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Question Number 106899 by abdomathmax last updated on 07/Aug/20

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{tsin}\left(2\mathrm{t}\right)}{{\mathrm{t}}^2+4}\mathrm{dt}$$

Answered by mathmax by abdo last updated on 08/Aug/20

$$\mathrm{A}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{t}\sin \left(2\mathrm{t}\right)}{{\mathrm{t}}^2+4}\mathrm{dt}\Rightarrow 2\mathrm{A}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{tsin}\left(2\mathrm{t}\right)}{{\mathrm{t}}^2+4}\mathrm{dt}$$$$=\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{t}{\mathrm{e}}^{2\mathrm{it}}}{{\mathrm{t}}^2+4}\mathrm{dt}\right)\mathrm{let}\phi \left(\mathrm{z}\right)=\frac{\mathrm{z}{\mathrm{e}}^{2\mathrm{iz}}}{{\mathrm{z}}^2+4}\Rightarrow \phi \left(\mathrm{z}\right)=\frac{\mathrm{z}{\mathrm{e}}^{2\mathrm{iz}}}{(\mathrm{z}-2\mathrm{i})(\mathrm{z}+2\mathrm{i})}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\phi \left(\mathrm{z}\right)\mathrm{dz}=2\mathrm{i}\pi \mathrm{Res}(\phi ,2\mathrm{i})=2\mathrm{i}\pi \times \frac{{2\mathrm{ie}}^{2\mathrm{i}\left(2\mathrm{i}\right)}}{4\mathrm{i}}=\mathrm{i}\pi {\mathrm{e}}^{-4}\Rightarrow$$$$2\mathrm{A}=\pi {\mathrm{e}}^{-4}\Rightarrow \mathrm{A}=\frac{\pi }{2}{\mathrm{e}}^{-4}$$

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