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Question Number 117088 by mnjuly1970 last updated on 09/Oct/20
... advanced calculus... evsluate :: I=∫_0 ^( ∞) ((xsin(2x))/(x^2 +4)) dx =? hint: Φ=∫_0 ^( ∞) ((cos(px))/(x^2 +1))dx =(π/2)e^(−p) (p>0) .m.n 1970
$$\:\:\:\:\:\:\:\:\:\:…\:{advanced}\:\:\:{calculus}… \\ $$$$ \\ $$$$\:\:\:\:\:\:\:{evsluate}\::: \\ $$$$\:\: \\ $$$$\:\:\:\:\:\:\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\infty} \frac{{xsin}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{2}} +\mathrm{4}}\:{dx}\:=? \\ $$$$\:\:\:\:\:{hint}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\Phi=\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left({px}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\:=\frac{\pi}{\mathrm{2}}{e}^{−{p}} \:\:\:\:\left({p}>\mathrm{0}\right)\:\:\: \\ $$$$ \\ $$$$\:\:\:\:\:.{m}.{n}\:\mathrm{1970} \\ $$$$ \\ $$
Answered by mnjuly1970 last updated on 09/Oct/20
Φ =∫_0 ^( ∞) ((cos(px))/(x^2 +1)) =_(respect to p) ^((diff both sides)) (π/2)e^(−p ) =−∫_0 ^( ∞) ((xsin(px))/(x^2 +1))dx=−(π/2)e^(−p) =∫_0 ^( ∞) ((xsin(px))/(x^2 +1))dx=(π/2)e^p I =^(x=2t) ∫_0 ^( ∞) ((tsin(4t))/(t^2 +1))dt =^Φ (π/2)e^(−4) ..m.n.1970...
$$\Phi\:=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{cos}\left({px}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}\:\underset{{respect}\:\:{to}\:\:{p}} {\overset{\left({diff}\:\:{both}\:{sides}\right)} {=}}\frac{\pi}{\mathrm{2}}{e}^{−{p}\:} \\ $$$$=−\int_{\mathrm{0}} ^{\:\infty} \:\frac{{xsin}\left({px}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=−\frac{\pi}{\mathrm{2}}{e}^{−{p}} \\ $$$$=\int_{\mathrm{0}} ^{\:\infty} \:\frac{{xsin}\left({px}\right)}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}=\frac{\pi}{\mathrm{2}}{e}^{{p}} \\ $$$$\mathrm{I}\:\overset{{x}=\mathrm{2}{t}} {=}\int_{\mathrm{0}} ^{\:\infty} \frac{{tsin}\left(\mathrm{4}{t}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}\:\overset{\Phi} {=}\:\frac{\pi}{\mathrm{2}}{e}^{−\mathrm{4}} \\ $$$$\:\:\:..{m}.{n}.\mathrm{1970}… \\ $$$$\: \\ $$$$ \\ $$
Answered by Bird last updated on 10/Oct/20
2I =∫_(−∞) ^(+∞ ) ((xsin(2x))/(x^(2 ) +4))dx =Im(∫_(−∞) ^(+∞) ((xe^(2ix) )/(x^2 +4))dx) let ϕ(z) =((z e^(2iz) )/(z^2 +4)) ⇒ϕ(z) =((ze^(2iz) )/((z−2i)(z+2i))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{Res(ϕ,2i)} =2iπ ×((2i e^(−4) )/(4i)) =((−4π e^(−4) )/(4i)) =iπ e^(−4) ⇒2I =π e^(−4) ⇒I =(π/2)e^(−4)
$$\mathrm{2}{I}\:=\int_{−\infty} ^{+\infty\:} \:\frac{{xsin}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{2}\:} +\mathrm{4}}{dx} \\ $$$$={Im}\left(\int_{−\infty} ^{+\infty} \:\frac{{xe}^{\mathrm{2}{ix}} }{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx}\right)\:{let} \\ $$$$\varphi\left({z}\right)\:=\frac{{z}\:{e}^{\mathrm{2}{iz}} }{{z}^{\mathrm{2}} \:+\mathrm{4}}\:\Rightarrow\varphi\left({z}\right)\:=\frac{{ze}^{\mathrm{2}{iz}} }{\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{{Res}\left(\varphi,\mathrm{2}{i}\right)\right\} \\ $$$$=\mathrm{2}{i}\pi\:×\frac{\mathrm{2}{i}\:{e}^{−\mathrm{4}} }{\mathrm{4}{i}}\:=\frac{−\mathrm{4}\pi\:{e}^{−\mathrm{4}} }{\mathrm{4}{i}}\:={i}\pi\:{e}^{−\mathrm{4}} \\ $$$$\Rightarrow\mathrm{2}{I}\:=\pi\:{e}^{−\mathrm{4}} \:\Rightarrow{I}\:=\frac{\pi}{\mathrm{2}}{e}^{−\mathrm{4}} \\ $$
Commented by mnjuly1970 last updated on 10/Oct/20
grateful mr bird very nice as always..
$${grateful}\:{mr}\:{bird}\: \\ $$$${very}\:{nice}\:{as}\:{always}.. \\ $$

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