Question-107107

Question Number 107107 by mohammad17 last updated on 08/Aug/20

Commented by Dwaipayan Shikari last updated on 08/Aug/20

\int_{0}^{\infty} \sqrt{y} e^{- t^{2}} y^{3} = t^{2}, y^{\frac{3}{2}} = t \frac{3}{2} \sqrt{y} = \frac{dt}{dy}

\frac{2}{3} \int_{0}^{\infty} \frac{3}{2} \sqrt{y} e^{- y^{3}} dy = \frac{2}{3} \int_{0}^{\infty} e^{- t^{2}} dt = \frac{2}{3} . \frac{\sqrt{π}}{2} = \frac{\sqrt{π}}{3}

Commented by PRITHWISH SEN 2 last updated on 08/Aug/20

put y^{3} = t \Rightarrow y = t^{\frac{1}{3}} \Rightarrow dy = \frac{1}{3} t^{- \frac{2}{3}} dt

= \frac{1}{3} \int_{0}^{\infty} t^{\frac{1}{2} - 1} e^{- t} dt = \frac{1}{3} Γ (\frac{1}{2}) = \frac{\sqrt{π}}{3}

Answered by mathmax by abdo last updated on 08/Aug/20

let I = \int_{0}^{\infty} \sqrt{y} e^{- y^{3}} dy we do the changement y^{3} = t \Rightarrow y = t^{\frac{1}{3}} \Rightarrow

I = \int_{0}^{\infty} t^{\frac{1}{6}} e^{- t} \frac{1}{3} t^{\frac{1}{3} - 1} dt = \frac{1}{3} \int_{0}^{\infty} t^{\frac{1}{6} + \frac{1}{3} - 1} e^{- t} dt

= \frac{1}{3} \int_{0}^{\infty} t^{\frac{1}{2} - 1} e^{- t} dt = \frac{1}{3} \int_{0}^{\infty} \frac{e^{- t}}{\sqrt{t}} dt =_{\sqrt{t} = u} \frac{1}{3} \int_{0}^{\infty} \frac{e^{- u^{2}}}{u} (2 u) du

= \frac{2}{3} \int_{0}^{\infty} e^{- u^{2}} du = \frac{2}{3} . \frac{\sqrt{π}}{2} = \frac{\sqrt{π}}{3} \Rightarrow I = \frac{\sqrt{π}}{3}