^{2} + a^{2})}^{2} {(x^{2} + b^{2})}^{2}}$ $\frac{\partial^{2} f (a, b)}{\partial a \partial b} = 4 a b f (a, b) (3)$ $(2) : \frac{\partial f (a, b)}{\partial a} = - \frac{π (2 a + b)}{2 a^{2} b {(a + b)}^{2}}$ $\frac{\partial f^{2} (a, b)}{\partial a \partial b} = \frac{π}{a^{2} b^{2}} [\frac{a^{3} + 4 a^{2} b + 4 {a b}^{2} + b^{3}}{{(a + b)}^{4}}] (4)$ $(3) and (4) :$ $f (a, b) = \frac{π}{4 a^{3} b^{3}} [\frac{a^{3} + 4 a^{2} b + 4 {a b}^{2} + b^{3}}{{(a + b)}^{4}}]$ $Ω = \int_{0}^{\infty} \frac{d x}{(x^{2} + 3) (x^{2} + 4)} = f (\sqrt{3}, 2)$ $\Rightarrow Ω = \frac{π}{96 \sqrt{3}} [\frac{19 \sqrt{3} + 32}{{(\sqrt{3} + 2)}^{4}}]$ $Ω = \frac{π}{96 \sqrt{3}} (\frac{32 + 19 \sqrt{3}}{97 + 56 \sqrt{3}})$ $Ω = \frac{π}{96 \sqrt{3}} (51 \sqrt{3} - 88)$ $Ω = \frac{π}{4} (\frac{17}{8} - \frac{11 \sqrt{3}}{9})$
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