calculate ∫_0 ^∞ ((x^2 cos(πx))/((x^2 +4)^2 ))dx


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Question Number 38724 by maxmathsup by imad last updated on 28/Jun/18

$c a l c u l a t e \int_{0}^{\infty} \frac{x^{2} c o s (π x)}{{(x^{2} + 4)}^{2}} d x$
Commented by maxmathsup by imad last updated on 30/Jun/18
$let I =∫_0 ^∞ ((x^2 cos(πx))/((x^2 +4)^2 ))dx 2I = ∫_(−∞) ^(+∞) ((x^2 cos(πx))/((x^2 +4)^2 ))dx =Re( ∫_(−∞) ^(+∞) ((x^2 e^(iπx) )/((x^2 +4)^2 ))) let ϕ(z)=((z^2 e^(iπz) )/((z^2 +4)^2 )) we have ϕ(z) =((z^2 e^(iπz) )/((z−2i)^2 (z+2i)^2 )) so the poles of ϕ are 2i and −2i ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,2i) but Res(ϕ,2i) =lim_(z→2i) (1/((2−1)!)) { (z−2i)^2 ϕ(z)}^((1)) =lim_(z→2i) { ((z^2 e^(iπz) )/((z+2i)^2 ))}^((1)) =lim_(z→2i) { (((2z e^(iπz) +iπz^2 e^(iπz) )(z+2i)^2 −2(z+2i)z^2 e^(iπz) )/((z+2i)^4 ))} lim_(z→2i) (((2z+iπz^2 )e^(iπz) (z+2i)−2z^2 e^(iπz) )/((z+2i)^3 )) = (((4i −4iπ)e^(−2π) (4i) +8 e^(−2π) )/((4i)^3 )) =(((−16 +16 π +8)e^(−2π) )/(−64i)) =(((8−16π)e^(−2π) )/(64i)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (((8−16π)e^(−2π) )/(64i)) =π (((8−16π)e^(−2π) )/(32)) =(π/(32)) 8(1−2π)e^(−2π) =(π/4)(1−2π)e^(−2π) ⇒$
$l e t I = \int_{0}^{\infty} \frac{x^{2} c o s (π x)}{{(x^{2} + 4)}^{2}} d x$ $2 I = \int_{- \infty}^{+ \infty} \frac{x^{2} c o s (π x)}{{(x^{2} + 4)}^{2}} d x = R e (\int_{- \infty}^{+ \infty} \frac{x^{2} e^{i π x}}{{(x^{2} + 4)}^{2}}) l e t$ $φ (z) = \frac{z^{2} e^{i π z}}{{(z^{2} + 4)}^{2}} w e h a v e φ (z) = \frac{z^{2} e^{i π z}}{{(z - 2 i)}^{2} {(z + 2 i)}^{2}} s o t h e p o l e s o f φ a r e 2 i a n d - 2 i$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, 2 i) b u t$ $R e s (φ, 2 i) = {l i m}_{z \to 2 i} \frac{1}{(2 - 1)!} {{(z - 2 i)}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to 2 i} {\frac{z^{2} e^{i π z}}{{(z + 2 i)}^{2}}}^{(1)} = {l i m}_{z \to 2 i} {\frac{(2 z e^{i π z} + i π z^{2} e^{i π z}) {(z + 2 i)}^{2} - 2 (z + 2 i) z^{2} e^{i π z}}{{(z + 2 i)}^{4}}}$ ${l i m}_{z \to 2 i} \frac{(2 z + i π z^{2}) e^{i π z} (z + 2 i) - 2 z^{2} e^{i π z}}{{(z + 2 i)}^{3}}$ $= \frac{(4 i - 4 i π) e^{- 2 π} (4 i) + 8 e^{- 2 π}}{{(4 i)}^{3}} = \frac{(- 16 + 16 π + 8) e^{- 2 π}}{- 64 i} = \frac{(8 - 16 π) e^{- 2 π}}{64 i}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{(8 - 16 π) e^{- 2 π}}{64 i} = π \frac{(8 - 16 π) e^{- 2 π}}{32}$ $= \frac{π}{32} 8 (1 - 2 π) e^{- 2 π} = \frac{π}{4} (1 - 2 π) e^{- 2 π} \Rightarrow$
Commented by maxmathsup by imad last updated on 30/Jun/18

$2 I = \frac{π}{4} (1 - 2 π) e^{- 2 π} \Rightarrow I = \frac{π}{8} (1 - 2 π) e^{- 2 π} .$ $π$
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