calculate I = ∫_(−∞) ^(+∞) ((x+1)/((x^2 +1)^2 ))dx .


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Question Number 36432 by prof Abdo imad last updated on 02/Jun/18

$c a l c u l a t e I = \int_{- \infty}^{+ \infty} \frac{x + 1}{{(x^{2} + 1)}^{2}} d x .$
Commented by abdo mathsup 649 cc last updated on 03/Jun/18
$let introduce the complex function ϕ(z) = ((z+1)/((z^2 +1)^2 )) we have ϕ(z) = ((z+1)/( (z−i)^2 (z+i)^2 )) so the poles of ϕ are i and−i (doubles) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((2−1)!)){ (z−i)^2 ϕ(z)}^′ =lim_(z→i) { ((z+1)/((z+i)^2 ))}^′ = lim_(z→i) (((z+i)^2 −2(z+i)(z+1))/((z+i)^4 )) =lim_(z→i) ((z+i −2z−2)/((z+i)^3 )) = ((−2)/((2i)^3 )) = ((−2)/(−8i)) = (1/(4i)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (1/(4i)) = (π/2) so ★ I = (π/2) ★$
$l e t i n t r o d u c e t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{z + 1}{{(z^{2} + 1)}^{2}} w e h a v e$ $φ (z) = \frac{z + 1}{{(z - i)}^{2} {(z + i)}^{2}} s o t h e p o l e s o f φ a r e i a n d - i$ $(d o u b l e s)$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{^{'}}$ $= {l i m}_{z \to i} {\frac{z + 1}{{(z + i)}^{2}}}^{^{'}}$ $= {l i m}_{z \to i} \frac{{(z + i)}^{2} - 2 (z + i) (z + 1)}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{z + i - 2 z - 2}{{(z + i)}^{3}}$ $= \frac{- 2}{{(2 i)}^{3}} = \frac{- 2}{- 8 i} = \frac{1}{4 i}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{1}{4 i} = \frac{π}{2} s o$ $★ I = \frac{π}{2} ★$
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Jun/18
	$x=tan(θ dx=sec^2 θdθ ∫_((−Π)/2) ^(Π/2) (((tanθ +1)×sec^2 θ)/(sec^2 θ×sec^2 θ)) dθ ∫_(−(Π/2)) ^(Π/2) { ((sinθ)/(cosθ))×cos^2 θ+cos^2 θ}dθ ∫_(−(Π/2)) ^(Π/2) sinθd(sinθ)+∫_(−(Π/2)) ^(Π/2) ((1+cos2θ)/2)dθ =∣((sin^2 θ)/2)∣_(−(Π/2)) ^(Π/2) +(1/2)∣θ∣_(−(Π/2)) ^(Π/2) +(1/2)×(1/2)∣sin2θ∣_(−(Π/2)) ^(Π/2) =0+(1/2)(Π)+(1/4)×0 =(Π/2)$
	$x = t a n (θ$ $d x = {s e c}^{2} θ d θ$ $\int_{\frac{- Π}{2}}^{\frac{Π}{2}} \frac{(t a n θ + 1) \times {s e c}^{2} θ}{{s e c}^{2} θ \times {s e c}^{2} θ} d θ$ $\int_{- \frac{Π}{2}}^{\frac{Π}{2}} {\frac{s i n θ}{c o s θ} \times {c o s}^{2} θ + {c o s}^{2} θ} d θ$ $\int_{- \frac{Π}{2}}^{\frac{Π}{2}} s i n θ d (s i n θ) + \int_{- \frac{Π}{2}}^{\frac{Π}{2}} \frac{1 + c o s 2 θ}{2} d θ$ $=∣ \frac{{s i n}^{2} θ}{2} ∣_{- \frac{Π}{2}}^{\frac{Π}{2}} + \frac{1}{2} ∣ θ ∣_{- \frac{Π}{2}}^{\frac{Π}{2}} + \frac{1}{2} \times \frac{1}{2} ∣ s i n 2 θ ∣_{- \frac{Π}{2}}^{\frac{Π}{2}}$ $= 0 + \frac{1}{2} (Π) + \frac{1}{4} \times 0$ $= \frac{Π}{2}$
	Commented by abdo mathsup 649 cc last updated on 03/Jun/18

	$c o r r e c t a n s w e r s i r T a n m a y t h a n k s . .$
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