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Question Number 107596 by Rio Michael last updated on 11/Aug/20

$$\mathrm{Given}$$ $$I_{m,n}=\underset{1}{\stackrel{e}{\int }}{x}^m{\left(\ln x\right)}^{n}dx\mathrm{where}m,n\in {\mathbb{N}}^{\ast }$$ $$\mathrm{Show}\mathrm{that}(1+m)I_{m,n}=e^{m+1}-{nI}_{m,n-1}\mathrm{for}m>0\mathrm{and}n>0$$ $$\mathrm{also},\mathrm{evaluate}{I}_{2,3}$$

Answered by Ar Brandon last updated on 11/Aug/20

$$\mathrm{I}=\underset{1}{\stackrel{e}{\int }}{x}^m{\left(\ln x\right)}^{n}dx$$ $$={[{\left(\mathrm{lnx}\right)}^{\mathrm{n}}\int {\mathrm{x}}^{\mathrm{m}}-\int \{\frac{\mathrm{d}{\left(\mathrm{lnx}\right)}^{\mathrm{n}}}{\mathrm{dx}}\cdot \int {\mathrm{x}}^{\mathrm{m}}\mathrm{dx}\}\mathrm{dx}]}_1^{\mathrm{e}}$$ $$={\left[\frac{{\left(\mathrm{lnx}\right)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{m}+1}}{\mathrm{m}+1}\right]}_1^{\mathrm{e}}-\frac{\mathrm{n}}{\mathrm{m}+1}{\int }_1^{\mathrm{e}}{\mathrm{x}}^{\mathrm{m}}{\left(\mathrm{lnx}\right)}^{\mathrm{n}-1}\mathrm{dx}$$ $$=\frac{{\mathrm{e}}^{\mathrm{m}+1}}{\mathrm{m}+1}-\frac{\mathrm{n}}{\mathrm{m}+1}{\mathrm{I}}_{(\mathrm{m},\mathrm{n}-1)}$$ $$\Rightarrow (\mathrm{m}+1){\mathrm{I}}_{\mathrm{m},\mathrm{n}}={\mathrm{e}}^{\mathrm{m}+1}-{\mathrm{nI}}_{(\mathrm{m},\mathrm{n}-1)}$$

Answered by hgrocks last updated on 11/Aug/20

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