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Question Number 161443 by smallEinstein last updated on 18/Dec/21

Answered by mathmax by abdo last updated on 18/Dec/21

$$\mathrm{I}={\int }_0^1\frac{{\log }^2(1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}\mathrm{letf}\left(\mathrm{a}\right)={\int }_0^1\frac{{(1+\mathrm{x})}^{\mathrm{a}}}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{a}\right)={\int }_0^1\frac{{\mathrm{e}}^{\mathrm{alog}(1+\mathrm{x})}}{1+{\mathrm{x}}^2}\mathrm{dx}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)={\int }_0^1\frac{\log (1+\mathrm{x}){(1+\mathrm{x})}^{\mathrm{a}}}{1+{\mathrm{x}}^2}\mathrm{dx}\Rightarrow$$$${\mathrm{f}}^{\left(2\right)}\left(\mathrm{a}\right)={\int }_0^1\frac{{\log }^2(1+\mathrm{x})}{1+{\mathrm{x}}^2}{(1+\mathrm{x})}^{\mathrm{a}}\mathrm{dx}\Rightarrow$$$${\mathrm{f}}^{\left(2\right)}\left(0\right)={\int }_0^1\frac{{\log }^2(1+\mathrm{x})}{1+{\mathrm{x}}^2}\mathrm{dx}$$$$\mathrm{f}\left(\mathrm{a}\right)={\int }_0^1{(\mathrm{x}+1)}^{\mathrm{a}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{2\mathrm{n}}{(\mathrm{x}+1)}^{\mathrm{a}}\mathrm{dx}\mathrm{but}$$$${\int }_0^1{\mathrm{x}}^{2\mathrm{n}}{(\mathrm{x}+1)}^{\mathrm{a}}\mathrm{dx}={\int }_0^1{\mathrm{x}}^{2\mathrm{n}}\sum _{\mathrm{p}=0}^{\mathrm{a}}{\mathrm{C}}_{\mathrm{a}}^{\mathrm{p}}{\mathrm{x}}^{\mathrm{p}}\mathrm{dx}$$$$=\sum _{\mathrm{p}=0}^{\mathrm{a}}{\mathrm{C}}_{\mathrm{a}}^{\mathrm{p}}{\int }_0^1{\mathrm{x}}^{2\mathrm{n}+\mathrm{p}}\mathrm{dx}=\sum _{\mathrm{p}=0}^{\mathrm{a}}\frac{{\mathrm{C}}_{\mathrm{a}}^{\mathrm{p}}}{2\mathrm{n}+\mathrm{p}+1}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}\sum _{\mathrm{p}=0}^{\mathrm{a}}\frac{{\mathrm{C}}_{\mathrm{a}}^{\mathrm{p}}}{2\mathrm{n}+\mathrm{p}+1}\mathrm{rest}\mathrm{calculus}\mathrm{of}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)....\mathrm{be}\mathrm{continued}...$$

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