calculate-0-dx-x-2-3-2-x-2-4-2-

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 = \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})