0-x-x-2-4x-4-

Question Number 162525 by mathlove last updated on 30/Dec/21

\underset{0}{\int^{\infty}} \frac{\sqrt{x}}{(x^{2} + 4 x + 4)} = ?

Answered by Ar Brandon last updated on 24/Mar/22

I = \int_{0}^{\infty} \frac{\sqrt{x}}{x^{2} + 4 x + 4} d x, x = t^{2} \Rightarrow d x = 2 t d t

= 2 \int_{0}^{\infty} \frac{t^{2}}{t^{4} + 4 t^{2} + 4} d t = 2 \int_{0}^{\infty} \frac{t^{2}}{{(t^{2} + 2)}^{2}} d t = \int_{- \infty}^{+ \infty} \frac{t^{2}}{{(t^{2} + 2)}^{2}} d t

φ (z) = \frac{z^{2}}{{(z^{2} + 2)}^{2}}, poles : z_{1} = - i \sqrt{2}, z_{2} = i \sqrt{2}

I = 2 i π R e s (φ, z_{2})

R e s (φ, z_{2}) = \lim_{z \to z_{2}} \frac{d}{d z} {\frac{z^{2}}{{(z - z_{1})}^{2}}} = \lim_{z \to z_{2}} {\frac{2 z {(z - z_{1})}^{2} - 2 z^{2} (z - z_{1})}{{(z - z_{1})}^{4}}}

= \frac{2 z_{2} {(z_{2} - z_{1})}^{2} - 2 z_{2}^{2} (z_{2} - z_{1})}{{(z_{2} - z_{1})}^{4}} = \frac{2 \sqrt{2} i {(2 \sqrt{2} i)}^{2} - 2 {(i \sqrt{2})}^{2} (2 \sqrt{2} i)}{64}

= \frac{- 16 \sqrt{2} i + 8 \sqrt{2} i}{64} = - \frac{\sqrt{2}}{8} i \Rightarrow I = 2 i π (- \frac{\sqrt{2}}{8} i) = \frac{π}{2 \sqrt{2}}

Answered by cortano last updated on 30/Dec/21

\int_{0}^{\infty} \frac{\sqrt{x}}{x^{2} + 4 x + 4} d x = \int_{0}^{\infty} \frac{2 u^{2}}{u^{4} + 4 u^{2} + 4} d u

[u = \sqrt{x}]

= \int_{0}^{\infty} \frac{2 u^{2}}{{(u^{2} + 2)}^{2}} d u = - \frac{u}{u^{2} + 2} ∣_{0}^{\infty} + \int \frac{d u}{u^{2} + 2}

= 0 + \frac{1}{\sqrt{2}} {[\tan^{- 1} (\frac{u}{\sqrt{2}})]}_{0}^{\infty} = \frac{π}{2 \sqrt{2}} = \frac{π \sqrt{2}}{4}

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