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Question Number 66082 by aliesam last updated on 08/Aug/19

Commented by mathmax by abdo last updated on 09/Aug/19
$let I =∫_(−∞) ^(+∞) xsin(x^3 )dx ⇒I =2∫_0 ^∞ x sin(x^3 )dx =−2 Im(∫_0 ^∞ x e^(−ix^3 ) dx) changement ix^3 =t give x^3 =−it ⇒ x =(−it)^(1/3) ⇒∫_0 ^∞ x e^(−ix^3 ) dx =∫_0 ^∞ (−it)^(1/3) e^(−t) (−i)^(1/3) (1/3)t^((1/3)−1) dt =(1/3)(−i)^(2/3) ∫_0 ^∞ t^((2/3)−1) e^(−t) dt =(1/3)(e^(−((iπ)/2)) )^(2/3) Γ((2/3)) =(1/3) e^(−((iπ)/3)) Γ((2/3)) =(1/3)Γ((2/3)){(1/2)−((√3)/2)i} =(1/6)Γ((2/3))−((Γ((2/3)))/(2(√3)))i ⇒ I =−2×(−(1/(2(√3)))Γ((2/3))) ⇒ I =((Γ((2/3)))/(√3)) .$
$l e t I = \int_{- \infty}^{+ \infty} x s i n (x^{3}) d x \Rightarrow I = 2 \int_{0}^{\infty} x s i n (x^{3}) d x$ $= - 2 I m (\int_{0}^{\infty} x e^{- {i x}^{3}} d x) c h a n g e m e n t {i x}^{3} = t g i v e x^{3} = - i t \Rightarrow$ $x = {(- i t)}^{\frac{1}{3}} \Rightarrow \int_{0}^{\infty} x e^{- {i x}^{3}} d x = \int_{0}^{\infty} {(- i t)}^{\frac{1}{3}} e^{- t} {(- i)}^{\frac{1}{3}} \frac{1}{3} t^{\frac{1}{3} - 1} d t$ $= \frac{1}{3} {(- i)}^{\frac{2}{3}} \int_{0}^{\infty} t^{\frac{2}{3} - 1} e^{- t} d t = \frac{1}{3} {(e^{- \frac{i π}{2}})}^{\frac{2}{3}} Γ (\frac{2}{3})$ $= \frac{1}{3} e^{- \frac{i π}{3}} Γ (\frac{2}{3}) = \frac{1}{3} Γ (\frac{2}{3}) {\frac{1}{2} - \frac{\sqrt{3}}{2} i} = \frac{1}{6} Γ (\frac{2}{3}) - \frac{Γ (\frac{2}{3})}{2 \sqrt{3}} i \Rightarrow$ $I = - 2 \times (- \frac{1}{2 \sqrt{3}} Γ (\frac{2}{3})) \Rightarrow I = \frac{Γ (\frac{2}{3})}{\sqrt{3}} .$
Commented by aliesam last updated on 09/Aug/19

$g o d b l e s s y o u$
Commented by mathmax by abdo last updated on 09/Aug/19

$y o u a r e w e l c o m e s i r .$
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