calculate ∫_0 ^∞ ((cos(2πx))/((x^2 +3)^2 ))dx


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Question Number 74349 by mathmax by abdo last updated on 22/Nov/19

$c a l c u l a t e \int_{0}^{\infty} \frac{c o s (2 π x)}{{(x^{2} + 3)}^{2}} d x$
Commented by abdomathmax last updated on 23/Nov/19
$let I =∫_0 ^∞ ((cos(2πx))/((x^2 +3)^2 ))dx changement x=(√3)t give I =∫_0 ^∞ ((cos(2π(√3)t))/(9(t^2 +1)^2 ))×(√3)dt =((√3)/9) ∫_0 ^∞ ((cos(2π(√3)t))/((t^2 +1)^2 ))dt ⇒2I=((√3)/9) ∫_(−∞) ^(+∞) ((cos(2π(√3)t))/((t^2 +1)^2 ))dt= ((√3)/9) Re(∫_(−∞) ^(+∞) (e^(i(2π(√3))t) /((t^2 +1)^2 ))dx) let W(z)=(e^(i(2π(√3))z) /((z^2 +1)^2 )) ⇒ W(z)=(e^(i(2π(√3))z) /((z−i)^2 (z+i)^2 )) ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,i) Res(W,i)= lim_(z→i) (1/((2−1)!)){ (z−i)^2 W(z)}^((1)) =lim_(z→i) { (e^(i(2π(√3))z) /((z+i)^2 ))}^((1)) =lim_(z→i) ((2iπ(√3)e^(i(2π(√3))z) (z+i)^2 −2(z+i)e^(2iπ(√3)z) )/((z+i)^4 )) =lim_(z→i) ((2iπ(√3)e^(i(2π(√3))z) (z+i)−2 e^(2iπ(√3)z) )/((z+i)^3 )) =((2iπ(√3)e^(−2π(√3)) (2i)−2e^(−2π(√3)) )/(−8i)) =(((−4π(√3)−2)e^(−2π(√3)) )/(+8i)) =(((2π(√3)+1)e^(−2π(√3)) )/(4i)) ⇒ ∫_(−∞) ^(+∞) W(z)dz =(π/2)(2π(√3)+1)e^(−2π(√3)) ⇒ I=((√3)/(18))×(π/2)(2π(√3)+1)e^(−2π(√3))$
$l e t I = \int_{0}^{\infty} \frac{c o s (2 π x)}{{(x^{2} + 3)}^{2}} d x c h a n g e m e n t x = \sqrt{3} t g i v e$ $I = \int_{0}^{\infty} \frac{c o s (2 π \sqrt{3} t)}{9 {(t^{2} + 1)}^{2}} \times \sqrt{3} d t = \frac{\sqrt{3}}{9} \int_{0}^{\infty} \frac{c o s (2 π \sqrt{3} t)}{{(t^{2} + 1)}^{2}} d t$ $\Rightarrow 2 I = \frac{\sqrt{3}}{9} \int_{- \infty}^{+ \infty} \frac{c o s (2 π \sqrt{3} t)}{{(t^{2} + 1)}^{2}} d t = \frac{\sqrt{3}}{9} R e (\int_{- \infty}^{+ \infty} \frac{e^{i (2 π \sqrt{3}) t}}{{(t^{2} + 1)}^{2}} d x)$ $l e t W (z) = \frac{e^{i (2 π \sqrt{3}) z}}{{(z^{2} + 1)}^{2}} \Rightarrow W (z) = \frac{e^{i (2 π \sqrt{3}) z}}{{(z - i)}^{2} {(z + i)}^{2}}$ $\int_{- \infty}^{+ \infty} W (z) d z = 2 i π R e s (W, i)$ $R e s (W, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} W (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{e^{i (2 π \sqrt{3}) z}}{{(z + i)}^{2}}}^{(1)}$ $= {l i m}_{z \to i} \frac{2 i π \sqrt{3} e^{i (2 π \sqrt{3}) z} {(z + i)}^{2} - 2 (z + i) e^{2 i π \sqrt{3} z}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{2 i π \sqrt{3} e^{i (2 π \sqrt{3}) z} (z + i) - 2 e^{2 i π \sqrt{3} z}}{{(z + i)}^{3}}$ $= \frac{2 i π \sqrt{3} e^{- 2 π \sqrt{3}} (2 i) - 2 e^{- 2 π \sqrt{3}}}{- 8 i}$ $= \frac{(- 4 π \sqrt{3} - 2) e^{- 2 π \sqrt{3}}}{+ 8 i} = \frac{(2 π \sqrt{3} + 1) e^{- 2 π \sqrt{3}}}{4 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} W (z) d z = \frac{π}{2} (2 π \sqrt{3} + 1) e^{- 2 π \sqrt{3}} \Rightarrow$ $I = \frac{\sqrt{3}}{18} \times \frac{π}{2} (2 π \sqrt{3} + 1) e^{- 2 π \sqrt{3}}$
Commented by abdomathmax last updated on 23/Nov/19

$I = \frac{π \sqrt{3}}{36} (2 π \sqrt{3} + 1) e^{- 2 π \sqrt{3}}$
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