prove that Ω =∫_(−∞) ^( +∞) (( cos (x))/((2+ 2x +x^( 2) )^( 2) )) dx = (π/e) cos(1)


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Question Number 162073 by mnjuly1970 last updated on 26/Dec/21

$p r o v e t h a t$ $Ω = \int_{- \infty}^{+ \infty} \frac{c o s (x)}{{(2 + 2 x + x^{2})}^{2}} d x = \frac{π}{e} c o s (1)$
	Answered by mindispower last updated on 26/Dec/21

	$Ω = \int_{- \infty}^{\infty} \frac{c o s (x)}{{(1 + {(1 + x)}^{2})}^{2}} d x$ $1 + x = t$ $= \int_{- \infty}^{\infty} \frac{c o s (t - 1)}{{(1 + t^{2})}^{2}} d t = \int_{- \infty}^{\infty} \frac{c o s (t) c o s (1)}{{(1 + t^{2})}^{2}} d t + s i n (1) \int_{- \infty}^{\infty} \frac{s i n (t)}{{(1 + t^{2})}^{2}} {d t}_{= 0}$ $= c o s (1) \int_{- \infty}^{\infty} \frac{c o s (t) d t}{{(1 + t^{2})}^{2}}$ $f (a) = \int_{- \infty}^{\infty} \frac{c o s (t)}{a^{2} + t^{2}} = R e \int_{- \infty}^{\infty} \frac{e^{i t}}{a^{2} + t^{2}} = R e (2 i π . e^{- a} . \frac{1}{2 i a})$ $= \frac{π}{a} e^{- a} = f (a)$ $f^{'} (a) = \int_{- \infty}^{\infty} \frac{- 2 a c o s (t)}{{(a^{2} + t^{2})}^{2}} d t$ $f^{'} (1) = - 2 \int_{- \infty}^{\infty} \frac{c o s (t) d t}{{(1 + t^{2})}^{2}}$ $f^{'} (a) = (- \frac{π}{a^{2}} e^{- a} - \frac{π}{a} e^{- a})$ $\int_{- \infty}^{\infty} \frac{c o s (t) d t}{{(1 + t^{2})}^{2}} = \frac{1}{2} . \frac{2 π}{e} = \frac{π}{e}$ $\int_{- \infty}^{\infty} \frac{c o s t)}{{(2 + 2 t + t^{2})}^{2}} d t = c o s (1) \int \frac{c o s (x)}{{(1 + x^{2})}^{2}} d x = c o s (1) \frac{π}{e}$
	Commented by Tawa11 last updated on 26/Dec/21

	$Great sir$
	Commented by mnjuly1970 last updated on 26/Dec/21

	$e x c e l l e n t s o l u t i o n s i r p o w e r . . .$
	Answered by Mathspace last updated on 26/Dec/21
	$Υ=∫_(−∞) ^(+∞) ((cosx)/((x^2 +2x+2)^2 ))dx Υ=Re(∫_(−∞) ^(+∞) (e^(ix) /((x^2 +2x+2)^2 ))dx) let f(z)=(e^(iz) /((z^2 +2z+2)^2 )) poles of f? z^(2 ) +2z+2=0→Δ^′ =1−2=−1 ⇒ z_1 =−1+i and z_2 =−1−i ∫_R f(z)dz =2iπRes(f,z_1 ) z_1 pole double ⇒ Res(f,z_1 )=lim_(z→z_1 ) (1/((2−1)!)){(z−z_1 )^2 f(z)}^((1)) =lim_(z→z_1 ) { (e^(iz) /((z−z_2 )^2 ))}^((1)) =lim_(z→z_1 ) ((ie^(iz) (z−z_2 )^2 −2(z−z_2 )e^(iz) )/((z−z_2 )^4 )) =lim_(z→z_1 ) (({i(z−z_2 )−2}e^(iz) )/((z−z_2 )^3 )) =(({i(z_1 −z_2 )−2}e^(iz1) )/((z_1 −z_2 )^3 )) =(({i(2i)−2}e^(i(−1+i)) )/((2i)^3 )) =(({−2−2}e^(−1) {cos(1)−isin(1)})/(−8i)) =(1/(4i))e^(−1) {cos(1)−isin(1)} ⇒ ∫_(−∞) ^(+∞) f(z)dz=((2iπ)/(4i))e^(−1) {cos(1)−isin(1)} =(π/2)e^(−1) {cos(1)−isin(1)} and Υ=Re(...) =(π/(2e))cos(1)$
	$Υ = \int_{- \infty}^{+ \infty} \frac{c o s x}{{(x^{2} + 2 x + 2)}^{2}} d x$ $Υ = R e (\int_{- \infty}^{+ \infty} \frac{e^{i x}}{{(x^{2} + 2 x + 2)}^{2}} d x)$ $l e t f (z) = \frac{e^{i z}}{{(z^{2} + 2 z + 2)}^{2}} p o l e s o f f ?$ $z^{2} + 2 z + 2 = 0 \to Δ^{^{'}} = 1 - 2 = - 1 \Rightarrow$ $z_{1} = - 1 + i a n d z_{2} = - 1 - i$ $\int_{R} f (z) d z = 2 i π R e s (f, z_{1})$ $z_{1} p o l e d o u b l e \Rightarrow$ $R e s (f, z_{1}) = {l i m}_{z \to z_{1}} \frac{1}{(2 - 1)!} {{(z - z_{1})}^{2} f (z)}^{(1)}$ $= {l i m}_{z \to z_{1}} {\frac{e^{i z}}{{(z - z_{2})}^{2}}}^{(1)}$ $= {l i m}_{z \to z_{1}} \frac{{i e}^{i z} {(z - z_{2})}^{2} - 2 (z - z_{2}) e^{i z}}{{(z - z_{2})}^{4}}$ $= {l i m}_{z \to z_{1}} \frac{{i (z - z_{2}) - 2} e^{i z}}{{(z - z_{2})}^{3}}$ $= \frac{{i (z_{1} - z_{2}) - 2} e^{i z 1}}{{(z_{1} - z_{2})}^{3}}$ $= \frac{{i (2 i) - 2} e^{i (- 1 + i)}}{{(2 i)}^{3}}$ $= \frac{{- 2 - 2} e^{- 1} {c o s (1) - i s i n (1)}}{- 8 i}$ $= \frac{1}{4 i} e^{- 1} {c o s (1) - i s i n (1)} \Rightarrow$ $\int_{- \infty}^{+ \infty} f (z) d z = \frac{2 i π}{4 i} e^{- 1} {c o s (1) - i s i n (1)}$ $= \frac{π}{2} e^{- 1} {c o s (1) - i s i n (1)}$ $a n d Υ = R e (. . .)$ $= \frac{π}{2 e} c o s (1)$
	Commented by mnjuly1970 last updated on 26/Dec/21

