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Question Number 146977 by tabata last updated on 16/Jul/21
find (1) ∫_C (e^z^2 /(z^2 +4z+3))dz ,C:∣z−2∣=5 (2)∫_(−∞) ^( ∞) ((cosx)/(x^2 +2x+2))dx
$${find} \\ $$$$\left(\mathrm{1}\right)\:\int_{{C}} \:\frac{{e}^{{z}^{\mathrm{2}} } }{{z}^{\mathrm{2}} +\mathrm{4}{z}+\mathrm{3}}{dz}\:\:,{C}:\mid{z}−\mathrm{2}\mid=\mathrm{5} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\int_{−\infty} ^{\:\infty} \frac{{cosx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}{dx} \\ $$
Answered by mathmax by abdo last updated on 17/Jul/21
1) Φ=∫_C (e^z^2 /(z^2 +4z+3))dz withC→∣z−2∣=5 Δ^′ =4−3=1 ⇒z_1 =−2+1=−1 and z_2 =−2−1=−3 f(z)=(e^z^2 /((z−z_1 )(z−z_2 ))) residus ⇒∫_C f(z)dz =2iπ{Res(f,z_1 )+Res(f,z_2 )} =2iπ{(e^z_1 ^2 /(z_1 −z_2 ))+(e^z_2 ^2 /(z_2 −z_1 ))} =2iπ{(e/2) +(e^9 /(−2))}=iπ{e−e^9 }
$$\left.\mathrm{1}\right)\:\Phi=\int_{\mathrm{C}} \:\frac{\mathrm{e}^{\mathrm{z}^{\mathrm{2}} } }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{4z}+\mathrm{3}}\mathrm{dz}\:\:\mathrm{withC}\rightarrow\mid\mathrm{z}−\mathrm{2}\mid=\mathrm{5} \\ $$$$\Delta^{'} \:=\mathrm{4}−\mathrm{3}=\mathrm{1}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =−\mathrm{2}+\mathrm{1}=−\mathrm{1}\:\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =−\mathrm{2}−\mathrm{1}=−\mathrm{3} \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{z}^{\mathrm{2}} } }{\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)}\:\:\mathrm{residus}\:\Rightarrow\int_{\mathrm{C}} \mathrm{f}\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\left\{\mathrm{Res}\left(\mathrm{f},\mathrm{z}_{\mathrm{1}} \right)+\mathrm{Res}\left(\mathrm{f},\mathrm{z}_{\mathrm{2}} \right)\right\} \\ $$$$=\mathrm{2i}\pi\left\{\frac{\mathrm{e}^{\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} } }{\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} }+\frac{\mathrm{e}^{\mathrm{z}_{\mathrm{2}} ^{\mathrm{2}} } }{\mathrm{z}_{\mathrm{2}} −\mathrm{z}_{\mathrm{1}} }\right\}\:=\mathrm{2i}\pi\left\{\frac{\mathrm{e}}{\mathrm{2}}\:+\frac{\mathrm{e}^{\mathrm{9}} }{−\mathrm{2}}\right\}=\mathrm{i}\pi\left\{\mathrm{e}−\mathrm{e}^{\mathrm{9}} \right\} \\ $$
Answered by mathmax by abdo last updated on 17/Jul/21
Ψ=∫_(−∞) ^(+∞) ((cosx)/(x^2 +2x+2))dx =Re(∫_(−∞) ^(+∞) (e^(ix) /(x^2 +2x+2))dx) let ϕ(z)=(e^(iz) /(z^2 +2x+2)) poles of ϕ? Δ^′ =1−2=−1 ⇒z_1 =−1+i and z_2 =−1−i ⇒ϕ(z)=(e^(iz) /((z−z_1 )(z−z_2 ))) residus ⇒∫_R ϕ(z)dz =2iπRes(ϕ/z_1 )=2iπ.(e^(iz_1 ) /(z_1 −z_2 )) =2iπ.(e^(i(−1+i)) /(2i))=π e^(−1) .e^(−i) =(π/e)(cos(1)−isin(1)) ⇒ Ψ=(π/e)cos(1) (1=1radian)
$$\Psi=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{cosx}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{2x}+\mathrm{2}}\mathrm{dx}\:=\mathrm{Re}\left(\int_{−\infty} ^{+\infty} \:\frac{\mathrm{e}^{\mathrm{ix}} }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{2x}+\mathrm{2}}\mathrm{dx}\right) \\ $$$$\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iz}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{2x}+\mathrm{2}}\:\:\mathrm{poles}\:\mathrm{of}\:\varphi? \\ $$$$\Delta^{'} \:=\mathrm{1}−\mathrm{2}=−\mathrm{1}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =−\mathrm{1}+\mathrm{i}\:\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =−\mathrm{1}−\mathrm{i}\:\Rightarrow\varphi\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{iz}} }{\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)} \\ $$$$\mathrm{residus}\:\Rightarrow\int_{\mathrm{R}} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\mathrm{Res}\left(\varphi/\mathrm{z}_{\mathrm{1}} \right)=\mathrm{2i}\pi.\frac{\mathrm{e}^{\mathrm{iz}_{\mathrm{1}} } }{\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} } \\ $$$$=\mathrm{2i}\pi.\frac{\mathrm{e}^{\mathrm{i}\left(−\mathrm{1}+\mathrm{i}\right)} }{\mathrm{2i}}=\pi\:\mathrm{e}^{−\mathrm{1}} .\mathrm{e}^{−\mathrm{i}} \:=\frac{\pi}{\mathrm{e}}\left(\mathrm{cos}\left(\mathrm{1}\right)−\mathrm{isin}\left(\mathrm{1}\right)\right)\:\Rightarrow \\ $$$$\Psi=\frac{\pi}{\mathrm{e}}\mathrm{cos}\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}=\mathrm{1radian}\right) \\ $$

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