prove that ∫_(−∞) ^(+∞) x^2 e^(−x^2 ) cos(x^2 )sin(x^2 ) dx =(((√π)sin[(((√3)tan^(−1) (2))/2)])/(4 ((125))^(1/4) ))


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Question Number 129764 by Eric002 last updated on 18/Jan/21

$p r o v e t h a t$ $\int_{- \infty}^{+ \infty} x^{2} e^{- x^{2}} c o s (x^{2}) s i n (x^{2}) d x$ $= \frac{\sqrt{π} s i n [\frac{\sqrt{3} {t a n}^{- 1} (2)}{2}]}{4 \sqrt[4]{125}}$
	Answered by Dwaipayan Shikari last updated on 18/Jan/21

	$\int_{- \infty}^{\infty} x^{2} e^{- x^{2}} c o s (x^{2}) s i n (x^{2}) d x$ $= \frac{1}{4 i} \int_{- \infty}^{\infty} x^{2} e^{- x^{2}} e^{2 {i x}^{2}} - x^{2} e^{- x^{2}} e^{- 2 {i x}^{2}} d x x^{2} (1 + 2 i) = j$ $= \frac{1}{4 i} \int_{- \infty}^{\infty} x^{2} e^{- x^{2} (1 - 2 i)} - x^{2} e^{- x^{2} (1 + 2 i)} d x x^{2} (1 - 2 i) = u$ $= \frac{1}{8 i (1 - 2 i)} \int_{- \infty}^{\infty} {(\frac{u}{1 - 2 i})}^{\frac{1}{2}} e^{- u} d u - \frac{1}{8 i (1 + 2 i)} \int_{- \infty}^{\infty} {(\frac{j}{1 - 2 j})}^{\frac{1}{2}} e^{- j} d j$ $= \frac{1}{4 i \sqrt{{(1 - 2 i)}^{3}}} Γ (\frac{3}{2}) - \frac{1}{4 i \sqrt{{(1 + 2 i)}^{3}}} Γ (\frac{3}{2})$ $= \frac{\sqrt{π}}{8 i} ({(1 - 2 i)}^{- \frac{3}{2}} - {(1 + 2 i)}^{- \frac{3}{2}}) 8 i$ $= \frac{\sqrt{π}}{8 i} (5^{- \frac{3}{4}} e^{\frac{3 i}{2} {t a n}^{- 1} (2)} - 5^{- \frac{3}{4}} e^{- \frac{3 i}{2} {t a n}^{- 1} (2)})$ $= \frac{\sqrt{π}}{8 i} (2 i s i n (\frac{3}{2} {t a n}^{- 1} (2)) \frac{1}{\sqrt[4]{125}} = \frac{\sqrt{π}}{4 \sqrt[4]{125}} s i n (\frac{3}{2} {t a n}^{- 1} (2))$
	Commented by Eric002 last updated on 18/Jan/21

	$t h a n k y o u s i r$
	Answered by mathmax by abdo last updated on 18/Jan/21
	$A=∫_(−∞) ^(+∞) x^2 e^(−x^2 ) cos(x^2 )sin(x^2 )dx we have cos(x^2 )sin(x^2 )=cos(x^2 )cos((π/2)−x^2 )=(1/2){cos((π/2))+cos(2x^2 −(π/2))} =(1/2)sin(2x^2 ) ⇒A=(1/2)∫_(−∞) ^(+∞) x^2 e^(−x^2 ) sin(2x^2 )dx ⇒2A=Im(∫_(−∞) ^(+∞) x^2 e^(−x^2 +2ix^2 ) dx) but ∫_(−∞) ^(+∞) x^2 e^((−1+2i)x^2 ) dx =2∫_0 ^∞ x^2 e^((−1+2i)x^2 ) dx =_(x=(√t)) 2∫_0 ^∞ t .e^((−1+2i)t) (dt/(2(√t))) =∫_0 ^∞ t^(1/2) .e^((−1+2i)t) dt =_((1−2i)t=z) ∫_0 ^∞ (−(z/(−1+2i)))^(1/2) e^(−z) ×((−dz)/(−1+2i)) =∫_0 ^∞ (z^(1/2) /( (√(1−2i))(1−2i)))e^(−z) dz =(1/((1−2i)^(3/2) ))×Γ((3/2)) 1−2i=(√5)e^(iarctan(−2)) =(√5)e^(−iarctan(2)) ⇒ (1−2i)^(3/2) =((√5))^(3/2) e^(−i(3/2)arctan(2)) ⇒ ∫_(−∞) ^(+∞) x^2 e^((−1+2i)x^2 ) dx =(e^((3/2)iarctan(2)) /5^(3/4) )×((√π)/2) ⇒ 2A=(1/5^(3/4) ) sin((3/2)arctan(2))×((√π)/2) ⇒A=(1/(4(^4 (√(125)))))sin((3/2)arctan(2))$
	$A = \int_{- \infty}^{+ \infty} x^{2} e^{- x^{2}} \cos (x^{2}) \sin (x^{2}) dx we have$ $\cos (x^{2}) \sin (x^{2}) = \cos (x^{2}) \cos (\frac{π}{2} - x^{2}) = \frac{1}{2} {\cos (\frac{π}{2}) + \cos ({2 x}^{2} - \frac{π}{2})}$ $= \frac{1}{2} \sin ({2 x}^{2}) \Rightarrow A = \frac{1}{2} \int_{- \infty}^{+ \infty} x^{2} e^{- x^{2}} \sin ({2 x}^{2}) dx$ $\Rightarrow 2 A = Im (\int_{- \infty}^{+ \infty} x^{2} e^{- x^{2} + {2 ix}^{2}} dx) but$ $\int_{- \infty}^{+ \infty} x^{2} e^{(- 1 + 2 i) x^{2}} dx = 2 \int_{0}^{\infty} x^{2} e^{(- 1 + 2 i) x^{2}} dx$ $=_{x = \sqrt{t}} 2 \int_{0}^{\infty} t . e^{(- 1 + 2 i) t} \frac{dt}{2 \sqrt{t}}$ $= \int_{0}^{\infty} t^{\frac{1}{2}} . e^{(- 1 + 2 i) t} dt =_{(1 - 2 i) t = z} \int_{0}^{\infty} {(- \frac{z}{- 1 + 2 i})}^{\frac{1}{2}} e^{- z} \times \frac{- dz}{- 1 + 2 i}$ $= \int_{0}^{\infty} \frac{z^{\frac{1}{2}}}{\sqrt{1 - 2 i} (1 - 2 i)} e^{- z} dz = \frac{1}{{(1 - 2 i)}^{\frac{3}{2}}} \times Γ (\frac{3}{2})$ $1 - 2 i = \sqrt{5} e^{iarctan (- 2)} = \sqrt{5} e^{- iarctan (2)} \Rightarrow$ ${(1 - 2 i)}^{\frac{3}{2}} = {(\sqrt{5})}^{\frac{3}{2}} e^{- i \frac{3}{2} \arctan (2)} \Rightarrow$ $\int_{- \infty}^{+ \infty} x^{2} e^{(- 1 + 2 i) x^{2}} dx = \frac{e^{\frac{3}{2} iarctan (2)}}{5^{\frac{3}{4}}} \times \frac{\sqrt{π}}{2} \Rightarrow$ $2 A = \frac{1}{5^{\frac{3}{4}}} \sin (\frac{3}{2} \arctan (2)) \times \frac{\sqrt{π}}{2} \Rightarrow A = \frac{1}{4 (^{4} \sqrt{125})} \sin (\frac{3}{2} \arctan (2))$
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