Question Number 144348 by ArielVyny last updated on 24/Jun/21
Commented by Dwaipayan Shikari last updated on 24/Jun/21
$${See}\:{Q}\mathrm{144057} \\ $$
Commented by ArielVyny last updated on 24/Jun/21
$${exercise}\:\mathrm{3}\:{question}\mathrm{2} \\ $$
Commented by ArielVyny last updated on 24/Jun/21
$${thank}\:{a}\:{lot} \\ $$
Answered by mathmax by abdo last updated on 25/Jun/21
$$\:\mathrm{Residus}\:\mathrm{method}\rightarrow\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{tsin}\left(\mathrm{tx}\right)}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\Rightarrow\mathrm{I}=_{\mathrm{t}=\mid\mathrm{a}\mid\mathrm{y}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mid\mathrm{a}\mid\mathrm{ysin}\left(\mid\mathrm{a}\mid\mathrm{xy}\right)}{\mathrm{a}^{\mathrm{2}} \left(\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}\right)}\mid\mathrm{a}\mid\:\mathrm{dy} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ysin}\left(\mid\mathrm{a}\mid\mathrm{xy}\right)}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dy}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{ysin}\left(\mid\mathrm{a}\mid\mathrm{xy}\right)}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{y}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{e}^{\mathrm{i}\mid\mathrm{a}\mid\mathrm{xy}} \mathrm{dy}\right)\:\:\:\mathrm{we}\:\mathrm{hsve} \\ $$$$\int_{−\infty} ^{+\infty} \:\frac{\mathrm{y}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{e}^{\mathrm{i}\mid\mathrm{a}\mid\mathrm{xy}} \mathrm{dy}\:=\mathrm{2i}\pi\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:\:\:\:\left(\varphi\left(\mathrm{z}\right)=\frac{\mathrm{z}}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{e}^{\mathrm{i}\mid\mathrm{a}\mid\mathrm{xz}} \right) \\ $$$$=\mathrm{2i}\pi×\frac{\mathrm{i}}{\mathrm{2i}}\mathrm{e}^{−\mid\mathrm{a}\mid\mathrm{x}} \:=\mathrm{i}\pi\:\mathrm{e}^{−\mid\mathrm{a}\mid\mathrm{x}} \:\Rightarrow\bigstar\mathrm{I}=\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mid\mathrm{a}\mid\mathrm{x}} \bigstar \\ $$