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Question Number 164163 by Zaynal last updated on 15/Jan/22

$$\mathrm{Prove}\mathrm{the};$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{1+{\mathit{x}}^2}\mathit{d}\mathit{x}=\mathit{\pi }$$$${}^{\{\mathrm{Z}.\mathrm{A}\}}$$

Answered by mathmax by abdo last updated on 15/Jan/22

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}={\lim }_{\xi \to +\mathrm{\infty }}{\int }_{-\xi }^{\xi }\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}={\lim }_{\xi \to +\mathrm{\infty }}{\left[\mathrm{arctanx}\right]}_{-\xi }^{\xi }$$$$={\lim }_{\xi \to +\mathrm{\infty }}2\arctan \xi =2.\frac{\pi }{2}=\pi$$$$\mathrm{or}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{dx}}{1+{\mathrm{x}}^2}{=}_{\mathrm{x}=\tan \theta }{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1+{\tan }^2\theta }{1+{\tan }^2\theta }\mathrm{d}\theta ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\mathrm{d}\theta =\pi$$

Commented by Zaynal last updated on 15/Jan/22

$$\mathrm{thank}\mathrm{you}\mathrm{sir},\mathrm{very}\mathrm{good}$$

Commented by Mathspace last updated on 15/Jan/22

$$youarewelcome$$

Answered by smallEinstein last updated on 16/Jan/22

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