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


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Question Number 80451 by abdomathmax last updated on 03/Feb/20

$c a l c u l a t e \int_{0}^{\infty} \frac{c o s (π x)}{{(x^{2} + 3)}^{2}} d x$
Commented by abdomathmax last updated on 04/Feb/20
$let I=∫_0 ^∞ ((cos(πx))/((x^2 +3)^2 ))dx changement x=(√3)t give I=∫_0 ^∞ ((cos(π(√3)t))/(9(t^2 +1)))×(√3)dt=((√3)/9)∫_0 ^∞ ((cos(π(√3)t))/((t^2 +1)^2 ))dt =((√3)/(18))∫_(−∞) ^(+∞) ((cos(π(√3)t))/((t^2 +1)^2 ))dt =((√3)/(18))Re( ∫_(−∞) ^(+∞) (e^(iπ(√3)t) /((t^2 +1)^2 ))dt) letϕ(z)=(e^(iπ(√3)z) /((z^2 +1)^2 )) ⇒ϕ(z)=(e^(iπ(√3)z) /((z−i)^2 (z+i)^2 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i)=lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { (e^(iπ(√3)z) /((z+i)^2 ))}^((1)) =lim_(z→i) ((iπ(√3)e^(iπ(√3)z) (z+i)^2 −2(z+i)e^(iπ(√3)z) )/((z+i)^4 )) =lim_(z→i) ((iπ(√3)(z+i)e^(iπ(√3)z) −2e^(iπ(√3)z) )/((z+i)^3 )) =lim_(z→i) (({iπ(√3)(z+i)−2}e^(iπ(√3)z) )/((z+i)^3 )) =(({(2i)iπ(√3)−2}e^(−π(√3)) )/((2i)^3 )) =(({−2π(√3)−2}e^(−π(√3)) )/(−8i)) =(((π(√3)+1)e^(−π(√3)) )/(4i)) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ×(((π(√3)+1)e^(−π(√3)) )/(4i)) =(π/2)(1+π(√3))e^(−π(√3)) ⇒I=((√3)/(18))×(π/2)(1+π(√3))e^(−π(√3)) =((π(√3))/(36))(1+π(√3))e^(−π(√3))$
$l e t I = \int_{0}^{\infty} \frac{c o s (π x)}{{(x^{2} + 3)}^{2}} d x c h a n g e m e n t x = \sqrt{3} t g i v e$ $I = \int_{0}^{\infty} \frac{c o s (π \sqrt{3} t)}{9 (t^{2} + 1)} \times \sqrt{3} d t = \frac{\sqrt{3}}{9} \int_{0}^{\infty} \frac{c o s (π \sqrt{3} t)}{{(t^{2} + 1)}^{2}} d t$ $= \frac{\sqrt{3}}{18} \int_{- \infty}^{+ \infty} \frac{c o s (π \sqrt{3} t)}{{(t^{2} + 1)}^{2}} d t = \frac{\sqrt{3}}{18} R e (\int_{- \infty}^{+ \infty} \frac{e^{i π \sqrt{3} t}}{{(t^{2} + 1)}^{2}} d t)$ $l e t φ (z) = \frac{e^{i π \sqrt{3} z}}{{(z^{2} + 1)}^{2}} \Rightarrow φ (z) = \frac{e^{i π \sqrt{3} z}}{{(z - i)}^{2} {(z + i)}^{2}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{e^{i π \sqrt{3} z}}{{(z + i)}^{2}}}^{(1)}$ $= {l i m}_{z \to i} \frac{i π \sqrt{3} e^{i π \sqrt{3} z} {(z + i)}^{2} - 2 (z + i) e^{i π \sqrt{3} z}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{i π \sqrt{3} (z + i) e^{i π \sqrt{3} z} - 2 e^{i π \sqrt{3} z}}{{(z + i)}^{3}}$ $= {l i m}_{z \to i} \frac{{i π \sqrt{3} (z + i) - 2} e^{i π \sqrt{3} z}}{{(z + i)}^{3}}$ $= \frac{{(2 i) i π \sqrt{3} - 2} e^{- π \sqrt{3}}}{{(2 i)}^{3}}$ $= \frac{{- 2 π \sqrt{3} - 2} e^{- π \sqrt{3}}}{- 8 i} = \frac{(π \sqrt{3} + 1) e^{- π \sqrt{3}}}{4 i} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \times \frac{(π \sqrt{3} + 1) e^{- π \sqrt{3}}}{4 i}$ $= \frac{π}{2} (1 + π \sqrt{3}) e^{- π \sqrt{3}}$ $\Rightarrow I = \frac{\sqrt{3}}{18} \times \frac{π}{2} (1 + π \sqrt{3}) e^{- π \sqrt{3}}$ $= \frac{π \sqrt{3}}{36} (1 + π \sqrt{3}) e^{- π \sqrt{3}}$
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