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Question Number 117121 by mnjuly1970 last updated on 09/Oct/20

...nice mathematics.. please evaluate... Ω =∫_(−∞) ^( +∞) (((x^2 −4)/(x^2 +4))∗ ((sin(2x))/x)) dx =??? m.n.1970

$$\:\:\:\:\:\:\:\:...{nice}\:\:{mathematics}.. \\ $$$$ \\ $$$$\:\:{please}\:\:{evaluate}... \\ $$$$\: \\ $$$$\:\Omega\:=\int_{−\infty} ^{\:+\infty} \left(\frac{{x}^{\mathrm{2}} −\mathrm{4}}{{x}^{\mathrm{2}} +\mathrm{4}}\ast\:\frac{{sin}\left(\mathrm{2}{x}\right)}{{x}}\right)\:{dx}\:=???\:\: \\ $$$$\:{m}.{n}.\mathrm{1970} \\ $$$$ \\ $$

Answered by Bird last updated on 10/Oct/20

I =∫_(−∞) ^(+∞) (((x^2 −4)sin(2x))/(x(x^2 +4)))dx ⇒ I=Im(∫_(−∞) ^(+∞) (((x^2 −4)e^(2ix) )/(x(x^2 +4))) dx) let consider the complex function ϕ(z) =(((z^2 −4)e^(2iz) )/(z(z^2 +4))) ⇒ ϕ(z) =(((z^2 −4)e^(2iz) )/(z(z−2i)(z+2i))) residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,o)+Res(ϕ,2i)} Res(ϕ,o) =lim_(z→0) zϕ(z) =lim_(z→0) ((−4)/4)=−1 Res(ϕ,2i) =lim_(z→2i) (z−2i)ϕ(z) =lim_(z→2i) (((z^2 −4)e^(2iz) )/(z(z+2i))) =((−8 e^(−4) )/((2i)(4i))) =e^(−4) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{−1 +e^(−4) } ⇒ I =2π{e^(−4) −1}

$${I}\:=\int_{−\infty} ^{+\infty} \:\frac{\left({x}^{\mathrm{2}} −\mathrm{4}\right){sin}\left(\mathrm{2}{x}\right)}{{x}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx}\:\Rightarrow \\ $$$${I}={Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\left({x}^{\mathrm{2}} −\mathrm{4}\right){e}^{\mathrm{2}{ix}} }{{x}\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:{dx}\right) \\ $$$${let}\:{consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\frac{\left({z}^{\mathrm{2}} −\mathrm{4}\right){e}^{\mathrm{2}{iz}} }{{z}\left({z}^{\mathrm{2}} +\mathrm{4}\right)}\:\Rightarrow \\ $$$$\varphi\left({z}\right)\:=\frac{\left({z}^{\mathrm{2}} −\mathrm{4}\right){e}^{\mathrm{2}{iz}} }{{z}\left({z}−\mathrm{2}{i}\right)\left({z}+\mathrm{2}{i}\right)}\:{residus} \\ $$$${theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,{o}\right)+{Res}\left(\varphi,\mathrm{2}{i}\right)\right\} \\ $$$${Res}\left(\varphi,{o}\right)\:={lim}_{{z}\rightarrow\mathrm{0}} \:{z}\varphi\left({z}\right) \\ $$$$={lim}_{{z}\rightarrow\mathrm{0}} \:\:\:\frac{−\mathrm{4}}{\mathrm{4}}=−\mathrm{1} \\ $$$${Res}\left(\varphi,\mathrm{2}{i}\right)\:={lim}_{{z}\rightarrow\mathrm{2}{i}} \left({z}−\mathrm{2}{i}\right)\varphi\left({z}\right) \\ $$$$={lim}_{{z}\rightarrow\mathrm{2}{i}} \:\:\:\frac{\left({z}^{\mathrm{2}} −\mathrm{4}\right){e}^{\mathrm{2}{iz}} }{{z}\left({z}+\mathrm{2}{i}\right)}\:\:\: \\ $$$$=\frac{−\mathrm{8}\:{e}^{−\mathrm{4}} }{\left(\mathrm{2}{i}\right)\left(\mathrm{4}{i}\right)}\:={e}^{−\mathrm{4}} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{−\mathrm{1}\:+{e}^{−\mathrm{4}} \right\}\:\Rightarrow \\ $$$${I}\:=\mathrm{2}\pi\left\{{e}^{−\mathrm{4}} −\mathrm{1}\right\} \\ $$

Commented by AbduraufKodiriy last updated on 10/Oct/20

This solution is wrong. Because, according to Cauchy′s formula, if singularity points of the function are not on the boundary of any set D, the formula that you wrote is correct. Sorry for my mistakes in English.

