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Question Number 145911 by Mrsof last updated on 09/Jul/21
Commented by Mrsof last updated on 09/Jul/21
help me sir please by complex number
$${help}\:{me}\:{sir}\:{please}\:{by}\:{complex}\:\:{number} \\ $$
Answered by Dwaipayan Shikari last updated on 09/Jul/21
∫_(−∞) ^∞ ((sin (x))/((x+1)^2 +1))dx =∫_(−∞) ^∞ ((sin (u−1))/(u^2 +1))du =∫_(−∞) ^∞ ((sin (u)cos (1)−sin (1)cos (u))/(u^2 +1))du =∫_(−∞) ^∞ odd fuction du−∫_(−∞) ^∞ ((cos (u))/(u^2 +1))du =0−2∫_0 ^∞ ((cos (u))/(u^2 +1))du =−2(π/2)e^(−1) =−(π/e)
$$\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left({x}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$=\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left({u}−\mathrm{1}\right)}{{u}^{\mathrm{2}} +\mathrm{1}}{du} \\ $$$$=\int_{−\infty} ^{\infty} \frac{\mathrm{sin}\:\left({u}\right)\mathrm{cos}\:\left(\mathrm{1}\right)−\mathrm{sin}\:\left(\mathrm{1}\right)\mathrm{cos}\:\left({u}\right)}{{u}^{\mathrm{2}} +\mathrm{1}}{du} \\ $$$$=\int_{−\infty} ^{\infty} {odd}\:{fuction}\:{du}−\int_{−\infty} ^{\infty} \frac{\mathrm{cos}\:\left({u}\right)}{{u}^{\mathrm{2}} +\mathrm{1}}{du} \\ $$$$=\mathrm{0}−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\:\left({u}\right)}{{u}^{\mathrm{2}} +\mathrm{1}}{du} \\ $$$$=−\mathrm{2}\frac{\pi}{\mathrm{2}}{e}^{−\mathrm{1}} =−\frac{\pi}{{e}} \\ $$
Commented by Mrsof last updated on 09/Jul/21
sir can you help me by resideo
$${sir}\:{can}\:{you}\:{help}\:{me}\:{by}\:{resideo}\: \\ $$
Answered by Mathspace last updated on 09/Jul/21
are you student or teacher msof..
$${are}\:{you}\:{student}\:{or}\:{teacher}\:{msof}.. \\ $$
Answered by mathmax by abdo last updated on 09/Jul/21
Υ=∫_(−∞) ^(+∞) ((sinx)/(x^2 +2x+2))dx ⇒Υ=Im(∫_(−∞) ^(+∞) (e^(ix) /(x^2 +2x+2))dx) let ϕ(z)=(e^(iz) /(z^2 +2z+2)) poles of ϕ? Δ^′ =1−2=−1 ⇒z_1 =−1+i and z_2 =−1−i residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ Res(ϕ,z_1 ) ϕ(z)=(e^(iz) /((z−z_1 )(z−z_2 ))) ⇒Res(ϕ,z_1 )=lim_(z→z_1 ) (z−z_1 )ϕ(z) =lim_(z→z_1 ) (e^(iz_ ) /(z_1 −z_2 ))=(e^(iz_1 ) /(2i)) ⇒∫_(−∞) ^(+∞) ϕ(z)dz=2iπ.(e^(iz_1 ) /(2i)) =πe^(i(−1+i)) =πe^(−1−i) =πe^(−1) {cos(1)−isin(1)} ⇒ =πe^(−1) cos(1)−iπe^(−1) sin(1) ⇒Υ=−(π/e)sin(1)
$$\Upsilon=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{sinx}}{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{2}}\mathrm{dx}\:\Rightarrow\Upsilon=\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{ix}} }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{2x}+\mathrm{2}}\mathrm{dx}\right) \\ $$$$\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iz}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{2z}+\mathrm{2}}\:\:\mathrm{poles}\:\mathrm{of}\:\varphi? \\ $$$$\Delta^{'} \:=\mathrm{1}−\mathrm{2}=−\mathrm{1}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =−\mathrm{1}+\mathrm{i}\:\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =−\mathrm{1}−\mathrm{i} \\ $$$$\mathrm{residus}\:\mathrm{theorem}\:\mathrm{give}\:\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{1}} \right) \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iz}} }{\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)}\:\Rightarrow\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{1}} \right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{z}_{\mathrm{1}} } \left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)\varphi\left(\mathrm{z}\right) \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{z}_{\mathrm{1}} } \:\:\:\:\frac{\mathrm{e}^{\mathrm{iz}_{} } }{\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} }=\frac{\mathrm{e}^{\mathrm{iz}_{\mathrm{1}} } }{\mathrm{2i}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi.\frac{\mathrm{e}^{\mathrm{iz}_{\mathrm{1}} } }{\mathrm{2i}} \\ $$$$=\pi\mathrm{e}^{\mathrm{i}\left(−\mathrm{1}+\mathrm{i}\right)} \:=\pi\mathrm{e}^{−\mathrm{1}−\mathrm{i}} \:=\pi\mathrm{e}^{−\mathrm{1}} \left\{\mathrm{cos}\left(\mathrm{1}\right)−\mathrm{isin}\left(\mathrm{1}\right)\right\}\:\Rightarrow \\ $$$$=\pi\mathrm{e}^{−\mathrm{1}} \mathrm{cos}\left(\mathrm{1}\right)−\mathrm{i}\pi\mathrm{e}^{−\mathrm{1}} \mathrm{sin}\left(\mathrm{1}\right)\:\Rightarrow\Upsilon=−\frac{\pi}{\mathrm{e}}\mathrm{sin}\left(\mathrm{1}\right) \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 09/Jul/21
Ψ=∫_0 ^∞ ((cosx)/(x^4 +1))dx ⇒2Ψ=∫_(−∞) ^(+∞) ((cosx)/(x^4 +1))dx=Re(∫_(−∞) ^(+∞) (e^(ix) /(x^4 +1))dx) ϕ(z)=(e^(iz) /(z^4 +1)) ⇒ϕ(z)=(e^(iz) /((z^2 −i)(z^2 +i)))=(e^(iz) /((z−e^((iπ)/4) )(z+e^((iπ)/4) )(z−e^(−((iπ)/4)) )(z+e^(−((iπ)/4)) ))) the poles of ϕ are +^− e^((iπ)/4) and +^− e^(−((iπ)/4)) ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ{Res(ϕ,e^((iπ)/4) )+Res(ϕ,−e^(−((iπ)/4)) )} Res(ϕ,e^((iπ)/4) )=(e^(ie^((iπ)/4) ) /(2e^((iπ)/4) (2i)))=(1/(4i))e^(−((iπ)/4)) e^(i((1/( (√2)))+(i/( (√2))))) =(1/(4i))e^(−((iπ)/4)) e^(i/( (√2))) e^(−(1/( (√2)))) =(e^(−(1/( (√2)))) /(4i))e^(i((1/( (√2)))−(π/4))) Res(ϕ,−e^(−((iπ)/4)) ) =(e^(−e^(−((iπ)/4)) ) /(−2e^(−((iπ)/4)) (−2i)))=(1/(4i))e^((iπ)/4) e^(−((1/( (√2)))−(i/( (√2))))) =(e^(−(1/( (√2)))) /(4i)) e^(i((1/( (√2)))+(π/4))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz=2iπ{ (e^(−(1/( (√2)))) /(4i))e^(i((1/( (√2)))−(π/4))) +(e^(−(1/( (√2)))) /(4i))e^(i((1/( (√2)))+(π/4))) } =(π/2)e^(−(1/( (√2)))) {e^(i/( (√2))) (e^((iπ)/4) +e^(−((iπ)/4)) )} =(π/2)e^(−(1/( (√2)))) (2(1/( (√2))))e^(i/( (√2))) =(π/(2(√2)))e^(−(1/( (√2)))) (cos((1/( (√2))))+isin((1/( (√2))))) ⇒ 2∫_0 ^∞ ((cosx)/(1+x^4 ))dx =(π/(2(√2)))e^(−(1/( (√2)))) cos((1/( (√2)))) ⇒ ∫_0 ^∞ ((cosx)/(1+x^4 ))dx =(π/(4(√2)))e^(−(1/( (√2)))) cos((1/( (√2))))
