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Question Number 33350 by caravan msup abdo. last updated on 14/Apr/18

$$provethat{\int }_0^{\mathrm{\infty }}\frac{cos\left(\alpha x\right)}{chx}dx=$$$$2\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n\frac{2n+1}{{(2n+1)}^2+{\alpha }^2}.$$

Commented by abdo imad last updated on 17/Apr/18

$$letputI={\int }_0^{\mathrm{\infty }}\frac{cos\left(\alpha x\right)}{chx}dx$$$$I=2{\int }_0^{\mathrm{\infty }}\frac{cos\left(\alpha x\right)}{e^x+e^{-x}}dx=2{\int }_0^{\mathrm{\infty }}\frac{e^{-x}cos\left(\alpha x\right)}{1+e^{-2x}}dx$$$$=2{\int }_0^{\mathrm{\infty }}\left(\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{e}^{-2nx}\right).e^{-x}cos\left(\alpha x\right)dx$$$$=2\sum _{n=0}^{\mathrm{\infty }}{(-1)}^n{\int }_0^{\mathrm{\infty }}{e}^{-(2n+1)x}cos\left(\alpha x\right)dxbut$$$$A_n={\int }_0^{\mathrm{\infty }}{e}^{-(2n+1)x}cos\left(\alpha x\right)dx=Re\left({\int }_0^{\mathrm{\infty }}{e}^{-(2n+1)x}{e}^{-i\alpha x}dx\right)$$$$=Re\left({\int }_0^{\mathrm{\infty }}{e}^{-(2n+1+i\alpha )x}dx\right)butwehave$$$${\int }_0^{\mathrm{\infty }}{e}^{-(2n+1+i\alpha )x}dx={[-\frac{1}{2n+1+i\alpha }{e}^{-(2n+1+i\alpha )}]}_0^{+\mathrm{\infty }}$$$$=\frac{1}{2n+1+i\alpha }=\frac{2n+1-i\alpha }{{(2n+1)}^2+{\alpha }^2}\Rightarrow A_n=\frac{2n+1}{{(2n+1)}^2+{\alpha }^2}\Rightarrow$$$$I=2\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n(2n+1)}{{(2n+1)}^2+{\alpha }^2}.$$

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