prove that ∫_(−∞) ^∞ x^2 e^(−x^2 ) cos(x^2 )sin(x^2 ) dx = (((√π)sin((3/2)tan^(−1) (2)))/(4 ((125))^(1/4) )) anyone can help me with this please


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Question Number 60611 by aliesam last updated on 22/May/19

$p r o v e t h a t \underset{- \infty}{\int^{\infty}} x^{2} e^{- x^{2}} c o s (x^{2}) s i n (x^{2}) d x = \frac{\sqrt{π} s i n (\frac{3}{2} {t a n}^{- 1} (2))}{4 \sqrt[4]{125}}$ $a n y o n e c a n h e l p m e w i t h t h i s p l e a s e$
	Answered by Smail last updated on 22/May/19

	$A = \int_{- \infty}^{\infty} x^{2} e^{- x^{2}} c o s (x^{2}) s i n (x^{2}) d x$ $= 2 \int_{0}^{\infty} x^{2} e^{- x^{2}} \frac{1}{2} s i n (2 x^{2}) d x$ $= I m (- \int_{0}^{\infty} x^{2} e^{- x^{2}} \times e^{- 2 {i x}^{2}} d x)$ $\int_{0}^{\infty} x^{2} e^{- x^{2} (2 i + 1)} d x = \int_{0}^{\infty} x \times {x e}^{- x^{2} (2 i + 1)} d x$ $B y p a r t s$ $u = x \Rightarrow u^{'} = 1$ $v^{'} = {x e}^{- x^{2} (2 i + 1)} \Rightarrow v = \frac{- 1}{2 (2 i + 1)} e^{- x^{2} (2 i + 1)}$ $\int_{0}^{\infty} x^{2} e^{- x^{2} (2 i + 1)} d x = \frac{- 1}{2 (2 i + 1)} {[{x e}^{- x^{2} (2 i + 1)}]}_{0}^{\infty} + \frac{1}{2 (2 i + 1)} \int_{0}^{\infty} e^{- x^{2} (2 i + 1)} d x$ $l e t t = \sqrt{2 i + 1} x \Rightarrow d x = \frac{d t}{\sqrt{2 i + 1}}$ $\int_{0}^{\infty} x^{2} e^{- x^{2} (2 i + 1)} d x = \frac{1}{2 {(2 i + 1)}^{3 / 2}} \int_{0}^{\infty} e^{- t^{2}} d t$ $= \frac{1}{2 {(2 i + 1)}^{3 / 2}} \times \frac{\sqrt{π}}{2} = \frac{\sqrt{π}}{4 {(\sqrt{5} (\frac{1}{\sqrt{5}} + i \frac{2}{\sqrt{5}}))}^{3 / 2}}$ $= \frac{\sqrt{π}}{4 \times {(\sqrt{5})}^{3 / 2} {(e^{{i t a n}^{- 1} (2)})}^{3 / 2}} = \frac{\sqrt{π}}{4 \times 5^{3 / 4} e^{i \frac{3}{2} {t a n}^{- 1} (2)}}$ $= \frac{\sqrt{π}}{4 \sqrt[4]{125}} e^{- i \frac{3}{2} {t a n}^{- 1} (2)}$ $I m (- \frac{\sqrt{π}}{4 \sqrt[4]{125}} e^{- i \frac{3}{2} {t a n}^{- 1} (2)}) = \frac{\sqrt{π}}{4 \sqrt[4]{125}} s i n (\frac{3}{2} {t a n}^{- 1} (2))$ $\int_{- \infty}^{\infty} x^{2} e^{- x^{2}} c o s (x^{2}) s i n (x^{2}) d x = \frac{\sqrt{π}}{4 \sqrt[5]{125}} s i n (\frac{3}{2} {t a n}^{- 1} (2))$
	Commented by aliesam last updated on 22/May/19

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