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Question Number 147287 by mnjuly1970 last updated on 19/Jul/21

$$$$$$...\mathrm{Advanced}\mathrm{Calculus}...$$$$$$$$\mathrm{C}alculate::\{\begin{array}{l}\mathrm{i}::\mathrm{I}:={\int }_0^1\ln \left(\mathrm{x}\right).\ln (1+\mathrm{x})\mathrm{dx}\\ \mathrm{ii}::\mathrm{J}:={\int }_0^1{\mathrm{Li}}_2(1-{\mathrm{x}}^2)=?\end{array}$$$$\mathrm{Note}::{\mathrm{Li}}_2\left(\mathrm{x}\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{x}}^n}{n^2}........◼$$$$....\mathrm{m}.\mathrm{n}....$$$$$$

Answered by mindispower last updated on 19/Jul/21

$$I={\left[(xln\left(x\right)-x)ln(1+x)\right]}_0^1-{\int }_0^1\frac{xln\left(x\right)}{1+x}-\frac{x}{1+x}dx$$$$=-ln\left(2\right)-{\int }_0^1\frac{(x+1-1)ln\left(x\right)}{1+x}-1+\frac{1}{1+x}dx$$$$=-ln\left(2\right)-{\int }_0^{1}ln\left(x\right)dx+{\int }_0^1\frac{ln\left(x\right)}{1+x}dx+\int dx-\int \frac{dx}{1+x}$$$$=-ln\left(2\right)+1-{\int }_0^1\frac{ln(1+x)}{x}dx+1-ln\left(2\right)$$$$=-ln\left(2\right)-{\int }_0^1\frac{ln(1-(-x))}{-x}d(-x)+2$$$$=ln\left(2\right)+{\left[{Li}_2(-x)\right]}_0^1=-2ln\left(2\right)+{Li}_2(-1)+2=-ln\left(4\right)-\frac{{\pi }^2}{12}+2$$$$J=\left[{xli}_2(1-x^2)\right]-{\int }_0^1\frac{2x^{2}ln\left(x^2\right)}{1-x^2}$$$$=-4{\int }_0^1\frac{ln\left(x\right)}{1-x^2}dx+4{\int }_0^{1}ln\left(x\right)dx$$$$=-2{\int }_0^1\frac{ln\left(x\right)}{1-x}dx-2{\int }_0^1\frac{ln\left(x\right)}{1+x}dx-4$$$${=2{Li}_2(1-x)]}_0^1+2{\int }_0^1\frac{ln(1+x)}{x}dx-4$$$$=\frac{{\pi }^2}{3}-2Li(-1)-4$$$$=\frac{{\pi }^2}{3}+\frac{{\pi }^2}{6}-4=\frac{{\pi }^2}{2}-4=\frac{1}{2}({\pi }^2-8)$$

Answered by mathmax by abdo last updated on 19/Jul/21

$$\mathrm{{\rm Y}}={\int }_0^1\ln \left(\mathrm{x}\right)\ln (1+\mathrm{x})\mathrm{dx}\mathrm{we}\mathrm{have}\frac{\mathrm{d}}{\mathrm{dx}}\ln (1+\mathrm{x})=\frac{1}{1+\mathrm{x}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$$$$\Rightarrow \ln (1+\mathrm{x})=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}$$$$\Rightarrow \mathrm{{\rm Y}}={\int }_0^1\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}}\right)\ln \left(\mathrm{x}\right)\mathrm{dx}$$$$=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}{\int }_0^1{\mathrm{x}}^{\mathrm{n}}\ln \left(\mathrm{x}\right)\mathrm{dx}$$$${\mathrm{U}}_{\mathrm{n}}={\int }_0^1{\mathrm{x}}^{\mathrm{n}}\ln \left(\mathrm{x}\right)\mathrm{dx}={\left[\frac{{\mathrm{x}}^{\mathrm{n}+1}}{\mathrm{n}+1}\ln \left(\mathrm{x}\right)\right]}_0^1-{\int }_0^1\frac{{\mathrm{x}}^{\mathrm{n}}}{\mathrm{n}+1}\mathrm{dx}$$$$=-\frac{1}{{(\mathrm{n}+1)}^2}\Rightarrow \mathrm{{\rm Y}}=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}{(\mathrm{n}+1)}^2}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}{(\mathrm{n}+1)}^2}$$$$\frac{1}{\mathrm{x}{(\mathrm{x}+1)}^2}=\frac{\mathrm{a}}{\mathrm{x}}+\frac{\mathrm{b}}{\mathrm{x}+1}+\frac{\mathrm{c}}{{(\mathrm{x}+1)}^2}=\mathrm{h}\left(\mathrm{x}\right)$$$$\mathrm{a}=1,\mathrm{c}=-1,{\lim }_{\mathrm{x}\to +\mathrm{\infty }}\mathrm{xh}\left(\mathrm{x}\right)=0=\mathrm{a}+\mathrm{b}\Rightarrow \mathrm{b}=-1\Rightarrow$$$$\mathrm{h}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}}-\frac{1}{\mathrm{x}+1}-\frac{1}{{(\mathrm{x}+1)}^2}\Rightarrow \mathrm{{\rm Y}}=\sum _{\mathrm{n}.1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}$$$$-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(\mathrm{n}+1)}^2}\mathrm{we}\mathrm{have}$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}}=-\ln 2$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+1}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}=\ln \left(2\right)-1$$$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(\mathrm{n}+1)}^2}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{{\mathrm{n}}^2}=-\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}$$$$=-(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{\mathrm{n}}^2}+1)=-(2^{1-2}-1)\xi \left(2\right)-1=\frac{1}{2}\frac{{\pi }^2}{6}-1=\frac{{\pi }^2}{12}-1$$$$\Rightarrow \mathrm{{\rm Y}}=-\ln 2-\ln 2+1-\frac{{\pi }^2}{12}+1=2-2\ln 2-\frac{{\pi }^2}{12}$$

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