$$\boldsymbol{{This}}\:\boldsymbol{{solution}}\:\boldsymbol{{is}}\:\boldsymbol{{wrong}}.\:\boldsymbol{{Because}},\:\boldsymbol{{according}}\:\boldsymbol{{to}} \\ $$$$\boldsymbol{{Cauchy}}'\boldsymbol{{s}}\:\boldsymbol{{formula}},\:\boldsymbol{{if}}\:\boldsymbol{{singularity}}\:\boldsymbol{{points}} \\ $$$$\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{function}}\:\boldsymbol{{are}}\:\boldsymbol{{not}}\:\boldsymbol{{on}}\:\boldsymbol{{the}}\:\boldsymbol{{boundary}} \\ $$$$\boldsymbol{{of}}\:\boldsymbol{{any}}\:\boldsymbol{{set}}\:\boldsymbol{{D}},\:\boldsymbol{{the}}\:\boldsymbol{{formula}}\:\boldsymbol{{that}}\:\boldsymbol{{you}}\:\boldsymbol{{wrote}} \\ $$$$\boldsymbol{{is}}\:\boldsymbol{{correct}}. \\ $$$$\boldsymbol{{Sorry}}\:\boldsymbol{{for}}\:\boldsymbol{{my}}\:\boldsymbol{{mistakes}}\:\boldsymbol{{in}}\:\boldsymbol{{English}}. \\ $$

Answered by AbduraufKodiriy last updated on 09/Oct/20

𝛀=∫_(−∞) ^( ∞) ((x^2 −4)/(x^2 +4))∙((sin2x)/x)dx=2∫_0 ^( ∞) (1−(8/(x^2 +4)))((sin2x)/x)dx= =2∫_0 ^( ∞) ((sin2x)/x)dx−16∫_0 ^( ∞) ((sin2x)/(x(x^2 +4)))dx=π−16∫_0 ^( ∞) ((sin2x)/(x(x^2 +4)))dx; 𝛒=∫_0 ^( ∞) ((sin2x)/(x(x^2 +4)))dx; 𝛒(t)=∫_0 ^( ∞) ((sin(tx))/(x(x^2 +4)))dx; L (𝛒(t))=L (∫_0 ^( ∞) ((sin(tx))/(x(x^2 +4)))dx)=∫_0 ^( ∞) e^(−st) ∫_0 ^( ∞) ((sin(tx))/(x(x^2 +4)))dxdt= =∫_0 ^( ∞) (1/(x(x^2 +4)))∫_0 ^( ∞) e^(−st) sin(tx)dtdx; I=∫_0 ^( ∞) e^(−st) sin(tx)dt=−(1/s)e^(−st) sin(tx)∣_0 ^∞ −(x/s^2 )e^(−st) cos(tx)∣_0 ^∞ −(x^2 /s^2 )I ⇒ ⇒ (1+(x^2 /s^2 ))I=(x/s^2 ) ⇒ I=(x/(s^2 +x^2 )) L (𝛒(t))=∫_0 ^( ∞) (1/((x^2 +4)(x^2 +s^2 )))dx=(1/(s^2 −4))∫((1/(x^2 +4))−(1/(x^2 +s^2 )))dx= =(1/(s^2 −4))((1/2)arctan((x/2))−(1/s)arctan((x/s)))∣_0 ^∞ = =(1/(s^2 −4))∙((1/2)−(1/s))(π/2)=(π/4)∙(1/(s(s+2)))=(π/8)((1/s)−(1/(s+2))) L^(−1) (L (𝛒(t)))=L^( −1) ((π/8)∙((1/s)−(1/(s+2)))) ⇒ ⇒ 𝛒(t)=(π/8)(1−e^(−2t) ); 𝛒=𝛒(t=2) ⇒ 𝛒=(π/8)(1−e^(−4) ) 𝛀=π−16𝛒=π−2π(1−e^(−4) )=−π(1−2e^(−4) )

