Question Number 60611 by aliesam last updated on 22/May/19
$${prove}\:{that}\:\underset{−\infty} {\overset{\infty} {\int}}{x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } {cos}\left({x}^{\mathrm{2}} \right){sin}\left({x}^{\mathrm{2}} \right)\:{dx}\:=\:\frac{\sqrt{\pi}{sin}\left(\frac{\mathrm{3}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)\right)}{\mathrm{4}\:\sqrt[{\mathrm{4}}]{\mathrm{125}}} \\ $$$${anyone}\:{can}\:{help}\:{me}\:{with}\:{this}\:{please} \\ $$$$ \\ $$
Answered by Smail last updated on 22/May/19
![A=∫_(−∞) ^∞ x^2 e^(−x^2 ) cos(x^2 )sin(x^2 )dx =2∫_0 ^∞ x^2 e^(−x^2 ) (1/2)sin(2x^2 )dx =Im(−∫_0 ^∞ x^2 e^(−x^2 ) ×e^(−2ix^2 ) dx) ∫_0 ^∞ x^2 e^(−x^2 (2i+1)) dx=∫_0 ^∞ x×xe^(−x^2 (2i+1)) dx By parts u=x⇒u′=1 v′=xe^(−x^2 (2i+1)) ⇒v=((−1)/(2(2i+1)))e^(−x^2 (2i+1)) ∫_0 ^∞ x^2 e^(−x^2 (2i+1)) dx=((−1)/(2(2i+1)))[xe^(−x^2 (2i+1)) ]_0 ^∞ +(1/(2(2i+1)))∫_0 ^∞ e^(−x^2 (2i+1)) dx let t=(√(2i+1))x⇒dx=(dt/( (√(2i+1)))) ∫_0 ^∞ x^2 e^(−x^2 (2i+1)) dx=(1/(2(2i+1)^(3/2) ))∫_0 ^∞ e^(−t^2 ) dt =(1/(2(2i+1)^(3/2) ))×((√π)/2)=((√π)/(4((√5)((1/( (√5)))+i(2/( (√5)))))^(3/2) )) =((√π)/(4×((√5))^(3/2) (e^(itan^(−1) (2)) )^(3/2) ))=((√π)/(4×5^(3/4) e^(i(3/2)tan^(−1) (2)) )) =((√π)/(4((125))^(1/4) ))e^(−i(3/2)tan^(−1) (2)) Im(−((√π)/(4((125))^(1/4) ))e^(−i(3/2)tan^(−1) (2)) )=((√π)/(4((125))^(1/4) ))sin((3/2)tan^(−1) (2)) ∫_(−∞) ^∞ x^2 e^(−x^2 ) cos(x^2 )sin(x^2 )dx=((√π)/(4((125))^(1/5) ))sin((3/2)tan^(−1) (2))](https://www.tinkutara.com/question/Q60619.png)
$${A}=\int_{−\infty} ^{\infty} {x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } {cos}\left({x}^{\mathrm{2}} \right){sin}\left({x}^{\mathrm{2}} \right){dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } \frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}^{\mathrm{2}} \right){dx} \\ $$$$={Im}\left(−\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } ×{e}^{−\mathrm{2}{ix}^{\mathrm{2}} } {dx}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} \left(\mathrm{2}{i}+\mathrm{1}\right)} {dx}=\int_{\mathrm{0}} ^{\infty} {x}×{xe}^{−{x}^{\mathrm{2}} \left(\mathrm{2}{i}+\mathrm{1}\right)} {dx} \\ $$$${By}\:{parts} \\ $$$${u}={x}\Rightarrow{u}'=\mathrm{1} \\ $$$${v}'={xe}^{−{x}^{\mathrm{2}} \left(\mathrm{2}{i}+\mathrm{1}\right)} \Rightarrow{v}=\frac{−\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{i}+\mathrm{1}\right)}{e}^{−{x}^{\mathrm{2}} \left(\mathrm{2}{i}+\mathrm{1}\right)} \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} \left(\mathrm{2}{i}+\mathrm{1}\right)} {dx}=\frac{−\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{i}+\mathrm{1}\right)}\left[{xe}^{−{x}^{\mathrm{2}} \left(\mathrm{2}{i}+\mathrm{1}\right)} \right]_{\mathrm{0}} ^{\infty} +\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{i}+\mathrm{1}\right)}\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} \left(\mathrm{2}{i}+\mathrm{1}\right)} {dx} \\ $$$${let}\:\:{t}=\sqrt{\mathrm{2}{i}+\mathrm{1}}{x}\Rightarrow{dx}=\frac{{dt}}{\:\sqrt{\mathrm{2}{i}+\mathrm{1}}} \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} \left(\mathrm{2}{i}+\mathrm{1}\right)} {dx}=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{i}+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} {e}^{−{t}^{\mathrm{2}} } {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{i}+\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }×\frac{\sqrt{\pi}}{\mathrm{2}}=\frac{\sqrt{\pi}}{\mathrm{4}\left(\sqrt{\mathrm{5}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}+{i}\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)\right)^{\mathrm{3}/\mathrm{2}} } \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{4}×\left(\sqrt{\mathrm{5}}\right)^{\mathrm{3}/\mathrm{2}} \left({e}^{{itan}^{−\mathrm{1}} \left(\mathrm{2}\right)} \right)^{\mathrm{3}/\mathrm{2}} }=\frac{\sqrt{\pi}}{\mathrm{4}×\mathrm{5}^{\mathrm{3}/\mathrm{4}} {e}^{{i}\frac{\mathrm{3}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)} } \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{125}}}{e}^{−{i}\frac{\mathrm{3}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)} \\ $$$${Im}\left(−\frac{\sqrt{\pi}}{\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{125}}}{e}^{−{i}\frac{\mathrm{3}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)} \right)=\frac{\sqrt{\pi}}{\mathrm{4}\sqrt[{\mathrm{4}}]{\mathrm{125}}}{sin}\left(\frac{\mathrm{3}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)\right) \\ $$$$\int_{−\infty} ^{\infty} {x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } {cos}\left({x}^{\mathrm{2}} \right){sin}\left({x}^{\mathrm{2}} \right){dx}=\frac{\sqrt{\pi}}{\mathrm{4}\sqrt[{\mathrm{5}}]{\mathrm{125}}}{sin}\left(\frac{\mathrm{3}}{\mathrm{2}}{tan}^{−\mathrm{1}} \left(\mathrm{2}\right)\right) \\ $$$$ \\ $$
Commented by aliesam last updated on 22/May/19
$${thank}\:{yoy}\:{verry}\:{much}\:{sir} \\ $$