calculate ∫_(−∞) ^(+∞) ((2x)/((x^2 +mx +1)^2 ))dx with ∣m∣<2


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Question Number 36056 by abdo mathsup 649 cc last updated on 28/May/18

$c a l c u l a t e \int_{- \infty}^{+ \infty} \frac{2 x}{{(x^{2} + m x + 1)}^{2}} d x w i t h ∣ m ∣< 2$
Commented by abdo mathsup 649 cc last updated on 30/May/18
$let consider the complex function ϕ(z) = ((2z)/((z^2 +mz +1)^2 )) poles of ϕ! roots of z^(2 ) +mz +1 Δ =m^2 −4 <0 because ∣m∣<2 ⇒ Δ =−(4−m^2 ) ={i(√(4−m^2 )) }^2 z_1 = ((−m +i(√(4−m^2 )))/2) and z_2 = ((−m−i(√(4−m^2 )))/2) the poles of ϕ are z_1 and z_2 (doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ, z_1 ) ϕ(z) = ((2z)/((z−z_1 )^2 (z−z_2 )^2 )) Res(ϕ,z_1 ) =lim_(z→z_1 ) (1/((2−1):)){ (z−z_1 )^2 ϕ(z)}^′$
$l e t c o n s i d e r t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{2 z}{{(z^{2} + m z + 1)}^{2}} p o l e s o f φ!$ $r o o t s o f z^{2} + m z + 1$ $Δ = m^{2} - 4 < 0 b e c a u s e ∣ m ∣< 2 \Rightarrow$ $Δ = - (4 - m^{2}) = {i \sqrt{4 - m^{2}}}^{2}$ $z_{1} = \frac{- m + i \sqrt{4 - m^{2}}}{2} a n d z_{2} = \frac{- m - i \sqrt{4 - m^{2}}}{2}$ $t h e p o l e s o f φ a r e z_{1} a n d z_{2} (d o u b l e s)$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, z_{1})$ $φ (z) = \frac{2 z}{{(z - z_{1})}^{2} {(z - z_{2})}^{2}}$ $R e s (φ, z_{1}) = {l i m}_{z \to z_{1}} \frac{1}{(2 - 1) :} {{(z - z_{1})}^{2} φ (z)}^{^{'}}$
Commented by abdo mathsup 649 cc last updated on 30/May/18
$Res(ϕ ,z_1 ) =lim_(z→z_1 ) { ((2z)/((z−z_2 )^2 ))}^′ =lim_(z→z_1 ) ((2(z−z_2 )^2 −2z 2(z−z_2 ))/((z−z_2 )^4 )) =lim_(z→z_1 ) ((2(z−z_2 ) −4z)/((z−z_2 )^3 )) = 2 ((z_1 −z_2 −2z_1 )/((z_1 −z_2 )^3 )) =2((i(√(4−m^2 )) −2((−m+i(√(4−m^2 )))/2))/((i(√(4−m^2 )))^3 )) =2 (m/(−i(4−m^2 )(√(4−m^( )))) = ((−2m)/(i(4−m^2 )(√(4−m^2 )))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((−2m)/(i(4−m^2 )(√(4−m^2 )))) = ((−4mπ)/((4−m^2 )(√(4−m^2 )))) so I = ((−4mπ)/((4−m^2 )(√(4−m^2 )))) .$
$R e s (φ, z_{1}) = {l i m}_{z \to z_{1}} {\frac{2 z}{{(z - z_{2})}^{2}}}^{^{'}}$ $= {l i m}_{z \to z_{1}} \frac{2 {(z - z_{2})}^{2} - 2 z 2 (z - z_{2})}{{(z - z_{2})}^{4}}$ $= {l i m}_{z \to z_{1}} \frac{2 (z - z_{2}) - 4 z}{{(z - z_{2})}^{3}} = 2 \frac{z_{1} - z_{2} - 2 z_{1}}{{(z_{1} - z_{2})}^{3}}$ $= 2 \frac{i \sqrt{4 - m^{2}} - 2 \frac{- m + i \sqrt{4 - m^{2}}}{2}}{{(i \sqrt{4 - m^{2}})}^{3}}$ $= 2 \frac{m}{- i (4 - m^{2}) \sqrt{4 - m^{(}}} = \frac{- 2 m}{i (4 - m^{2}) \sqrt{4 - m^{2}}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{- 2 m}{i (4 - m^{2}) \sqrt{4 - m^{2}}}$ $= \frac{- 4 m π}{(4 - m^{2}) \sqrt{4 - m^{2}}} s o$ $I = \frac{- 4 m π}{(4 - m^{2}) \sqrt{4 - m^{2}}} .$
	Answered by sma3l2996 last updated on 28/May/18

	$I = \int_{- \infty}^{+ \infty} \frac{2 x}{{(x^{2} + m x + 1)}^{2}} d x = \int_{- \infty}^{+ \infty} \frac{2 x + m - m}{{(x^{2} + m x + 1)}^{2}} d x$ $= - {[\frac{1}{x^{2} + m x + 1}]}_{- \infty}^{+ \infty} - m \int_{- \infty}^{+ \infty} \frac{d x}{{(x^{2} + 2 \times \frac{m}{2} \times x + \frac{m^{2}}{4} - \frac{m^{2}}{4} + 1)}^{2}}$ $= 0 - m \int_{- \infty}^{+ \infty} \frac{d x}{{({(x + \frac{m}{2})}^{2} + \frac{4 - m^{2}}{4})}^{2}}$ $= - m \int_{- \infty}^{+ \infty} \frac{d x}{{(\frac{4 - m^{2}}{4} ({(\frac{2 x + m}{\sqrt{4 - m^{2}}})}^{2} + 1))}^{2}} / ∣ m ∣< 2$ $l e t u = \frac{2 x + m}{\sqrt{4 - m^{2}}} \Rightarrow d x = \frac{\sqrt{4 - m^{2}}}{2} d u$ $I = - \frac{8 m \sqrt{4 - m^{2}}}{{(4 - m^{2})}^{2}} \int_{- \infty}^{+ \infty} \frac{d u}{{(u^{2} + 1)}^{2}}$ $t a n t = u \Rightarrow d t = \frac{d u}{1 + u^{2}}$ $\frac{d u}{{(u^{2} + 1)}^{2}} = \frac{d t}{1 + {t a n}^{2} t} = {c o s}^{2} (t) d t = (\frac{1 + c o s (2 t)}{2}) d t$ $\int_{- π / 2}^{π / 2} (1 + c o s (2 t)) d t = {[t + \frac{1}{2} s i n 2 t]}_{- π / 2}^{π / 2} = π$ $I = - \frac{4 π m \sqrt{4 - m^{2}}}{{(4 - m^{2})}^{2}}$
	Commented by abdo mathsup 649 cc last updated on 30/May/18

	$c o r r e c t s i r s m a 3 l t h a n k s a l o t s .$
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