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Question Number 106899 by abdomathmax last updated on 07/Aug/20
calculate ∫_0 ^∞ ((tsin(2t))/(t^2 +4)) dt
$$\mathrm{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{tsin}\left(\mathrm{2t}\right)}{\mathrm{t}^{\mathrm{2}} +\mathrm{4}}\:\mathrm{dt} \\ $$
Answered by mathmax by abdo last updated on 08/Aug/20
A =∫_0 ^∞ ((t sin(2t))/(t^2 +4))dt ⇒2A =∫_(−∞) ^(+∞) ((tsin(2t))/(t^2 +4))dt =Im(∫_(−∞) ^(+∞) ((t e^(2it) )/(t^2 +4))dt) let ϕ(z) =((z e^(2iz) )/(z^2 +4)) ⇒ϕ(z) =((z e^(2iz) )/((z−2i)(z+2i))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ ,2i) =2iπ×((2ie^(2i(2i)) )/(4i)) =iπ e^(−4) ⇒ 2A =πe^(−4) ⇒ A =(π/2) e^(−4)
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}\:\mathrm{sin}\left(\mathrm{2t}\right)}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{4}}\mathrm{dt}\:\Rightarrow\mathrm{2A}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{tsin}\left(\mathrm{2t}\right)}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{4}}\mathrm{dt} \\ $$$$=\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{t}\:\mathrm{e}^{\mathrm{2it}} }{\mathrm{t}^{\mathrm{2}} \:+\mathrm{4}}\mathrm{dt}\right)\:\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{z}\:\mathrm{e}^{\mathrm{2iz}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow\varphi\left(\mathrm{z}\right)\:=\frac{\mathrm{z}\:\mathrm{e}^{\mathrm{2iz}} }{\left(\mathrm{z}−\mathrm{2i}\right)\left(\mathrm{z}+\mathrm{2i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi\:,\mathrm{2i}\right)\:=\mathrm{2i}\pi×\frac{\mathrm{2ie}^{\mathrm{2i}\left(\mathrm{2i}\right)} }{\mathrm{4i}}\:=\mathrm{i}\pi\:\mathrm{e}^{−\mathrm{4}} \:\Rightarrow \\ $$$$\mathrm{2A}\:=\pi\mathrm{e}^{−\mathrm{4}} \:\Rightarrow\:\mathrm{A}\:=\frac{\pi}{\mathrm{2}}\:\mathrm{e}^{−\mathrm{4}} \: \\ $$

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