calculate ∫_(−∞) ^(+∞) (dx/((x^2 +2x+2)^2 ))


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Question Number 161839 by mathmax by abdo last updated on 23/Dec/21

$calculate \int_{- \infty}^{+ \infty} \frac{dx}{{(x^{2} + 2 x + 2)}^{2}}$
	Answered by ArielVyny last updated on 23/Dec/21

	$I = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 2 x + 2)}^{2}} = \int_{- \infty}^{+ \infty} \frac{d x}{{[{(x + 1)}^{2} + 1]}^{2}}$ $l e t u s c o n s i d e r u = x + 1$ $\int_{- \infty}^{+ \infty} \frac{d u}{{(u^{2} + 1)}^{2}} = 2 \int_{0}^{\infty} \frac{1}{{(u^{2} + 1)}^{2}} d u$ $u^{2} = t \to 2 u d u = d t \to d u = \frac{d t}{2 \sqrt{t}}$ $I = \int_{0}^{\infty} \frac{t^{- \frac{1}{2}}}{{(1 + t)}^{2}} d t$ $k n o w i n g t h a t \int_{0}^{\infty} \frac{x^{m - 1}}{{(1 + x)}^{m + n}} = β (m, n)$ $m - 1 = - \frac{1}{2} \to m = \frac{1}{2}$ $m + n = 2 \to n = 2 - \frac{1}{2} = \frac{3}{2}$ $I = \int_{0}^{\infty} \frac{t^{- \frac{1}{2}}}{{(1 + t)}^{2}} d t = β (\frac{1}{2}, \frac{3}{2}) = Γ (\frac{3}{2}) Γ (\frac{1}{2})$ $I = \frac{1}{2} π$
	Answered by Ar Brandon last updated on 23/Dec/21

	$I = \int \frac{d x}{{(x^{2} + 2 x + 2)}^{2}}, Ostrogradsky gives;$ $= \frac{a x + b}{x^{2} + 2 x + 2} + \int \frac{c}{x^{2} + 2 x + 2} d x$ $= \frac{x + 1}{2 (x^{2} + 2 x + 2)} + \frac{1}{2} \int \frac{d x}{x^{2} + 2 x + 2}$ $= \frac{x + 1}{2 (x^{2} + 2 x + 2)} + \frac{1}{2} \int \frac{d x}{{(x + 1)}^{2} + 1}$ $= \frac{x + 1}{2 (x^{2} + 2 x + 2)} + \frac{\arctan (x + 1)}{2} + C$ $\Rightarrow \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 2 x + 2)}^{2}} = \frac{π}{2}$
	Answered by Ar Brandon last updated on 24/Mar/22
	$I=∫_(−∞) ^(+∞) (dx/((x^2 +2x+2)^2 )) , ϕ(z)=(1/((z^2 +2z+2)^2 )) Poles: z_1 =−1+i=(√2)e^(((3π)/4)i) , z_2 =(√2)e^(−(π/4)i) I=∫_(−∞) ^(+∞) ϕ(z)dz=2iπRes(ϕ, z_1 ) Res (ϕ, z_1 )=lim_(z→z_1 ) (d/dz){(z−z_1 )^2 ϕ(z)}=lim_(z→z_1 ) (d/dz){(1/((z−z_2 )^2 ))} =lim_(z→z_1 ) {−(2/((z−z_2 )^3 ))}=−(2/((z_1 −z_2 )^3 ))=−(2/((2i)^3 ))=(1/(4i)) ⇒∫_(−∞) ^(+∞) (dx/((x^2 +2x+2)^2 ))=2πi((1/(4i)))=(π/2)$
	$I = \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 2 x + 2)}^{2}}, φ (z) = \frac{1}{{(z^{2} + 2 z + 2)}^{2}}$ $Poles : z_{1} = - 1 + i = \sqrt{2} e^{\frac{3 π}{4} i}, z_{2} = \sqrt{2} e^{- \frac{π}{4} i}$ $I = \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{1})$ $R e s (φ, z_{1}) = \lim_{z \to z_{1}} \frac{d}{d z} {{(z - z_{1})}^{2} φ (z)} = \lim_{z \to z_{1}} \frac{d}{d z} {\frac{1}{{(z - z_{2})}^{2}}}$ $= \lim_{z \to z_{1}} {- \frac{2}{{(z - z_{2})}^{3}}} = - \frac{2}{{(z_{1} - z_{2})}^{3}} = - \frac{2}{{(2 i)}^{3}} = \frac{1}{4 i}$ $\Rightarrow \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 2 x + 2)}^{2}} = 2 π i (\frac{1}{4 i}) = \frac{π}{2}$
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