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Question Number 27187 by abdo imad last updated on 02/Jan/18

$$findI={\int }_0^{\propto }\frac{cosx}{cosh\left(x\right)}dx$$

Commented by abdo imad last updated on 08/Jan/18

$$I={\int }_0^{\mathrm{\infty }}\frac{cosx}{\frac{e^x+e^{-x}}{2}}dx=2{\int }_0^{\mathrm{\infty }}\frac{e^{-x}}{1+e^{-2x}}cosxdx$$$$=2{\int }_0^{\mathrm{\infty }}{e}^{-x}\left(\sum _{n=0}^{\propto }{e}^{-2nx}\right)cosxdx$$$$=2\sum _{n=0}^{\propto }{\int }_0^{\mathrm{\infty }}{e}^{-(2n+1)x}cosxdxbut$$$$letcalculate{\int }_0^{\mathrm{\infty }}{e}^{-ax}cosxdxwitha>0$$$${\int }_0^{\mathrm{\infty }}{e}^{-ax}cosxdx=Re\left({\int }_0^{\mathrm{\infty }}{e}^{(i-a)x}dx\right)and$$$${\int }_0^{\mathrm{\infty }}{e}^{(i-a)x}dx={\left[\frac{1}{i-a}{e}^{(i-a)x}\right]}_{x=0}^{x->\propto }=-\frac{1}{i-a}=\frac{1}{a-i}$$$$=\frac{a+i}{a^2+1}\Rightarrow {\int }_0^{\mathrm{\infty }}{e}^{-ax}cosxdx=\frac{a}{1+a^2}so$$$${\int }_0^{\mathrm{\infty }}{e}^{-(2n+1)x}cosxdx=\frac{2n+1}{1+{(2n+1)}^2}$$$$I=2\sum _{n=0}^{\propto }\frac{2n+1}{1+{(2n+1)}^2}andthissumiscalculatedbyfourier$$$$series...becontinued....$$$$$$

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