find the value of ∫_0 ^∞ ((sin(x^2 ))/(x^4 +4))dx


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Question Number 56698 by maxmathsup by imad last updated on 21/Mar/19

$f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{s i n (x^{2})}{x^{4} + 4} d x$
Commented by maxmathsup by imad last updated on 23/Mar/19
$let A =∫_0 ^∞ ((sin(x^2 ))/(x^4 +4)) dx ⇒2A = ∫_(−∞) ^(+∞) ((sin(x^2 ))/(x^4 +4))dx =Im (∫_(−∞) ^(+∞) (e^(ix^2 ) /(x^4 +4))dx) let but ∫_(−∞) ^(+∞) (e^(ix^2 ) /(x^4 +4)) dx =_(x=(√2)t) ∫_(−∞) ^(+∞) (e^(i2t^2 ) /(4t^4 +4)) (√2)dt =((√2)/4) ∫_(−∞) ^(+∞) (e^(2it^2 ) /(t^4 +1)) dt let ϕ(z) = (e^(2iz^2 ) /(z^4 +1)) ⇒ϕ(z) =(e^(2iz^2 ) /((z^2 −i)(z^2 +i))) =(e^(2iz^2 ) /((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) so the poles of ϕ are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ, e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^((iπ)/4) ) = (e^(2i(i)) /(2e^((iπ)/4) (2i))) =(e^(−2) /(4i)) e^(−((iπ)/4)) Res(ϕ,−e^(−((iπ)/4)) ) = (e^(2i(−i)) /(−2e^(−((iπ)/4)) (−2i))) =(e^2 /(4i)) e^((iπ)/4) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ {(e^(−2) /(4i)) e^(−((iπ)/4)) +(e^2 /(4i)) e^((iπ)/4) } =(π/2) { e^(−2) {(1/(√2)) −(i/(√2))} +e^2 {(1/(√2)) +(i/(√2))}} =(π/2){ (1/(√2))(e^2 +e^(−2) ) +(i/(√2))(e^2 −e^(−2) )} ⇒ 2A =(π/(2(√2)))(e^2 −e^(−2) ) ⇒ ★A =(π/(4(√2))) (e^2 −e^(−2) ) ★$
$l e t A = \int_{0}^{\infty} \frac{s i n (x^{2})}{x^{4} + 4} d x \Rightarrow 2 A = \int_{- \infty}^{+ \infty} \frac{s i n (x^{2})}{x^{4} + 4} d x = I m (\int_{- \infty}^{+ \infty} \frac{e^{{i x}^{2}}}{x^{4} + 4} d x) l e t$ $b u t \int_{- \infty}^{+ \infty} \frac{e^{{i x}^{2}}}{x^{4} + 4} d x =_{x = \sqrt{2} t} \int_{- \infty}^{+ \infty} \frac{e^{i 2 t^{2}}}{4 t^{4} + 4} \sqrt{2} d t = \frac{\sqrt{2}}{4} \int_{- \infty}^{+ \infty} \frac{e^{2 {i t}^{2}}}{t^{4} + 1} d t l e t$ $φ (z) = \frac{e^{2 {i z}^{2}}}{z^{4} + 1} \Rightarrow φ (z) = \frac{e^{2 {i z}^{2}}}{(z^{2} - i) (z^{2} + i)} = \frac{e^{2 {i z}^{2}}}{(z - e^{\frac{i π}{4}}) (z + e^{\frac{i π}{4}}) (z - e^{- \frac{i π}{4}}) (z + e^{- \frac{i π}{4}})}$ $s o t h e p o l e s o f φ a r e \bar{+} e^{\frac{i π}{4}} a n d \bar{+} e^{- \frac{i π}{4}} r e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {R e s (φ, e^{\frac{i π}{4}}) + R e s (φ, - e^{- \frac{i π}{4}})}$ $R e s (φ, e^{\frac{i π}{4}}) = \frac{e^{2 i (i)}}{2 e^{\frac{i π}{4}} (2 i)} = \frac{e^{- 2}}{4 i} e^{- \frac{i π}{4}}$ $R e s (φ, - e^{- \frac{i π}{4}}) = \frac{e^{2 i (- i)}}{- 2 e^{- \frac{i π}{4}} (- 2 i)} = \frac{e^{2}}{4 i} e^{\frac{i π}{4}} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π {\frac{e^{- 2}}{4 i} e^{- \frac{i π}{4}} + \frac{e^{2}}{4 i} e^{\frac{i π}{4}}} = \frac{π}{2} {e^{- 2} {\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}} + e^{2} {\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}}}$ $= \frac{π}{2} {\frac{1}{\sqrt{2}} (e^{2} + e^{- 2}) + \frac{i}{\sqrt{2}} (e^{2} - e^{- 2})} \Rightarrow 2 A = \frac{π}{2 \sqrt{2}} (e^{2} - e^{- 2}) \Rightarrow$ $★ A = \frac{π}{4 \sqrt{2}} (e^{2} - e^{- 2}) ★$
Commented by maxmathsup by imad last updated on 23/Mar/19

$e r r o r a t f i n a l l i n e s 2 A = \frac{\sqrt{2}}{4} \frac{π}{2 \sqrt{2}} (e^{2} - e^{- 2}) = \frac{π}{8} (e^{2} - e^{- 2}) \Rightarrow$ $A = \frac{π}{16} (e^{2} - e^{- 2}) .$
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