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


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Question Number 73333 by mathmax by abdo last updated on 10/Nov/19

$c a l c u l a t e \int_{0}^{\infty} \frac{c o s (π + 2 x^{2})}{{(x^{2} + 4)}^{2}} d x$
Commented by mathmax by abdo last updated on 11/Nov/19
$let A =∫_0 ^∞ ((cos(π+2x^2 ))/((x^2 +4)^2 ))dx ⇒−2A = ∫_(−∞) ^(+∞) ((cos(2x^2 ))/((x^2 +4)^2 ))dx =_(x=2t) ∫_(−∞) ^(+∞) ((cos(8t^2 ))/(16(t^2 +1)^2 ))(2dt) =(1/8) ∫_(−∞) ^(+∞) ((cos(8t^2 ))/((t^2 +1)^2 ))dt ⇒ A =−(1/(16)) ∫_(−∞) ^(+∞) ((cos(8t^2 ))/((t^2 +1)^2 ))dt =−(1/(16)) Re(∫_(−∞) ^(+∞) (e^(8it^2 ) /((t^2 +1)^2 ))dt) let ϕ(z)=(e^(8iz^2 ) /((z^2 +1)^2 )) ⇒ϕ(z)=(e^(8iz^2 ) /((z−i)^2 (z+i)^2 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i)=lim_(z→i) (1/((2−1)!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {(e^(i8z^2 ) /((z+i)^2 ))}^((1)) =lim_(z→i) ((16iz e^(i8z^2 ) (z+i)^2 −2(z+i)e^(8iz^2 ) )/((z+i)^4 )) =lim_(z→i) (({16iz(z+i)−2}e^(8iz^2 ) )/((z+i)^3 )) =(((−16(2i)−2}e^(−8i) )/(−8i)) =(((16i+1)e^(−8i) )/(4i)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ×(((16i+1)e^(−8i) )/(4i)) =(π/2)(1+16i)( cos(8)−isin(8)) =(π/2){cos(8)−isin(8)+16cos(8)i+16 sin(8)} ⇒ A =−(π/(32))( cos(8) +16 sin(8)}$
$l e t A = \int_{0}^{\infty} \frac{c o s (π + 2 x^{2})}{{(x^{2} + 4)}^{2}} d x \Rightarrow - 2 A = \int_{- \infty}^{+ \infty} \frac{c o s (2 x^{2})}{{(x^{2} + 4)}^{2}} d x$ $=_{x = 2 t} \int_{- \infty}^{+ \infty} \frac{c o s (8 t^{2})}{16 {(t^{2} + 1)}^{2}} (2 d t) = \frac{1}{8} \int_{- \infty}^{+ \infty} \frac{c o s (8 t^{2})}{{(t^{2} + 1)}^{2}} d t \Rightarrow$ $A = - \frac{1}{16} \int_{- \infty}^{+ \infty} \frac{c o s (8 t^{2})}{{(t^{2} + 1)}^{2}} d t = - \frac{1}{16} R e (\int_{- \infty}^{+ \infty} \frac{e^{8 {i t}^{2}}}{{(t^{2} + 1)}^{2}} d t)$ $l e t φ (z) = \frac{e^{8 {i z}^{2}}}{{(z^{2} + 1)}^{2}} \Rightarrow φ (z) = \frac{e^{8 {i z}^{2}}}{{(z - i)}^{2} {(z + i)}^{2}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i} {\frac{e^{i 8 z^{2}}}{{(z + i)}^{2}}}^{(1)} = {l i m}_{z \to i} \frac{16 i z e^{i 8 z^{2}} {(z + i)}^{2} - 2 (z + i) e^{8 {i z}^{2}}}{{(z + i)}^{4}}$ $= {l i m}_{z \to i} \frac{{16 i z (z + i) - 2} e^{8 {i z}^{2}}}{{(z + i)}^{3}} = \frac{(- 16 (2 i) - 2} e^{- 8 i}}{- 8 i}$ $= \frac{(16 i + 1) e^{- 8 i}}{4 i} \Rightarrow \int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \times \frac{(16 i + 1) e^{- 8 i}}{4 i}$ $= \frac{π}{2} (1 + 16 i) (c o s (8) - i s i n (8))$ $= \frac{π}{2} {c o s (8) - i s i n (8) + 16 c o s (8) i + 16 s i n (8)} \Rightarrow$ $A = - \frac{π}{32} (c o s (8) + 16 s i n (8)}$
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