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Question Number 222044 by wewji12 last updated on 15/Jun/25
How do you evaluate ∫_(−∞) ^( ∞) ((sin(z+1))/((z+1)(z^2 +1))) dz
$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{evaluate} \\ $$$$\int_{−\infty} ^{\:\:\infty} \:\:\frac{\mathrm{sin}\left({z}+\mathrm{1}\right)}{\left({z}+\mathrm{1}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)}\:\mathrm{d}{z} \\ $$
Answered by vnm last updated on 15/Jun/25
(1/((z+1)(z^2 +1)))=(a/(z+1))+((bz+c)/(z^2 +1))=(((a+b)z^2 +(b+c)z+(a+c))/((z+1)(z^2 +1))) a+b=0, b+c=0, a+c=1 a=c=(1/2), b=−(1/2) (1/2)∫_(−∞) ^(+∞) sin(z+1)∙((1/(z+1))+((−z+1)/(z^2 +1)))dz= (1/2)π+(1/2)∫_(−∞) ^(+∞) (((1−z)sin(z+1))/(z^2 +1))dz=(π/2)+(I/2) (((1−z)sin(z+1))/(z^2 +1))=(((1−z)(cos1∙sinz+sin1∙cosz))/(z^2 +1))= ((cos1∙sinz+sin1∙cosz−cos1∙zsinz−sin1∙zcosz)/(z^2 +1)) I=sin1∙∫_(−∞) ^(+∞) ((cosz)/(z^2 +1))dz−cos1∙∫_(−∞) ^(+∞) ((zsinz)/(z^2 +1))dz z∈R, w∈C ∫_(−∞) ^(+∞) ((cosz)/(z^2 +1))dz=Re∫_(−∞) ^(+∞) (e^(iz) /(z^2 +1))dz= Re{2πi∙res((e^(iw) /(w^2 +1)))_(w=i) }=Re{2πi(e^(iw) /(2w))∣_(w=i) }=(π/e) ∫_(−∞) ^(+∞) ((zsinz)/(z^2 +1))dz=Im∫_(−∞) ^(+∞) ((ze^(iz) )/(z^2 +1))dz= Im{2πi∙res(((we^(iw) )/(w^2 +1)))_(w=i) }=Im{2πi((we^(iw) )/(2w))∣_(w=i) }= Im{i(π/e)}=(π/e) I=(sin1−cos1)(π/e) answer: (π/2)+(1/2)(sin1−cos1)(π/e)=(π/2)(1+((sin1−cos1)/e))
$$ \\ $$$$\frac{\mathrm{1}}{\left({z}+\mathrm{1}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{a}}{{z}+\mathrm{1}}+\frac{{bz}+{c}}{{z}^{\mathrm{2}} +\mathrm{1}}=\frac{\left({a}+{b}\right){z}^{\mathrm{2}} +\left({b}+{c}\right){z}+\left({a}+{c}\right)}{\left({z}+\mathrm{1}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${a}+{b}=\mathrm{0},\:\:{b}+{c}=\mathrm{0},\:\:{a}+{c}=\mathrm{1} \\ $$$${a}={c}=\frac{\mathrm{1}}{\mathrm{2}},\:\:{b}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \mathrm{sin}\left({z}+\mathrm{1}\right)\centerdot\left(\frac{\mathrm{1}}{{z}+\mathrm{1}}+\frac{−{z}+\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}\right){dz}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\pi+\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \frac{\left(\mathrm{1}−{z}\right)\mathrm{sin}\left({z}+\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}{dz}=\frac{\pi}{\mathrm{2}}+\frac{{I}}{\mathrm{2}} \\ $$$$\frac{\left(\mathrm{1}−{z}\right)\mathrm{sin}\left({z}+\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}=\frac{\left(\mathrm{1}−{z}\right)\left(\mathrm{cos1}\centerdot\mathrm{sin}{z}+\mathrm{sin1}\centerdot\mathrm{cos}{z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$\frac{\mathrm{cos1}\centerdot\mathrm{sin}{z}+\mathrm{sin1}\centerdot\mathrm{cos}{z}−\mathrm{cos1}\centerdot{z}\mathrm{sin}{z}−\mathrm{sin1}\centerdot{z}\mathrm{cos}{z}}{{z}^{\mathrm{2}} +\mathrm{1}} \\ $$$${I}=\mathrm{sin1}\centerdot\int_{−\infty} ^{+\infty} \frac{\mathrm{cos}{z}}{{z}^{\mathrm{2}} +\mathrm{1}}{dz}−\mathrm{cos1}\centerdot\int_{−\infty} ^{+\infty} \frac{{z}\mathrm{sin}{z}}{{z}^{\mathrm{2}} +\mathrm{1}}{dz} \\ $$$${z}\in\mathbb{R},\:{w}\in\mathbb{C} \\ $$$$\int_{−\infty} ^{+\infty} \frac{\mathrm{cos}{z}}{{z}^{\mathrm{2}} +\mathrm{1}}{dz}=\mathrm{Re}\int_{−\infty} ^{+\infty} \frac{{e}^{{iz}} }{{z}^{\mathrm{2}} +\mathrm{1}}{dz}= \\ $$$$\mathrm{Re}\left\{\mathrm{2}\pi{i}\centerdot\mathrm{res}\left(\frac{{e}^{{iw}} }{{w}^{\mathrm{2}} +\mathrm{1}}\right)_{{w}={i}} \right\}=\mathrm{Re}\left\{\mathrm{2}\pi{i}\frac{{e}^{{iw}} }{\mathrm{2}{w}}\mid_{{w}={i}} \right\}=\frac{\pi}{{e}} \\ $$$$\int_{−\infty} ^{+\infty} \frac{{z}\mathrm{sin}{z}}{{z}^{\mathrm{2}} +\mathrm{1}}{dz}=\mathrm{Im}\int_{−\infty} ^{+\infty} \frac{{ze}^{{iz}} }{{z}^{\mathrm{2}} +\mathrm{1}}{dz}= \\ $$$$\mathrm{Im}\left\{\mathrm{2}\pi{i}\centerdot\mathrm{res}\left(\frac{{we}^{{iw}} }{{w}^{\mathrm{2}} +\mathrm{1}}\right)_{{w}={i}} \right\}=\mathrm{Im}\left\{\mathrm{2}\pi{i}\frac{{we}^{{iw}} }{\mathrm{2}{w}}\mid_{{w}={i}} \right\}= \\ $$$$\mathrm{Im}\left\{{i}\frac{\pi}{{e}}\right\}=\frac{\pi}{{e}} \\ $$$${I}=\left(\mathrm{sin1}−\mathrm{cos1}\right)\frac{\pi}{{e}} \\ $$$$\mathrm{answer}:\:\frac{\pi}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{sin1}−\mathrm{cos1}\right)\frac{\pi}{{e}}=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{sin1}−\mathrm{cos1}}{{e}}\right) \\ $$

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