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Question Number 101959 by MathQoutes last updated on 05/Jul/20

$$$$$$$$$$Startingfrom$$$$y=\frac{4}{\sqrt{\pi }}{t}^{\frac{3}{2}}{\int }_0^{\mathrm{\infty }}{x}^3{e}^{-{tx}^2}dx$$$$find\frac{\pi }{8}=?$$

Answered by mathmax by abdo last updated on 06/Jul/20

$$\mathrm{the}\mathrm{question}\mathrm{is}\mathrm{not}\mathrm{clear}\mathrm{let}\mathrm{find}\mathrm{y}\left(\mathrm{t}\right)=\frac{4}{\sqrt{\pi }}{\mathrm{t}}^{\frac{3}{2}}{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^3{\mathrm{e}}^{-{\mathrm{tx}}^2}\mathrm{dx}\mathrm{for}\mathrm{t}>0$$$$\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}\sqrt{\mathrm{t}}=\mathrm{u}\Rightarrow$$$$\mathrm{y}\left(\mathrm{t}\right)=\frac{4}{\sqrt{\pi }}{\mathrm{t}}^{\frac{3}{2}}{\int }_0^{\mathrm{\infty }}{\left(\frac{\mathrm{u}}{\sqrt{\mathrm{t}}}\right)}^3{\mathrm{e}}^{-{\mathrm{u}}^2}\frac{\mathrm{du}}{\sqrt{\mathrm{t}}}$$$$=\frac{4}{\sqrt{\pi \mathrm{t}}}{\int }_0^{\mathrm{\infty }}{\mathrm{u}}^3{\mathrm{e}}^{-{\mathrm{u}}^2}\mathrm{du}{=}_{\mathrm{u}=\sqrt{\mathrm{z}}}\frac{4}{\sqrt{\pi \mathrm{t}}}{\int }_0^{\mathrm{\infty }}{\mathrm{z}}^{\frac{3}{2}}{\mathrm{e}}^{-\mathrm{z}}\frac{\mathrm{dz}}{2\sqrt{\mathrm{z}}}$$$$=\frac{2}{\sqrt{\pi \mathrm{t}}}{\int }_0^{\mathrm{\infty }}\mathrm{z}{\mathrm{e}}^{-\mathrm{z}}\mathrm{dz}\mathrm{we}\mathrm{know}\mathrm{\Gamma }\left(\mathrm{z}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{z}}^{\mathrm{x}-1}{\mathrm{e}}^{-\mathrm{z}}\mathrm{dt}\Rightarrow {\int }_0^{\mathrm{\infty }}\mathrm{z}{\mathrm{e}}^{-\mathrm{z}}\mathrm{dz}=\mathrm{\Gamma }\left(2\right)=1!=1$$$$\Rightarrow \mathrm{y}\left(\mathrm{t}\right)=\frac{2}{\sqrt{\pi \mathrm{t}}}$$

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