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Question Number 60510 by aliesam last updated on 21/May/19

Commented by maxmathsup by imad last updated on 21/May/19

$$letA={\int }_0^{\mathrm{\infty }}\frac{cos\left(xt\right)}{ch\left(t\right)}dt\Rightarrow A={\int }_0^{\mathrm{\infty }}\frac{cos\left(xt\right)}{\frac{e^t+e^{-t}}{2}}dt$$$$=2{\int }_0^{\mathrm{\infty }}\frac{cos\left(xt\right)}{e^t+e^{-t}}dt=2{\int }_0^{\mathrm{\infty }}\frac{e^{-t}cos\left(xt\right)}{1+e^{-2t}}dt$$$$=2{\int }_0^{\mathrm{\infty }}{e}^{-t}cos\left(xt\right)\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{e}^{-2nt}\right)dt$$$$=2\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^{\mathrm{\infty }}{e}^{-(2n+1)t}cos\left(xt\right)dt=2\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{w}_n$$$$w_n=Re\left({\int }_0^{\mathrm{\infty }}{e}^{-(2n+1)t}{e}^{ixt}dt\right)=Re\left({\int }_0^{\mathrm{\infty }}{e}^{(ix-2n+1))t}dt\right)but$$$${\int }_0^{\mathrm{\infty }}{e}^{(ix-(2n+1))t}dt={\left[\frac{1}{ix-(2n+1)}{e}^{(ix-(2n+1))t}\right]}_{t=0}^{t\to +\mathrm{\infty }}$$$$=-\frac{1}{ix-(2n+1)}=\frac{1}{(2n+1)-ix}=\frac{2n+1+ix}{{(2n+1)}^2+x^2}\Rightarrow Re(...)=\frac{2n+1}{{(2n+1)}^2+x^2}\Rightarrow$$$$A=2\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{2n+1}{{(2n+1)}^2+x^2}.$$

Commented by aliesam last updated on 21/May/19

$$thankyousir$$$$$$

Commented by maxmathsup by imad last updated on 22/May/19

$$youarewelcomesir$$

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