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Question Number 147322 by Gbenga last updated on 19/Jul/21

Commented by Mrsof last updated on 19/Jul/21

$$I={\int }_0^{\mathrm{\infty }}\frac{cos\left(ax\right)}{x^2+b^2}dx{=}_{x=z}2I={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{cos\left(az\right)}{z^2+b^2}dz$$$$$$$$2I=Re\left({\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{aiz}}{(z-bi)(z+bi)}dz\right)$$$$$$$$thefunctionhasaresideoofz=bi$$$$$$$$Res(f,bi)={lim}_{z\to bi}(z-bi)\times \frac{e^{aiz}}{(z-bi)(z+bi)}=\frac{e^{-ab}}{2bi}$$$$$$$$2I=2\pi i\left(Res(f,bi)\right)\Rightarrow I=\pi i\times \frac{e^{-ab}}{2bi}=\frac{\pi e^{-ab}}{2b}$$$$$$$$⟨M.T⟩$$

Commented by Gbenga last updated on 19/Jul/21

$$thanks$$

Commented by Mrsof last updated on 20/Jul/21

$$youarewelcome$$

Answered by qaz last updated on 20/Jul/21

$${\int }_0^{\mathrm{\infty }}\frac{\mathcal{L}\left(\cos \left(\mathrm{ax}\right)\right)}{{\mathrm{x}}^2+{\mathrm{b}}^2}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{s}}{({\mathrm{x}}^2+{\mathrm{b}}^2)({\mathrm{x}}^2+{\mathrm{s}}^2)}\mathrm{dx}$$$$=\frac{\mathrm{s}}{{\mathrm{s}}^2-{\mathrm{b}}^2}{\int }_0^{\mathrm{\infty }}(\frac{1}{{\mathrm{x}}^2+{\mathrm{b}}^2}-\frac{1}{{\mathrm{x}}^2+{\mathrm{s}}^2})\mathrm{dx}$$$$=\frac{\pi }{2\mathrm{b}(\mathrm{s}+\mathrm{b})}$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{\cos \left(\mathrm{ax}\right)}{{\mathrm{x}}^2+{\mathrm{b}}^2}\mathrm{dx}={\mathcal{L}}^{-1}\left(\frac{\pi }{2\mathrm{b}(\mathrm{s}+\mathrm{b})}\right)=\frac{\pi }{2\mathrm{b}}{\mathrm{e}}^{-\mathrm{ba}}$$

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