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Question Number 98445 by mathmax by abdo last updated on 14/Jun/20

$$\mathrm{give}\mathrm{at}\mathrm{form}\mathrm{of}\mathrm{serie}{\mathrm{U}}_{\mathrm{n}}={\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}\ln \left(\mathrm{x}\right)}{{(1+\mathrm{x})}^2}\mathrm{dx}$$

Answered by mathmax by abdo last updated on 14/Jun/20

$$\mathrm{we}\mathrm{have}\frac{1}{1+\mathrm{x}}=\sum _{\mathrm{k}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}\mathrm{for}\mid \mathrm{x}\mid <1\Rightarrow$$$$-\frac{1}{(1+{\mathrm{x}}^2)}=\sum _{\mathrm{k}=1}^{\mathrm{\infty }}\mathrm{k}{(-1)}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}-1}=\sum _{\mathrm{k}=0}^{\mathrm{\infty }}(\mathrm{k}+1){(-1)}^{\mathrm{k}+1}{\mathrm{x}}^{\mathrm{k}}\Rightarrow$$$$\frac{1}{{(1+\mathrm{x})}^2}=\sum _{\mathrm{k}=0}^{\mathrm{\infty }}(\mathrm{k}+1){(-1)}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}\Rightarrow {\mathrm{U}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{\mathrm{n}}\ln \left(\mathrm{x}\right)\left(\sum _{\mathrm{k}=0}^{\mathrm{\infty }}(\mathrm{k}+1){(-1)}^{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}\right)\mathrm{dx}$$$$=\sum _{\mathrm{k}=0}^{\mathrm{\infty }}(\mathrm{k}+1){(-1)}^{\mathrm{k}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}+\mathrm{k}}\ln \left(\mathrm{x}\right)\mathrm{dx}\mathrm{and}\mathrm{by}\mathrm{parts}$$$${\int }_0^1{\mathrm{x}}^{\mathrm{n}+\mathrm{k}}\ln \left(\mathrm{x}\right)\mathrm{dx}={\left[\frac{{\mathrm{x}}^{\mathrm{n}+\mathrm{k}+1}}{\mathrm{n}+\mathrm{k}+1}\ln \left(\mathrm{x}\right)\right]}_0^1-{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}+\mathrm{k}}}{\mathrm{n}+\mathrm{k}+1}\mathrm{dx}$$$$=-\frac{1}{{(\mathrm{n}+\mathrm{k}+1)}^2}\Rightarrow {\mathrm{U}}_{\mathrm{n}}=-\sum _{\mathrm{k}=0}^{\mathrm{\infty }}(\mathrm{k}+1){(-1)}^{\mathrm{k}}\times \frac{1}{{(\mathrm{n}+\mathrm{k}+1)}^2}$$$$=\sum _{\mathrm{k}=0}^{\mathrm{\infty }}\frac{\mathrm{k}+1}{{(\mathrm{n}+\mathrm{k}+1)}^2}{(-1)}^{\mathrm{k}+1}\Rightarrow$$$${\mathrm{U}}_{\mathrm{n}}=-\frac{1}{{(\mathrm{n}+1)}^2}+\frac{2}{{(\mathrm{n}+2)}^2}-\frac{3}{{(\mathrm{n}+3)}^2}+....$$

Answered by maths mind last updated on 14/Jun/20

$$\frac{1}{{(1+x)}^2}=\frac{\partial }{\partial x}\left(\sum _{n⩾0}{(-1)}^{n+1}{x}^n\right)=\sum _{n⩾1}n{(-1)}^{n+1}{x}^{n-1}$$$$=\sum _{t⩾0}(t+1){(-x)}^t$$$$U_n={\int }_0^1\frac{x^{n}ln\left(x\right)}{{(1+x)}^2}dx={\int }_0^1{x}^{n}ln\left(x\right).\sum _{t⩾0}((t+1){(-x)}^{t}dx$$$$=\sum _{t⩾0}(t+1){(-1)}^{t+1}{\int }_0^{\mathrm{\infty }}{re}^{-(n+t+1)r}dr$$$$=\sum _{t⩾0}(t+1){(-1)}^{t+1}{\int }_0^{+\mathrm{\infty }}\frac{{ue}^{-u}}{{(n+t+1)}^2}du$$$$=\sum _{t⩾0}(t+1){(-1)}^{t+1}.\frac{\mathrm{\Gamma }\left(2\right)}{{(n++1+t)}^2}$$$$=\sum _{t⩾0}\frac{{(-1)}^{t+1}(t+1)}{{(n+t+1)}^2}$$

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