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Question Number 157665 by mnjuly1970 last updated on 26/Oct/21

Answered by Raxreedoroid last updated on 26/Oct/21

$$I={\int }_0^1{x}^{n-1}\ln (1-x)$$$$dw=x^{n-1},w=\frac{x^n}{n}$$$$t=\ln (1-x),dt=\frac{-1}{1-x}dx$$$$I={[\frac{x^n\ln (1-x)}{n}+\frac{1}{n}\int \frac{x^n}{1-x}dx]}_0^1$$$$u=1-x,x=1-u,dx=-du$$$$I={[\frac{{(1-u)}^n\ln \left(u\right)}{n}-\frac{1}{n}\int \frac{{(1-u)}^n}{u}du]}_1^0$$$$I={[\frac{{(1-u)}^n\ln \left(u\right)}{n}+\frac{1}{n}\int {(1-u)}^{n-1}+\frac{1}{u}du]}_1^0$$$$I={[\frac{{(1-u)}^n\ln \left(u\right)}{n}-\frac{1}{n}(\frac{{(1-u)}^n}{n}+\ln u)]}_1^0$$$$I={\left[\frac{{(1-u)}^{n}n\ln \left(u\right)-{(1-u)}^n-n\ln u}{n^2}\right]}_1^0$$$$I=\underset{u\to 0^{+}}{\lim }\frac{{(1-u)}^{n}n\ln \left(u\right)-{(1-u)}^n-n\ln u}{n^2}$$$$I=\underset{u\to 0^{+}}{\lim }\frac{n\ln u({(1-u)}^n-1)-{(1-u)}^n}{n^2}$$$$I=\underset{u\to 0^{+}}{\lim }\left(\frac{\ln u({(1-u)}^n-1)}{n}\right)-\frac{1}{n^2}$$$$I=-\frac{1}{n^2}$$$$\phi \left(n\right)=-\frac{1}{n^2}$$$$S=-\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n^3}=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^3}-2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{\left(2n\right)}^3}$$$$S=\zeta \left(3\right)-2\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{8n^3}=\zeta \left(3\right)-\frac{1}{4}\zeta \left(3\right)$$$$S=\frac{3}{4}\zeta \left(3\right)$$$$$$

Answered by qaz last updated on 26/Oct/21

$$\mathrm{S}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}-1}\ln (1-\mathrm{x})\mathrm{dx}$$$$=-{\int }_0^1\frac{\ln (1-\mathrm{x})\ln (1+\mathrm{x})}{\mathrm{x}}\mathrm{dx}$$$$=\frac{1}{4}{\int }_0^1\frac{{\ln }^2\frac{1-\mathrm{x}}{1+\mathrm{x}}}{\mathrm{x}}\mathrm{dx}-\frac{1}{4}{\int }_0^1\frac{{\ln }^2(1-{\mathrm{x}}^2)}{\mathrm{x}}\mathrm{dx}$$$$=\frac{1}{2}{\int }_0^1\frac{{\ln }^2\mathrm{x}}{1-{\mathrm{x}}^2}\mathrm{dx}-\frac{1}{8}{\int }_0^1\frac{{\ln }^2\mathrm{x}}{1-\mathrm{x}}\mathrm{dx}................\frac{1-\mathrm{x}}{1+\mathrm{x}}\to \mathrm{x}$$$$=\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\mathrm{x}}^{2\mathrm{n}}{\ln }^2\mathrm{xdx}-\frac{1}{8}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}}{\ln }^2\mathrm{xdx}$$$$=\frac{1}{2}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{2}2!}{{(2\mathrm{n}+1)}^3}-\frac{1}{8}\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{2}2!}{{(\mathrm{n}+1)}^3}$$$$=(1-2^{-3})\zeta \left(3\right)-\frac{1}{4}\zeta \left(3\right)$$$$=\frac{5}{8}\zeta \left(3\right)$$

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