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Question Number 188086 by universe last updated on 25/Feb/23

$$$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\tan }^{-1}\left(x/a\right)}{x(x^2+b^2)}dx$$

Answered by witcher3 last updated on 25/Feb/23

$$\mathrm{for}\mathrm{a}>0$$$$\frac{\mathrm{x}}{\mathrm{a}}=\mathrm{y}$$$$\iff {\int }_0^{\mathrm{\infty }}\frac{{\tan }^{-1}\left(\mathrm{y}\right)}{\mathrm{y}({\mathrm{a}}^2{\mathrm{y}}^2+{\mathrm{b}}^2)}\mathrm{dy}$$$$\frac{{\tan }^{-1}\left(\mathrm{y}\right)}{\mathrm{y}}={\int }_0^1\frac{\mathrm{dz}}{1+{\mathrm{z}}^2{\mathrm{y}}^2}$$$$\mathrm{I}={\int }_0^1{\int }_0^{\mathrm{\infty }}\frac{\mathrm{dydz}}{({\mathrm{a}}^2{\mathrm{y}}^2+{\mathrm{b}}^2)(1+{\mathrm{z}}^2{\mathrm{y}}^2)}$$$${\int }_0^{\mathrm{\infty }}\frac{\mathrm{dy}}{({\mathrm{a}}^2{\mathrm{y}}^2+{\mathrm{b}}^2)(1+{\mathrm{z}}^2{\mathrm{y}}^2)}=2\mathrm{i}\pi .\frac{{\mathrm{a}}^2}{2\mathrm{aib}({\mathrm{a}}^2-{\mathrm{b}}^2{\mathrm{z}}^2)}$$$$+2\mathrm{i}\pi .\frac{{\mathrm{z}}^2}{2\mathrm{iz}({\mathrm{b}}^2{\mathrm{z}}^2-{\mathrm{a}}^2)}$$$$=\frac{\pi \mathrm{a}}{\mathrm{b}({\mathrm{a}}^2-{\mathrm{b}}^2{\mathrm{z}}^2)}+\frac{\pi \mathrm{z}}{({\mathrm{b}}^2{\mathrm{z}}^2-{\mathrm{a}}^2)}=\pi \left(\frac{\mathrm{z}-\frac{\mathrm{a}}{\mathrm{b}}}{(\mathrm{bz}-\mathrm{a})(\mathrm{bz}+\mathrm{a})}\right)$$$$\mathrm{I}={\int }_0^1\frac{\pi }{\mathrm{b}(\mathrm{bz}+\mathrm{a})}=\frac{\pi }{{\mathrm{b}}^2}\ln (1+\frac{\mathrm{a}}{\mathrm{b}})$$

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