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Question Number 97492 by 675480065 last updated on 08/Jun/20

Answered by smridha last updated on 08/Jun/20

$$\mathit{l}\mathit{e}\mathit{t}\mathit{l}\mathit{n}\left(\mathit{x}\right)=-\mathit{k}\mathit{s}\mathit{o}\mathit{w}\mathit{e}\mathit{g}\mathit{e}\mathit{t}$$$${\int }_0^{\mathrm{\infty }}{\mathit{e}}^{-2021\mathit{k}}{\mathit{k}}^{2020}\mathit{d}\mathit{k}$$$$=\frac{\mathbf{\Gamma }\left(2021\right)}{{\left(2021\right)}^{2021}}\left[using\mathit{L}\mathit{a}\mathit{p}\mathit{l}\mathit{a}\mathit{c}\mathit{e}\mathit{T}\mathit{r}\mathit{a}\mathit{n}\mathit{s}\mathit{f}\mathit{o}\mathit{r}\mathit{m}\right]$$

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

$$\mathrm{I}={\int }_0^1{\left(\mathrm{xlnx}\right)}^{2020}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{lnx}=-\mathrm{t}\Rightarrow$$$$\mathrm{I}=-{\int }_0^{\mathrm{\infty }}{\left({\mathrm{e}}^{-\mathrm{t}}\right)}^{2020}{(-\mathrm{t})}^{2020}(-{\mathrm{e}}^{-\mathrm{t}})\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-2021\mathrm{t}}{\mathrm{t}}^{2020}\mathrm{dt}{=}_{2021\mathrm{t}=\mathrm{u}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{u}}{\left(\frac{\mathrm{u}}{2021}\right)}^{2020}\frac{\mathrm{du}}{2021}$$$$=\frac{1}{{\left(2021\right)}^{2021}}{\int }_0^{\mathrm{\infty }}{\mathrm{u}}^{2020}{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}\mathrm{we}\mathrm{have}\mathrm{\Gamma }\left(\mathrm{x}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{x}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\Rightarrow$$$$\mathrm{I}=\frac{1}{{\left(2021\right)}^{2021}}\mathrm{\Gamma }\left(2021\right)=\frac{\left(2\mathrm{o}20\right)!}{{\left(2021\right)}^{2021}}$$

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