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Question Number 103741 by mathmax by abdo last updated on 17/Jul/20

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{\arctan \left(\mathrm{ch}\left(\mathrm{x}\right)\right)}{{\mathrm{x}}^2+9}\mathrm{dx}$$

Answered by mathmax by abdo last updated on 17/Jul/20

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\arctan \left(\mathrm{chx}\right)}{{\mathrm{x}}^2+9}\mathrm{dx}\Rightarrow 2\mathrm{A}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\arctan \left(\mathrm{chx}\right)}{{\mathrm{x}}^2+9}\mathrm{dx}\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{\arctan \left(\mathrm{chz}\right)}{{\mathrm{z}}^2+9}$$$$\mathrm{we}\mathrm{have}{\lim }_{\mathrm{z}\to +\mathrm{\infty }}\mid \mathrm{z}\phi \left(\mathrm{z}\right)\mid =0\mathrm{and}\phi \left(\mathrm{z}\right)=\frac{\arctan \left(\mathrm{chz}\right)}{(\mathrm{z}-3\mathrm{i})(\mathrm{z}+3\mathrm{i})}$$$$\mathrm{residus}\mathrm{theorem}\mathrm{give}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,3\mathrm{i})$$$$=2\mathrm{i}\pi \times \frac{\arctan \left(\mathrm{ch}\left(3\mathrm{i}\right)\right)}{6\mathrm{i}}=\frac{\pi }{3}\arctan \left(\frac{{\mathrm{e}}^{3\mathrm{i}}+{\mathrm{e}}^{-3\mathrm{i}}}{2}\right)=\frac{\pi }{3}\arctan \left(\cos \left(3\right)\right)\Rightarrow$$$$\mathrm{A}=\frac{\pi }{6}\arctan \left(\cos 3\right)$$

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