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Question Number 54374 by maxmathsup by imad last updated on 02/Feb/19

$$calculateh\left(a\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(at\right)}{ch\left(\frac{t}{2}\right)}dt.$$

Commented by Abdo msup. last updated on 08/Feb/19

$$h\left(a\right){=}_{\frac{t}{2}=x}2{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{cos\left(2ax\right)}{ch\left(x\right)}dx$$$$=4{\int }_0^{\mathrm{\infty }}\frac{cos\left(2ax\right)}{\frac{e^x+e^{-x}}{2}}dx=8{\int }_0^{\mathrm{\infty }}\frac{e^{-x}cos\left(2ax\right)}{1+e^{-2x}}dx$$$$=8{\int }_0^{\mathrm{\infty }}{e}^{-x}cos\left(2ax\right)\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{e}^{-2nx}\right)dx$$$$=8\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^{\mathrm{\infty }}{e}^{-(2n+1)x}cos\left(2ax\right)dx$$$$let{A}_n={\int }_0^{\mathrm{\infty }}{e}^{-(2n+1)x}cos\left(2ax\right)dx\Rightarrow$$$$A_n=Re\left({\int }_0^{\mathrm{\infty }}{e}^{-(2n+1)x}{e}^{2iax}dx\right)but$$$${\int }_0^{\mathrm{\infty }}{e}^{(-(2n+1)+2ia)x}dx={\left[\frac{1}{-(2n+1)+2ia}{e}^{(-(2n+1)+2ia)x}\right]}_0^{+\mathrm{\infty }}$$$$=\frac{1}{-(2n+1)+2ia}=-\frac{1}{(2n+1)-2ia}$$$$=-\frac{2n+1+2ia}{{(2n+1)}^2+4a^2}\Rightarrow A_n=-\frac{2n+1}{{(2n+1)}^2+4a^2}\Rightarrow$$$$h\left(a\right)=8\sum _{n=0}^{\mathrm{\infty }}{(-1)}^{n+1}\frac{2n+1}{{(2n+1)}^2+4a^2}$$$$andthisseriecanbecalculatedbyfourierseries$$$$...becontinued...$$

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