$$\Psi=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cosx}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}}\mathrm{dx}\:\Rightarrow\mathrm{2}\Psi=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cosx}}{\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}}\mathrm{dx}=\mathrm{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{ix}} }{\mathrm{x}^{\mathrm{4}} \:+\mathrm{1}}\mathrm{dx}\right) \\ $$$$\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iz}} }{\mathrm{z}^{\mathrm{4}} \:+\mathrm{1}}\:\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iz}} }{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{i}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{i}\right)}=\frac{\mathrm{e}^{\mathrm{iz}} }{\left(\mathrm{z}−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\left(\mathrm{z}+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\mathrm{the}\:\mathrm{poles}\:\mathrm{of}\:\varphi\:\mathrm{are}\:\overset{−} {+}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\mathrm{and}\:\overset{−} {+}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\left\{\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)+\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)=\frac{\mathrm{e}^{\mathrm{ie}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} } }{\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(\mathrm{2i}\right)}=\frac{\mathrm{1}}{\mathrm{4i}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\mathrm{e}^{\mathrm{i}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4i}}\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\mathrm{e}^{\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}} \:\:\mathrm{e}^{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\:\:=\frac{\mathrm{e}^{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} }{\mathrm{4i}}\mathrm{e}^{\mathrm{i}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{4}}\right)} \\ $$$$\mathrm{Res}\left(\varphi,−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\:=\frac{\mathrm{e}^{−\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} } }{−\mathrm{2e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \left(−\mathrm{2i}\right)}=\frac{\mathrm{1}}{\mathrm{4i}}\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:\mathrm{e}^{−\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}\right)} \\ $$$$=\frac{\mathrm{e}^{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} }{\mathrm{4i}}\:\mathrm{e}^{\mathrm{i}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\pi}{\mathrm{4}}\right)} \:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}=\mathrm{2i}\pi\left\{\:\frac{\mathrm{e}^{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} }{\mathrm{4i}}\mathrm{e}^{\mathrm{i}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\frac{\pi}{\mathrm{4}}\right)} \:+\frac{\mathrm{e}^{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} }{\mathrm{4i}}\mathrm{e}^{\mathrm{i}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\pi}{\mathrm{4}}\right)} \right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\:\left\{\mathrm{e}^{\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}} \left(\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:+\mathrm{e}^{−\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$$=\frac{\pi}{\mathrm{2}}\mathrm{e}^{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\:\left(\mathrm{2}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\mathrm{e}^{\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}} \:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{e}^{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\:\left(\mathrm{cos}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{isin}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\right)\:\Rightarrow \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cosx}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx}\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{e}^{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\:\mathrm{cos}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{cosx}}{\mathrm{1}+\mathrm{x}^{\mathrm{4}} }\mathrm{dx}\:=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\mathrm{e}^{−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \:\:\mathrm{cos}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right) \\ $$

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