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Question Number 211857 by Frix last updated on 22/Sep/24

$$\underset{1}{\stackrel{\mathrm{\infty }}{\int }}\left(\frac{{\tan }^{-1}x}{x}\times \frac{\ln x}{x}\right)dx=?$$

Answered by Berbere last updated on 23/Sep/24

$${\int }_1^{\mathrm{\infty }}\frac{{\tan }^{-1}\left(x\right)ln\left(x\right)}{x^2}dx;u^{\prime }=\frac{1}{x^2};v=ln\left(x\right){\tan }^{-1}\left(x\right)$$$$={[-\frac{ln\left(x\right){\tan }^{-1}\left(x\right)}{x}]}_1^{\mathrm{\infty }}+{\int }_1^{\mathrm{\infty }}{[\frac{{\tan }^{-1}\left(x\right)}{x^2}+\frac{ln\left(x\right)}{x(1+x^2)}]}_1^{\mathrm{\infty }}dx$$$$={\int }_1^{\mathrm{\infty }}\frac{{\tan }^{-1}\left(x\right)}{x^2}dx+{\int }_1^{\mathrm{\infty }}\frac{ln\left(x\right)}{x(1+x^2)}dx=A+B$$$$A={[-\frac{{\tan }^{-1}\left(x\right)}{x}]}_1^{\mathrm{\infty }}+{\int }_1^{\mathrm{\infty }}\frac{1}{x(x^2+1)}dx;x\to \frac{1}{x}$$$$=\frac{\pi }{4}+{\int }_0^1\frac{x}{x^2+1}=\frac{\pi }{4}+\frac{ln\left(5\right)}{2}$$$$B=-{\int }_0^{1}x\frac{ln\left(x\right)}{1+x^2}dx=-\frac{1}{2}{\int }_0^1\frac{ln\left(x\right)}{1+x}dx...x\to x^2$$$$=\frac{1}{2}{\int }_0^1\frac{ln(1+x)}{x}dx=\frac{1}{2}{\int }_0^1\sum _{n⩾1}\frac{{(-1)}^{n-1}}{n}{x}^{n-1}dx$$$$=\frac{1}{2}\mathrm{\Sigma }\frac{{(-1)}^{n-1}}{n^2}=-\frac{1}{2}{Li}_2(-1)=\frac{{\pi }^2}{24}$$$$J=\frac{{\pi }^2}{24}+\frac{\pi }{4}+\frac{ln\left(5\right)}{2}$$$$$$

Commented by Frix last updated on 23/Sep/24

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