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Question Number 83558 by Jidda28 last updated on 03/Mar/20

Commented by Jidda28 last updated on 03/Mar/20

$$helpmeoutpls.$$

Commented by mathmax by abdo last updated on 03/Mar/20

$$A_n={\int }_0^1{x}^n{\left(lnx\right)}^{n}dxchangementlnx=-tgive$$$$A_n={\int }_{+\mathrm{\infty }}^0{\left(e^{-t}\right)}^n{(-t)}^n(-e^{-t})dt={\int }_0^{\mathrm{\infty }}{e}^{-(n+1)t}{t}^{n}dt$$$${=}_{(n+1)t=u}{\int }_0^{\mathrm{\infty }}{e}^{-u}{\left(\frac{u}{n+1}\right)}^n\frac{du}{n+1}=\frac{1}{{(n+1)}^{n+1}}{\int }_0^{\mathrm{\infty }}{u}^n{e}^{-u}du$$$$wehave\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt(x>0)\Rightarrow A_n=\frac{1}{{(n+1)}^{n+1}}\times \mathrm{\Gamma }(n+1)$$$$=\frac{n!}{{(n+1)}^{n+1}}$$

Commented by Jidda28 last updated on 03/Mar/20

$$thankyouverymuchsir,but{x}^{m\stackrel{}{}}$$

Commented by mathmax by abdo last updated on 03/Mar/20

$$youarewelcome$$

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