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Question Number 126788 by john_santu last updated on 24/Dec/20

$$\sigma =\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\sqrt{x}{e}^{-x/4}dx=?$$

Answered by Ar Brandon last updated on 24/Dec/20

$$\mathrm{x}={\mathrm{u}}^2\Rightarrow \mathrm{dx}=2\mathrm{udu}$$$$\sigma =2{\int }_0^{\mathrm{\infty }}{\mathrm{u}}^2{\mathrm{e}}^{-{\mathrm{u}}^2/4}\mathrm{du}$$$${\mathrm{e}}^{-{\mathrm{u}}^2/4}=\mathrm{t}\Rightarrow -\frac{1}{2}{\mathrm{ue}}^{-{\mathrm{u}}^2/4}\mathrm{du}=\mathrm{dt}$$$$\sigma =-4{\int }_0^{\mathrm{\infty }}\mathrm{u}\cdot (-\frac{1}{2}{\mathrm{ue}}^{-{\mathrm{u}}^2/4})\mathrm{du}$$$$=-4{\{{\mathrm{ue}}^{-{\mathrm{u}}^2/4}-\int {\mathrm{e}}^{-{\mathrm{u}}^2/4}\mathrm{du}\}}_0^{\mathrm{\infty }}$$$$=4{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{u}}^2/4}\mathrm{du}$$$$\{\begin{array}{l}\mathcal{I}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{x}}^2/4}\mathrm{dx}\\ \mathcal{I}={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{y}}^2/4}\mathrm{dy}\end{array}$$$${\mathcal{I}}^2={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-({\mathrm{x}}^2+{\mathrm{y}}^2)/4}\mathrm{dxdy}$$$$={\int }_0^{\frac{\pi }{2}}{\int }_0^{\mathrm{\infty }}{\mathrm{re}}^{-{\mathrm{r}}^2/4}\mathrm{drd}\theta =-2{\int }_0^{\frac{\pi }{2}}{\int }_0^{\mathrm{\infty }}-\frac{1}{2}{\mathrm{re}}^{-{\mathrm{r}}^2/4}\mathrm{drd}\theta$$$$=-2{\left[\frac{\theta }{1}\right]}_0^{\frac{\pi }{2}}{\left[{\mathrm{e}}^{-{\mathrm{r}}^2/4}\right]}_0^{\mathrm{\infty }}=-\left(\frac{2}{1}\right)\left(\frac{\pi }{2}\right)(0-1)=\pi$$$$\mathcal{I}=\sqrt{\pi }$$$$\sigma =4\mathcal{I}=4\sqrt{\pi }$$

Answered by john_santu last updated on 24/Dec/20

Answered by Dwaipayan Shikari last updated on 24/Dec/20

$${\int }_0^{\mathrm{\infty }}\sqrt{x}{e}^{-\frac{x}{4}}dx\frac{x}{4}=u$$$$=8{\int }_0^{\mathrm{\infty }}\sqrt{u}{e}^{-u}du=8\mathrm{\Gamma }\left(\frac{3}{2}\right)=4\sqrt{\pi }$$

Answered by mathmax by abdo last updated on 24/Dec/20

$$\sigma ={\int }_0^{\mathrm{\infty }}\sqrt{\mathrm{x}}{\mathrm{e}}^{-\frac{\mathrm{x}}{4}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\sqrt{\mathrm{x}}=\mathrm{t}\Rightarrow$$$$\sigma ={\int }_0^{\mathrm{\infty }}\mathrm{t}{\mathrm{e}}^{-\frac{{\mathrm{t}}^2}{4}}\left(2\mathrm{t}\right)\mathrm{dt}=2{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^2{\mathrm{e}}^{-\frac{{\mathrm{t}}^2}{4}}\mathrm{dt}{=}_{\frac{\mathrm{t}}{2}=\mathrm{z}}2{\int }_0^{\mathrm{\infty }}{4\mathrm{z}}^2{\mathrm{e}}^{-{\mathrm{z}}^2}2\mathrm{dz}$$$$=16{\int }_0^{\mathrm{\infty }}{\mathrm{z}}^2{\mathrm{e}}^{-{\mathrm{z}}^2}\mathrm{dz}{=}_{{\mathrm{z}}^2=\mathrm{u}}16{\int }_0^{\mathrm{\infty }}\mathrm{u}{\mathrm{e}}^{-\mathrm{u}}\frac{\mathrm{du}}{\sqrt{\mathrm{u}}}=8{\int }_0^{\mathrm{\infty }}{\mathrm{u}}^{\frac{1}{2}}{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}$$$$=8{\int }_0^{\mathrm{\infty }}{\mathrm{u}}^{\frac{3}{2}-1}{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}=8\times \mathrm{\Gamma }\left(\frac{3}{2}\right)=8\mathrm{\Gamma }(\frac{1}{2}+1)$$$$=4\mathrm{\Gamma }\left(\frac{1}{2}\right)=4\sqrt{\pi }$$$$$$

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