	$t h a n k s a l o t m a s t e r$
	Answered by Ar Brandon last updated on 24/Mar/22
	$Ω=∫_(−∞) ^(+∞) ((cosx)/((x^2 +2x+2)^2 ))dx ϕ(z)=(e^(iz) /((z^2 +2z+2)^2 )) , Poles: z_1 =((−2+2i)/2)=(√2)e^(((3π)/4)i) , z_2 =(√2)e^(−(π/4)i) Ω=Re∫_(−∞) ^(+∞) ϕ(z)dz=Re(2iπRes(ϕ, z_1 )) Res (ϕ, z_1 )=lim_(z→z_1 ) (d/dz){(z−z_1 )^2 ϕ(z)}=lim_(z→z_1 ) (d/dz){(e^(iz) /((z−z_2 )^2 ))} =lim_(z→z_1 ) {((ie^(iz) (z−z_2 )^2 −2e^(iz) (z−z_2 ))/((z−z_2 )^4 ))}=((ie^(iz_1 ) (z_1 −z_2 )^2 −2e^(iz_1 ) (z_1 −z_2 ))/((z_1 −z_2 )^4 )) =((ie^(i(−1+i)) (2i)^2 −2e^(i(−1+i)) (2i))/(16))=−((4ie^(−(1+i)) +4ie^(−(1+i)) )/(16)) =−(1/(2e))×e^(((π/2)−1)i) =−(1/(2e))(sin(1)+icos(1)) ⇒Re(2iπRes(ϕ, z_1 )=(π/e)cos(1) ⇒ determinant (((∫_(−∞) ^(+∞) ((cosx)/((x^2 +2x+2)^2 ))dx=(π/e)cos(1))))$
	$Ω = \int_{- \infty}^{+ \infty} \frac{\cos x}{{(x^{2} + 2 x + 2)}^{2}} d x$ $φ (z) = \frac{e^{i z}}{{(z^{2} + 2 z + 2)}^{2}}, Poles : z_{1} = \frac{- 2 + 2 i}{2} = \sqrt{2} e^{\frac{3 π}{4} i}, z_{2} = \sqrt{2} e^{- \frac{π}{4} i}$ $Ω = R e \int_{- \infty}^{+ \infty} φ (z) d z = R e (2 i π R e s (φ, z_{1}))$ $R e s (φ, z_{1}) = \lim_{z \to z_{1}} \frac{d}{d z} {{(z - z_{1})}^{2} φ (z)} = \lim_{z \to z_{1}} \frac{d}{d z} {\frac{e^{i z}}{{(z - z_{2})}^{2}}}$ $= \lim_{z \to z_{1}} {\frac{{i e}^{i z} {(z - z_{2})}^{2} - 2 e^{i z} (z - z_{2})}{{(z - z_{2})}^{4}}} = \frac{{i e}^{{i z}_{1}} {(z_{1} - z_{2})}^{2} - 2 e^{{i z}_{1}} (z_{1} - z_{2})}{{(z_{1} - z_{2})}^{4}}$ $= \frac{{i e}^{i (- 1 + i)} {(2 i)}^{2} - 2 e^{i (- 1 + i)} (2 i)}{16} = - \frac{4 {i e}^{- (1 + i)} + 4 {i e}^{- (1 + i)}}{16}$ $= - \frac{1}{2 e} \times e^{(\frac{π}{2} - 1) i} = - \frac{1}{2 e} (\sin (1) + i \cos (1))$ $\Rightarrow R e (2 i π R e s (φ, z_{1}) = \frac{π}{e} \cos (1)$ $\Rightarrow \begin{matrix} \int_{- \infty}^{+ \infty} \frac{\cos x}{{(x^{2} + 2 x + 2)}^{2}} d x = \frac{π}{e} \cos (1) \end{matrix}$
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