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

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

$${\int }_0^{\mathrm{\infty }}\sqrt{\mathrm{y}}{\mathrm{e}}^{-{\mathrm{t}}^2}{\mathrm{y}}^3={\mathrm{t}}^2,{\mathrm{y}}^{\frac{3}{2}}=\mathrm{t}\frac{3}{2}\sqrt{\mathrm{y}}=\frac{\mathrm{dt}}{\mathrm{dy}}$$$$\frac{2}{3}{\int }_0^{\mathrm{\infty }}\frac{3}{2}\sqrt{\mathrm{y}}{\mathrm{e}}^{-{\mathrm{y}}^3}\mathrm{dy}=\frac{2}{3}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}=\frac{2}{3}.\frac{\sqrt{\pi }}{2}=\frac{\sqrt{\pi }}{3}$$

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

$$\mathrm{put}{\mathrm{y}}^3=\mathrm{t}\Rightarrow \mathrm{y}={\mathrm{t}}^{\frac{1}{3}}\Rightarrow \mathrm{dy}=\frac{1}{3}{\mathrm{t}}^{-\frac{2}{3}}\mathrm{dt}$$$$=\frac{1}{3}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\frac{1}{2}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}=\frac{1}{3}\mathrm{\Gamma }\left(\frac{1}{2}\right)=\frac{\sqrt{\pi }}{3}$$

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

$$\mathrm{let}\mathrm{I}={\int }_0^{\mathrm{\infty }}\sqrt{\mathrm{y}}{\mathrm{e}}^{-{\mathrm{y}}^3}\mathrm{dy}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}{\mathrm{y}}^3=\mathrm{t}\Rightarrow \mathrm{y}={\mathrm{t}}^{\frac{1}{3}}\Rightarrow$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\frac{1}{6}}{\mathrm{e}}^{-\mathrm{t}}\frac{1}{3}{\mathrm{t}}^{\frac{1}{3}-1}\mathrm{dt}=\frac{1}{3}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\frac{1}{6}+\frac{1}{3}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$$$$=\frac{1}{3}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\frac{1}{2}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}=\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{t}}}{\sqrt{\mathrm{t}}}\mathrm{dt}{=}_{\sqrt{\mathrm{t}}=\mathrm{u}}\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{-{\mathrm{u}}^2}}{\mathrm{u}}\left(2\mathrm{u}\right)\mathrm{du}$$$$=\frac{2}{3}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}=\frac{2}{3}.\frac{\sqrt{\pi }}{2}=\frac{\sqrt{\pi }}{3}\Rightarrow \mathrm{I}=\frac{\sqrt{\pi }}{3}$$$$$$

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