Question

$\huge\displaystyle \sf \red{ \int^{ \infty}_{0} {e}^{ - {x}^{3} } sin( {x}^{3} ) \: dx }$​

Answer 1

$\displaystyle \sf \black{ Let \: I=\int^{ \infty}_{0} {e}^{ - {x}^{3} } sin( {x}^{3} ) \: dx }$

$\sf \black{{e}^{ i \theta} = \cos \theta + i \sin \theta}$

$\sf \black{ \implies{e}^{ i {x}^{3} } = \cos \: x ^{3} + i \sin {x}^{3} }$

$\displaystyle \sf \black{ I = Img\bigg[\int^{ \infty}_{0} {e}^{ - {x}^{3} } \cdot {e}^{i {x}^{3} } \: dx \bigg]}$

$\displaystyle \sf \black{ \implies I = Img\bigg[\int^{ \infty}_{0} {e}^{ - {(1 - i) {x}^{3} }^{} }\: dx \bigg]}$

$\sf \black{(1 - i) {x}^{3} = t}$

$\sf \black{ \implies(1 - i) 3{x}^{2} \: dx = dt}$

$\sf \black{ \implies dx = \frac{dt}{3(1 - i) {x}^{2} } }$

$\sf \black{\because \: \frac{1}{ {x}^{2} } = \bigg(\frac{1 - i}{t} \bigg)^{ \frac{2}{3} } }$

$\sf \black{ \implies dx = \frac{dt}{3(1 - i) {}^{} } } \: \: \: \: \: \sf \black{ \frac{(1 - i)^{ \frac{2}{3} } }{ {t}^{ \frac{2}{3} }} }$

$\displaystyle \sf \black{I=Img \bigg[ \int_{0}^{ \infty} \frac{1}{3(1 - i)^{ \frac{1}{3} } } \cdot {t}^{ \frac{ - 2}{3}} {e}^{ - t} \: dt\bigg]}$

$\displaystyle \sf \black{=Img \bigg[ \frac{1}{3 [2 \sqrt{2} {e}^{ i\frac{\pi}{4} } ] ^ { \frac{1}{3} } } \int_{0}^{ \infty} {t}^{ \frac{ - 2}{3}} {e}^{ - t} \: dt\bigg]}$

$\sf{\black{ \{{ \: (1 - i) = \sqrt{2} \: {e}^{ - i \frac{\pi}{4} } \}}}}$

$\displaystyle \sf \black{ \int_{0}^{ \infty} {t}^{ \frac{ - 2}{3} } {e}^{ - t} \: dt = \Gamma \bigg( \frac{1}{3} \bigg)}$

$\sf \black{I=Img \bigg[ \frac{ {e}^{i \frac{\pi}{12} } }{3( \sqrt{2})^{ \frac{1}{3} } } \: \: \Gamma( \frac{1}{3} ) \bigg ]}$

$\sf \black{ \implies I=Img \bigg[ \frac{\Gamma( \frac{1}{3} )}{ 3 \cdot2^{ \frac{1}{6} } } \: \: ( {cos}{ \frac{\pi}{12} } + sin \frac{\pi}{12} )\bigg ]}$

$\sf \black{I = \frac{\Gamma( \frac{1}{3}) \: \: sin( \frac{\pi}{12} ) }{3 \: \cdot \: {2}^{ \frac{1}{6} } } }$