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Question Number 127948 by rs4089 last updated on 03/Jan/21

Answered by mathmax by abdo last updated on 04/Jan/21

$${\mathrm{A}}_{\mathrm{N}}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{2\pi \mathrm{x}}-1}{{\mathrm{e}}^{2\pi \mathrm{x}}+1}(\frac{1}{\mathrm{x}}-\frac{\mathrm{x}}{{\mathrm{N}}^2+{\mathrm{x}}^2})\mathrm{dx}\Rightarrow$$$${\mathrm{A}}_{\mathrm{N}}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\pi \mathrm{x}}-{\mathrm{e}}^{-\pi \mathrm{x}}}{{\mathrm{e}}^{\pi \mathrm{x}}+{\mathrm{e}}^{-\pi \mathrm{x}}}\left(\frac{{\mathrm{N}}^2+{\mathrm{x}}^2-{\mathrm{x}}^2}{\mathrm{x}({\mathrm{N}}^2+{\mathrm{x}}^2)}\right)\mathrm{dx}={\mathrm{N}}^2{\int }_0^{\mathrm{\infty }}\frac{\mathrm{sh}\left(\mathrm{x}\right)}{\mathrm{xch}\left(\mathrm{x}\right)({\mathrm{x}}^2+{\mathrm{N}}^2)}\mathrm{dx}$$$$=\frac{{\mathrm{N}}^2}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{shx}}{\mathrm{xch}\left(\mathrm{x}\right)({\mathrm{x}}^2+{\mathrm{N}}^2)}\mathrm{dx}\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{\mathrm{shz}}{\mathrm{xch}\left(\mathrm{z}\right)({\mathrm{z}}^2+{\mathrm{N}}^2)}\Rightarrow$$$$\phi \left(\mathrm{z}\right)=\frac{\mathrm{th}\left(\mathrm{z}\right)}{\mathrm{z}({\mathrm{z}}^2+{\mathrm{N}}^2)}\mathrm{the}\mathrm{poles}\mathrm{of}\phi \mathrm{are}0,\stackrel{-}{+}\mathrm{iN}\mathrm{residus}\Rightarrow$$$${\int }_{\mathrm{R}}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \{\mathrm{Res}(\phi ,\mathrm{o})+\mathrm{Res}(\phi ,\mathrm{iN})\}$$$$\mathrm{Res}(\phi ,\mathrm{o})=0$$$$\mathrm{Res}(\phi ,\mathrm{iN})=\frac{\mathrm{th}\left(\mathrm{iN}\right)}{2\mathrm{iN}}=\frac{\mathrm{sh}\left(\mathrm{iN}\right)}{2\mathrm{iNch}\left(\mathrm{iN}\right)}$$$$\mathrm{sh}\left(\mathrm{iN}\right)=\frac{{\mathrm{e}}^{\mathrm{iN}}-{\mathrm{e}}^{-\mathrm{iN}}}{2}=\mathrm{isinN}$$$$\mathrm{ch}\left(\mathrm{iN}\right)=\frac{{\mathrm{e}}^{\mathrm{iN}}+{\mathrm{e}}^{-\mathrm{iN}}}{2}=\mathrm{cosN}\Rightarrow {\int }_{\mathrm{R}}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \times \frac{\mathrm{isinN}}{2\mathrm{iNcosN}}$$$$=\mathrm{i}\pi \tan \left(\mathrm{N}\right)\Rightarrow {\mathrm{A}}_{\mathrm{N}}=\frac{\pi {\mathrm{N}}^2}{2}\mathrm{itan}\left(\mathrm{N}\right)\mathrm{but}\mathrm{my}\mathrm{answer}\mathrm{is}\mathrm{not}\mathrm{sure}$$$$\mathrm{if}{\mathrm{A}}_{\mathrm{N}}\mathrm{is}\mathrm{real}...!$$

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