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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

See Q144057

$${See}\:{Q}\mathrm{144057} \\ $$

Commented by ArielVyny last updated on 24/Jun/21

exercise 3 question2

$${exercise}\:\mathrm{3}\:{question}\mathrm{2} \\ $$

Commented by ArielVyny last updated on 24/Jun/21

thank a lot

$${thank}\:{a}\:{lot} \\ $$

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

 Residus method→I=∫_0 ^∞  ((tsin(tx))/(a^2  +t^2 ))dt⇒I=_(t=∣a∣y)   ∫_0 ^∞   ((∣a∣ysin(∣a∣xy))/(a^2 (y^2  +1)))∣a∣ dy  =∫_0 ^∞  ((ysin(∣a∣xy))/(y^2  +1))dy =(1/2)∫_(−∞) ^(+∞)  ((ysin(∣a∣xy))/(y^2  +1))dy  =(1/2)Im(∫_(−∞) ^(+∞)  (y/(y^2  +1))e^(i∣a∣xy) dy)   we hsve  ∫_(−∞) ^(+∞)  (y/(y^2  +1))e^(i∣a∣xy) dy =2iπRes(ϕ,i)    (ϕ(z)=(z/(z^2  +1))e^(i∣a∣xz) )  =2iπ×(i/(2i))e^(−∣a∣x)  =iπ e^(−∣a∣x)  ⇒★I=(π/2)e^(−∣a∣x) ★

$$\:\mathrm{Residus}\:\mathrm{method}\rightarrow\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{tsin}\left(\mathrm{tx}\right)}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\Rightarrow\mathrm{I}=_{\mathrm{t}=\mid\mathrm{a}\mid\mathrm{y}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mid\mathrm{a}\mid\mathrm{ysin}\left(\mid\mathrm{a}\mid\mathrm{xy}\right)}{\mathrm{a}^{\mathrm{2}} \left(\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}\right)}\mid\mathrm{a}\mid\:\mathrm{dy} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ysin}\left(\mid\mathrm{a}\mid\mathrm{xy}\right)}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dy}\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{ysin}\left(\mid\mathrm{a}\mid\mathrm{xy}\right)}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{y}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{e}^{\mathrm{i}\mid\mathrm{a}\mid\mathrm{xy}} \mathrm{dy}\right)\:\:\:\mathrm{we}\:\mathrm{hsve} \\ $$$$\int_{−\infty} ^{+\infty} \:\frac{\mathrm{y}}{\mathrm{y}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{e}^{\mathrm{i}\mid\mathrm{a}\mid\mathrm{xy}} \mathrm{dy}\:=\mathrm{2i}\pi\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:\:\:\:\left(\varphi\left(\mathrm{z}\right)=\frac{\mathrm{z}}{\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{e}^{\mathrm{i}\mid\mathrm{a}\mid\mathrm{xz}} \right) \\ $$$$=\mathrm{2i}\pi×\frac{\mathrm{i}}{\mathrm{2i}}\mathrm{e}^{−\mid\mathrm{a}\mid\mathrm{x}} \:=\mathrm{i}\pi\:\mathrm{e}^{−\mid\mathrm{a}\mid\mathrm{x}} \:\Rightarrow\bigstar\mathrm{I}=\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\mid\mathrm{a}\mid\mathrm{x}} \bigstar \\ $$

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