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Question Number 98016 by bemath last updated on 11/Jun/20

$$\underset{0}{\stackrel{1}{\int }}\frac{{\ln }^2\left(x\right)}{x^2+1}dx?$$

Commented by bobhans last updated on 11/Jun/20

$$\mathrm{substitution}w=-\ln \left(x\right),x=e^{-w}$$$$I=\underset{\mathrm{\infty }}{\stackrel{0}{\int }}\frac{{(-\mathrm{w})}^2}{1+{\left({\mathrm{e}}^{-\mathrm{w}}\right)}^2}(-{\mathrm{e}}^{-\mathrm{w}}\mathrm{dw})$$$$\mathrm{I}=\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{w}}^2.\frac{{\mathrm{e}}^{-\mathrm{w}}}{1+{\mathrm{e}}^{-2\mathrm{w}}}\mathrm{dw}$$$$[\frac{{\mathrm{e}}^{-\mathrm{w}}}{1+{\mathrm{e}}^{-2\mathrm{w}}}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}-1}{\mathrm{e}}^{-(2\mathrm{n}+1)\mathrm{w}}]$$$$\mathrm{I}=\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\left(\frac{\mathrm{t}}{2\mathrm{n}+1}\right)}^2{\mathrm{e}}^{-\mathrm{t}}\frac{\mathrm{dt}}{2\mathrm{n}+1};$$$$\mathrm{where}\mathrm{t}=(2\mathrm{n}+1)\mathrm{w}$$$$\mathrm{I}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^3}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\mathrm{t}}^2{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}$$$$\mathrm{I}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^3}.2!=2\times \underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^3}$$$$\mathrm{I}=2\times \frac{{\pi }^3}{32}=\frac{{\pi }^3}{16}◼$$

Commented by  M±th+et+s last updated on 11/Jun/20

$$gotoQ.97369$$

Commented by bemath last updated on 13/Jun/20

$$\mathrm{greatt}$$

Answered by mathmax by abdo last updated on 11/Jun/20

$$\mathrm{I}={\int }_0^1\frac{{\ln }^2\mathrm{x}}{{\mathrm{x}}^2+1}\mathrm{dx}\Rightarrow \mathrm{I}={\int }_0^1{\ln }^2\left(\mathrm{x}\right)\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}\right)\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^1{\ln }^2\mathrm{x}{\mathrm{x}}^{2\mathrm{n}}\mathrm{dx}\mathrm{let}{\mathrm{A}}_{\mathrm{n}}={\int }_0^1{\ln }^2\mathrm{x}{\mathrm{x}}^{2\mathrm{n}}\mathrm{dx}\mathrm{changement}\mathrm{lnx}=-\mathrm{t}$$$$\mathrm{give}{\mathrm{A}}_{\mathrm{n}}=-{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^2{\left({\mathrm{e}}^{-\mathrm{t}}\right)}^{2\mathrm{n}}(-{\mathrm{e}}^{-\mathrm{t}})\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}{\mathrm{t}}^2{\mathrm{e}}^{-(2\mathrm{n}+1)\mathrm{t}}\mathrm{dt}{=}_{(2\mathrm{n}+1)\mathrm{t}=\mathrm{u}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^2}{{(2\mathrm{n}+1)}^2}{\mathrm{e}}^{-\mathrm{u}}\frac{\mathrm{du}}{(2\mathrm{n}+1)}$$$$=\frac{1}{{(2\mathrm{n}+1)}^3}{\int }_0^{\mathrm{\infty }}{\mathrm{u}}^2{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}\mathrm{but}{\int }_0^{\mathrm{\infty }}{\mathrm{t}}^{\mathrm{x}-1}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}=\mathrm{\Gamma }\left(\mathrm{x}\right)\Rightarrow {\int }_0^{\mathrm{\infty }}{\mathrm{u}}^2{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}=\mathrm{\Gamma }\left(3\right)=2!=2$$$$\Rightarrow {\mathrm{A}}_{\mathrm{n}}=\frac{2}{{(2\mathrm{n}+1)}^3}\Rightarrow \mathrm{I}=2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{(2\mathrm{n}+1)}^3}\mathrm{rest}\mathrm{to}\mathrm{calculate}\mathrm{this}\mathrm{serie}\mathrm{by}\mathrm{fourier}...$$

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