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Question Number 219361 by SdC355 last updated on 23/Apr/25
∫_(−∞) ^(+∞) ((cos(z)e^(−z) )/(z^2 +1)) dz=??
$$\int_{−\infty} ^{+\infty} \:\:\frac{\mathrm{cos}\left({z}\right){e}^{−{z}} }{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z}=?? \\ $$
Answered by breniam last updated on 24/Apr/25
∫_(−∞) ^0 ((cos(z)e^(−z) )/(z^2 +1))dz={z^− =−z}=∫_0 ^∞ ((cos(z)e^z )/(z^2 +1))dz For some ρ∈R^+ ∣∫_0 ^∞ ((cos(z)e^z )/(z^2 +1))∣dz≥∣∫_0 ^ρ ((cos(z)e^z )/(z^2 +1))dz+∫_ρ ^∞ cos(z)e^(z/2) dz∣ f. e. y∈(ρ,∞) ∫_ρ ^y cos(z)e^(z/2) dz=∫_ρ ^y (sin(z))′e^(z/2) dz=sin(y)e^(y/2) −sin(ρ)e^(ρ/2) −(1/2)∫_ρ ^y sin(z)e^(z/2) dz= sin(y)e^(y/2) −sin(ρ)e^(ρ/2) +(1/2)∫_ρ ^y (cos(z))′e^(z/2) dz= sin(y)e^(y/2) −sin(ρ)e^(ρ/2) +(1/2)cos(y)e^(y/2) −(1/2)cos(ρ)e^(ρ/2) −(1/4)∫_ρ ^y cos(z)e^(z/2) dz⇒ ⇒∫_ρ ^y cos(z)e^(z/2) dz=(4/5)(sin(y)e^(y/2) +(1/2)cos(y)e^(y/2) −sin(ρ)e^(ρ/2) −(1/2)cos(ρ)e^(ρ/2) ) Which, for some subsequences of y_n diverges to ∞ For the same subsequences, by comparison ∣∫_(−y_n ) ^0 ((cos(z)e^(−z) )/(z^2 +1))dz∣→^(n→∞) ∞ On the other side of the hand ∣∫_0 ^∞ ((e^(−z) cos(z))/(z^2 +1))dz∣≤∫_0 ^∞ (1/(z^2 +1))dz=(π/2) Thus integral diverges.
$$\underset{−\infty} {\overset{\mathrm{0}} {\int}}\frac{\mathrm{cos}\left({z}\right){e}^{−{z}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}=\left\{\overset{−} {{z}}=−{z}\right\}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{cos}\left({z}\right){e}^{{z}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z} \\ $$$$\mathrm{For}\:\mathrm{some}\:\rho\in\mathbb{R}^{+} \\ $$$$\mid\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{cos}\left({z}\right){e}^{{z}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mid\mathrm{d}{z}\geqslant\mid\underset{\mathrm{0}} {\overset{\rho} {\int}}\frac{\mathrm{cos}\left({z}\right){e}^{{z}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}+\underset{\rho} {\overset{\infty} {\int}}\mathrm{cos}\left({z}\right){e}^{\frac{{z}}{\mathrm{2}}} \mathrm{d}{z}\mid \\ $$$$\mathrm{f}.\:\mathrm{e}.\:{y}\in\left(\rho,\infty\right) \\ $$$$\underset{\rho} {\overset{{y}} {\int}}\mathrm{cos}\left({z}\right){e}^{\frac{{z}}{\mathrm{2}}} \mathrm{d}{z}=\underset{\rho} {\overset{{y}} {\int}}\left(\mathrm{sin}\left({z}\right)\right)'{e}^{\frac{{z}}{\mathrm{2}}} \mathrm{d}{z}=\mathrm{sin}\left({y}\right){e}^{\frac{{y}}{\mathrm{2}}} −\mathrm{sin}\left(\rho\right){e}^{\frac{\rho}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{2}}\underset{\rho} {\overset{{y}} {\int}}\mathrm{sin}\left({z}\right){e}^{\frac{{z}}{\mathrm{2}}} \mathrm{d}{z}= \\ $$$$\mathrm{sin}\left({y}\right){e}^{\frac{{y}}{\mathrm{2}}} −\mathrm{sin}\left(\rho\right){e}^{\frac{\rho}{\mathrm{2}}} +\frac{\mathrm{1}}{\mathrm{2}}\underset{\rho} {\overset{{y}} {\int}}\left(\mathrm{cos}\left({z}\right)\right)'{e}^{\frac{{z}}{\mathrm{2}}} \mathrm{d}{z}= \\ $$$$\mathrm{sin}\left({y}\right){e}^{\frac{{y}}{\mathrm{2}}} −\mathrm{sin}\left(\rho\right){e}^{\frac{\rho}{\mathrm{2}}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left({y}\right){e}^{\frac{{y}}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\rho\right){e}^{\frac{\rho}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{4}}\underset{\rho} {\overset{{y}} {\int}}\mathrm{cos}\left({z}\right){e}^{\frac{{z}}{\mathrm{2}}} \mathrm{d}{z}\Rightarrow \\ $$$$\Rightarrow\underset{\rho} {\overset{{y}} {\int}}\mathrm{cos}\left({z}\right){e}^{\frac{{z}}{\mathrm{2}}} \mathrm{d}{z}=\frac{\mathrm{4}}{\mathrm{5}}\left(\mathrm{sin}\left({y}\right){e}^{\frac{{y}}{\mathrm{2}}} +\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left({y}\right){e}^{\frac{{y}}{\mathrm{2}}} −\mathrm{sin}\left(\rho\right){e}^{\frac{\rho}{\mathrm{2}}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\left(\rho\right){e}^{\frac{\rho}{\mathrm{2}}} \right) \\ $$$$\mathrm{Which},\:\mathrm{for}\:\mathrm{some}\:\mathrm{subsequences}\:\mathrm{of}\:{y}_{{n}} \mathrm{diverges}\:\mathrm{to}\:\infty \\ $$$$\mathrm{For}\:\mathrm{the}\:\mathrm{same}\:\mathrm{subsequences},\:\mathrm{by}\:\mathrm{comparison} \\ $$$$\mid\underset{−{y}_{{n}} } {\overset{\mathrm{0}} {\int}}\frac{\mathrm{cos}\left({z}\right){e}^{−{z}} }{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}\mid\overset{{n}\rightarrow\infty} {\rightarrow}\infty \\ $$$$\mathrm{On}\:\mathrm{the}\:\mathrm{other}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hand} \\ $$$$\mid\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{e}^{−{z}} \mathrm{cos}\left({z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}\mid\leqslant\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{z}=\frac{\pi}{\mathrm{2}} \\ $$$${T}\mathrm{hus}\:\mathrm{integral}\:\mathrm{diverges}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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