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Question Number 144348 by ArielVyny last updated on 24/Jun/21

Commented by Dwaipayan Shikari last updated on 24/Jun/21

$$SeeQ144057$$

Commented by ArielVyny last updated on 24/Jun/21

$$exercise3question2$$

Commented by ArielVyny last updated on 24/Jun/21

$$thankalot$$

Answered by mathmax by abdo last updated on 25/Jun/21

$$\mathrm{Residus}\mathrm{method}\to \mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\mathrm{tsin}\left(\mathrm{tx}\right)}{{\mathrm{a}}^2+{\mathrm{t}}^2}\mathrm{dt}\Rightarrow \mathrm{I}{=}_{\mathrm{t}=\mid \mathrm{a}\mid \mathrm{y}}{\int }_0^{\mathrm{\infty }}\frac{\mid \mathrm{a}\mid \mathrm{ysin}(\mid \mathrm{a}\mid \mathrm{xy})}{{\mathrm{a}}^2({\mathrm{y}}^2+1)}\mid \mathrm{a}\mid \mathrm{dy}$$$$={\int }_0^{\mathrm{\infty }}\frac{\mathrm{ysin}(\mid \mathrm{a}\mid \mathrm{xy})}{{\mathrm{y}}^2+1}\mathrm{dy}=\frac{1}{2}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{ysin}(\mid \mathrm{a}\mid \mathrm{xy})}{{\mathrm{y}}^2+1}\mathrm{dy}$$$$=\frac{1}{2}\mathrm{Im}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{y}}{{\mathrm{y}}^2+1}{\mathrm{e}}^{\mathrm{i}\mid \mathrm{a}\mid \mathrm{xy}}\mathrm{dy}\right)\mathrm{we}\mathrm{hsve}$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\mathrm{y}}{{\mathrm{y}}^2+1}{\mathrm{e}}^{\mathrm{i}\mid \mathrm{a}\mid \mathrm{xy}}\mathrm{dy}=2\mathrm{i}\pi \mathrm{Res}(\phi ,\mathrm{i})(\phi \left(\mathrm{z}\right)=\frac{\mathrm{z}}{{\mathrm{z}}^2+1}{\mathrm{e}}^{\mathrm{i}\mid \mathrm{a}\mid \mathrm{xz}})$$$$=2\mathrm{i}\pi \times \frac{\mathrm{i}}{2\mathrm{i}}{\mathrm{e}}^{-\mid \mathrm{a}\mid \mathrm{x}}=\mathrm{i}\pi {\mathrm{e}}^{-\mid \mathrm{a}\mid \mathrm{x}}\Rightarrow ★\mathrm{I}=\frac{\pi }{2}{\mathrm{e}}^{-\mid \mathrm{a}\mid \mathrm{x}}★$$

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