$$\boldsymbol{\Omega}=\int_{−\infty} ^{\:\infty} \frac{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{4}}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}}\centerdot\frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{x}}}\boldsymbol{{dx}}=\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \left(\mathrm{1}−\frac{\mathrm{8}}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}}\right)\frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{x}}}\boldsymbol{{dx}}= \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{x}}}\boldsymbol{{dx}}−\mathrm{16}\int_{\mathrm{0}} ^{\:\infty} \frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\right)}\boldsymbol{{dx}}=\pi−\mathrm{16}\int_{\mathrm{0}} ^{\:\infty} \frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\right)}\boldsymbol{{dx}}; \\ $$$$\boldsymbol{\rho}=\int_{\mathrm{0}} ^{\:\infty} \frac{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\right)}\boldsymbol{{dx}};\:\boldsymbol{\rho}\left(\boldsymbol{{t}}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{\boldsymbol{{sin}}\left(\boldsymbol{{tx}}\right)}{\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\right)}\boldsymbol{{dx}}; \\ $$$$\mathscr{L}\:\left(\boldsymbol{\rho}\left(\boldsymbol{{t}}\right)\right)=\mathscr{L}\:\left(\int_{\mathrm{0}} ^{\:\infty} \frac{\boldsymbol{{sin}}\left(\boldsymbol{{tx}}\right)}{\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\right)}\boldsymbol{{dx}}\right)=\int_{\mathrm{0}} ^{\:\infty} \boldsymbol{{e}}^{−\boldsymbol{{st}}} \int_{\mathrm{0}} ^{\:\infty} \frac{\boldsymbol{{sin}}\left(\boldsymbol{{tx}}\right)}{\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\right)}\boldsymbol{{dxdt}}= \\ $$$$=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{\boldsymbol{{x}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\right)}\int_{\mathrm{0}} ^{\:\infty} \boldsymbol{{e}}^{−\boldsymbol{{st}}} \boldsymbol{{sin}}\left(\boldsymbol{{tx}}\right)\boldsymbol{{dtdx}}; \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\:\infty} \boldsymbol{{e}}^{−\boldsymbol{{st}}} \boldsymbol{{sin}}\left(\boldsymbol{{tx}}\right)\boldsymbol{{dt}}=−\frac{\mathrm{1}}{\boldsymbol{{s}}}\boldsymbol{{e}}^{−\boldsymbol{{st}}} \boldsymbol{{sin}}\left(\boldsymbol{{tx}}\right)\mid_{\mathrm{0}} ^{\infty} −\frac{\boldsymbol{{x}}}{\boldsymbol{{s}}^{\mathrm{2}} }\boldsymbol{{e}}^{−\boldsymbol{{st}}} \boldsymbol{{cos}}\left(\boldsymbol{{tx}}\right)\mid_{\mathrm{0}} ^{\infty} −\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\boldsymbol{{s}}^{\mathrm{2}} }\boldsymbol{{I}}\:\Rightarrow \\ $$$$\Rightarrow\:\left(\mathrm{1}+\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\boldsymbol{{s}}^{\mathrm{2}} }\right)\boldsymbol{{I}}=\frac{\boldsymbol{{x}}}{\boldsymbol{{s}}^{\mathrm{2}} }\:\Rightarrow\:\boldsymbol{{I}}=\frac{\boldsymbol{{x}}}{\boldsymbol{{s}}^{\mathrm{2}} +\boldsymbol{{x}}^{\mathrm{2}} } \\ $$$$\mathscr{L}\:\left(\boldsymbol{\rho}\left(\boldsymbol{{t}}\right)\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}}{\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\right)\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{s}}^{\mathrm{2}} \right)}\boldsymbol{{dx}}=\frac{\mathrm{1}}{\boldsymbol{{s}}^{\mathrm{2}} −\mathrm{4}}\int\left(\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}}−\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{s}}^{\mathrm{2}} }\right)\boldsymbol{{dx}}= \\ $$$$=\frac{\mathrm{1}}{\boldsymbol{{s}}^{\mathrm{2}} −\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{arctan}}\left(\frac{\boldsymbol{{x}}}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\boldsymbol{{s}}}\boldsymbol{{arctan}}\left(\frac{\boldsymbol{{x}}}{\boldsymbol{{s}}}\right)\right)\mid_{\mathrm{0}} ^{\infty} = \\ $$$$=\frac{\mathrm{1}}{\boldsymbol{{s}}^{\mathrm{2}} −\mathrm{4}}\centerdot\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\boldsymbol{{s}}}\right)\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}\centerdot\frac{\mathrm{1}}{\boldsymbol{{s}}\left(\boldsymbol{{s}}+\mathrm{2}\right)}=\frac{\pi}{\mathrm{8}}\left(\frac{\mathrm{1}}{\boldsymbol{{s}}}−\frac{\mathrm{1}}{\boldsymbol{{s}}+\mathrm{2}}\right) \\ $$$$\mathscr{L}\:^{−\mathrm{1}} \left(\mathscr{L}\:\left(\boldsymbol{\rho}\left(\boldsymbol{{t}}\right)\right)\right)=\mathscr{L}^{\:−\mathrm{1}} \left(\frac{\pi}{\mathrm{8}}\centerdot\left(\frac{\mathrm{1}}{\boldsymbol{{s}}}−\frac{\mathrm{1}}{\boldsymbol{{s}}+\mathrm{2}}\right)\right)\:\Rightarrow \\ $$$$\Rightarrow\:\boldsymbol{\rho}\left(\boldsymbol{{t}}\right)=\frac{\pi}{\mathrm{8}}\left(\mathrm{1}−\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{t}}} \right);\:\boldsymbol{\rho}=\boldsymbol{\rho}\left(\boldsymbol{{t}}=\mathrm{2}\right)\:\Rightarrow\:\boldsymbol{\rho}=\frac{\pi}{\mathrm{8}}\left(\mathrm{1}−\boldsymbol{{e}}^{−\mathrm{4}} \right) \\ $$$$\boldsymbol{\Omega}=\pi−\mathrm{16}\boldsymbol{\rho}=\pi−\mathrm{2}\pi\left(\mathrm{1}−\boldsymbol{{e}}^{−\mathrm{4}} \right)=−\pi\left(\mathrm{1}−\mathrm{2}\boldsymbol{{e}}^{−\mathrm{4}} \right) \\ $$

Commented by mnjuly1970 last updated on 09/Oct/20

bravo bravo mr abdurauf..thank you

$${bravo}\:{bravo}\:{mr} \\ $$$${abdurauf}..{thank}\:{you} \\ $$

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