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Question Number 122186 by physicstutes last updated on 14/Nov/20

$$\mathrm{Obtain}\mathrm{a}\mathrm{reduction}\mathrm{formulae}\mathrm{for}$$$$I_n=\underset{0}{\stackrel{1}{\int }}{\left(\ln x\right)}^{n}dx$$$$\mathrm{find}{I}_2=\underset{0}{\stackrel{1}{\int }}{\left(\ln x\right)}^{2}dx$$

Answered by Olaf last updated on 14/Nov/20

$${\mathrm{I}}_n={\int }_0^1{\ln }^{n}xdx$$$${\mathrm{I}}_0=1\mathrm{and}\mathrm{for}n⩾1:$$$${\mathrm{I}}_n={\left[x{\ln }^{n}x\right]}_0^1-{\int }_0^{1}x\left(n\frac{1}{x}{\ln }^{n-1}x\right)dx$$$${\mathrm{I}}_n=-n{\int }_0^1{\ln }^{n-1}xdx=-n{\mathrm{I}}_{n-1}$$$$\Rightarrow {\mathrm{I}}_n={(-1)}^{n}n!{\mathrm{I}}_0={(-1)}^{n}n!$$$$\mathrm{If}n=2,{\mathrm{I}}_2={(-1)}^{2}2!=2$$

Commented by physicstutes last updated on 14/Nov/20

$$\mathrm{This}\mathrm{is}\mathrm{incredible}.\mathrm{Thank}\mathrm{you}\mathrm{sir}\mathrm{Olaf}.$$

Answered by Dwaipayan Shikari last updated on 14/Nov/20

$${\int }_{\mathrm{\infty }}^0{t}^{n}xdtlogx=t\Rightarrow \frac{1}{x}=\frac{dt}{dx}$$$$={\int }_{\mathrm{\infty }}^0{t}^n{e}^{t}dt$$$$={\int }_0^{\mathrm{\infty }}{(-1)}^n{u}^n{e}^{-t}dtt=-u$$$$={(-1)}^n\mathrm{\Gamma }(n+1)={(-1)}^{n}n!$$$$asperquestion$$$$Itis{(-1)}^{2}2!=2$$

Commented by physicstutes last updated on 14/Nov/20

$$\mathrm{thats}\mathrm{right}!$$

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