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Question Number 145547 by physicstutes last updated on 05/Jul/21

$$I_{m,n}={\int }_0^1{(1-x^m)}^{n}dx$$$$\mathrm{Show}\mathrm{that}{I}_{m,n}(mn+1)=I_{m,n-1}$$

Answered by qaz last updated on 06/Jul/21

$${\mathrm{I}}_{\mathrm{m},\mathrm{n}}={\int }_0^1{(1-{\mathrm{x}}^{\mathrm{m}})}^{\mathrm{n}}\mathrm{dx}=\mathrm{x}{(1-{\mathrm{x}}^{\mathrm{m}})}^{\mathrm{n}}{\mid }_0^1+{\int }_0^1\mathrm{nm}\cdot {\mathrm{x}}^{\mathrm{m}}{(1-{\mathrm{x}}^{\mathrm{m}})}^{\mathrm{n}-1}\mathrm{dx}$$$$=\mathrm{nm}{\int }_0^1{\mathrm{x}}^{\mathrm{m}}{(1-{\mathrm{x}}^{\mathrm{m}})}^{\mathrm{n}-1}\mathrm{dx}$$$$=\mathrm{nm}{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{m}}{(1-{\mathrm{x}}^{\mathrm{m}})}^{\mathrm{n}}}{1-{\mathrm{x}}^{\mathrm{m}}}\mathrm{dx}$$$$=\mathrm{nm}{\int }_0^1(\frac{1}{1-{\mathrm{x}}^{\mathrm{m}}}-1){(1-{\mathrm{x}}^{\mathrm{m}})}^{\mathrm{n}}\mathrm{dx}$$$$=\mathrm{nm}[{\int }_0^1{(1-{\mathrm{x}}^{\mathrm{m}})}^{\mathrm{n}-1}\mathrm{dx}-{\int }_0^1{(1-{\mathrm{x}}^{\mathrm{m}})}^{\mathrm{n}}\mathrm{dx}]$$$$=\mathrm{nm}({\mathrm{I}}_{\mathrm{m},\mathrm{n}-1}-{\mathrm{I}}_{\mathrm{m},\mathrm{n}})$$$$\Rightarrow {\mathrm{I}}_{\mathrm{m},\mathrm{n}}(1+\mathrm{nm})=\mathrm{nm}\cdot {\mathrm{I}}_{\mathrm{m},\mathrm{n}-1}$$

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