prove-that-x-2-e-x-2-cos-x-2-sin-x-2-dx-pi-sin-3-tan-1-2-2-4-125-1-4-

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 = \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))