Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 131969 by mnjuly1970 last updated on 10/Feb/21

... math analysis... φ= ∫_(−∞) ^( +∞) ((xsin(x))/(x^2 +2x+2))dx=? φ=∫_(−∞) ^( +∞) ((xsin(x))/((x+1)^2 +1))dx =^(x+1=t) ∫_(−∞) ^( +∞) (((t−1)sin(t−1))/(t^2 +1))dt =∫_(−∞) ^( +∞) ((tsin(t)cos(1)−tcos(t)sin(1)−sin(t)cos(1)+cos(t)sin(1))/(t^2 +1))dt =2cos(1)∫_0 ^( ∞) ((tsin(t))/(t^2 +1))dt+2sin(1)∫_0 ^( ∞) ((cos(t))/(t^2 +1))dt =2cos(1).(π/(2e))+2sin(1).(π/(2e)) =(π/e)(cos(1)+sin(1))....

$$\:\:\:\:\:\:\:\:\:\:\:\:...\:{math}\:\:{analysis}... \\ $$$$\:\:\:\:\phi=\:\:\int_{−\infty} ^{\:+\infty} \frac{{xsin}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}{dx}=? \\ $$$$\:\:\:\:\:\phi=\int_{−\infty} ^{\:+\infty} \frac{{xsin}\left({x}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\:\:\:\:\:\:\:\overset{{x}+\mathrm{1}={t}} {=}\int_{−\infty} ^{\:+\infty} \frac{\left({t}−\mathrm{1}\right){sin}\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\:\:\:\:\:\:\:\:=\int_{−\infty} ^{\:+\infty} \frac{{tsin}\left({t}\right){cos}\left(\mathrm{1}\right)−{tcos}\left({t}\right){sin}\left(\mathrm{1}\right)−{sin}\left({t}\right){cos}\left(\mathrm{1}\right)+{cos}\left({t}\right){sin}\left(\mathrm{1}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$=\mathrm{2}{cos}\left(\mathrm{1}\right)\int_{\mathrm{0}} ^{\:\infty} \frac{{tsin}\left({t}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}{dt}+\mathrm{2}{sin}\left(\mathrm{1}\right)\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left({t}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$=\mathrm{2}{cos}\left(\mathrm{1}\right).\frac{\pi}{\mathrm{2}{e}}+\mathrm{2}{sin}\left(\mathrm{1}\right).\frac{\pi}{\mathrm{2}{e}} \\ $$$$\:=\frac{\pi}{{e}}\left({cos}\left(\mathrm{1}\right)+{sin}\left(\mathrm{1}\right)\right).... \\ $$$$\:\:\: \\ $$

Answered by mathmax by abdo last updated on 10/Feb/21

another way Φ=∫_(−∞) ^(+∞) ((xsinx)/(x^2 +2x+2)) ⇒Φ=Im(∫_(−∞) ^(+∞) ((xe^(ix) )/(x^2 +2x+2))) let ϕ(z)=((ze^(iz) )/(z^2 +2z+2)) ,poles of ϕ z^2 +2z+2=0 →Δ^′ =−1 ⇒z_1 =−1+i and z_2 =−1−i and ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,z_1 ) but ϕ(z)=((ze^(iz) )/((z−z_1 )(z−z_2 ))) Res(ϕ,z_1 )=lim_(z→z_1 ) (z−z_1 )ϕ(z)=lim_(z→z_1 ) ((ze^(iz) )/(z−z_2 )) =((z_1 e^(iz_1 ) )/(z_1 −z_2 )) =(((−1+i)e^(i(−1+i)) )/(2i)) =((((−1+i)e^(−i−1) )/(2i)) =((e^(−1) (−1+i)(cos1−isin1))/(2i)) =(e^(−1) /(2i)){−cos1+isin1 +icos1+sin1} ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ.(e^(−1) /(2i)){sin1−cos1 +i(cos1+sin1)} ⇒ Φ=(π/e)(cos(1)+sin(1))

$$\mathrm{another}\:\mathrm{way}\:\:\Phi=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{xsinx}}{\mathrm{x}^{\mathrm{2}} \:+\mathrm{2x}+\mathrm{2}}\:\Rightarrow\Phi=\mathrm{Im}\left(\int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{xe}^{\mathrm{ix}} }{\mathrm{x}^{\mathrm{2}} \:+\mathrm{2x}+\mathrm{2}}\right) \\ $$$$\mathrm{let}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{ze}^{\mathrm{iz}} }{\mathrm{z}^{\mathrm{2}} \:+\mathrm{2z}+\mathrm{2}}\:\:,\mathrm{poles}\:\mathrm{of}\:\varphi \\ $$$$\mathrm{z}^{\mathrm{2}} \:+\mathrm{2z}+\mathrm{2}=\mathrm{0}\:\rightarrow\Delta^{'} \:=−\mathrm{1}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =−\mathrm{1}+\mathrm{i}\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =−\mathrm{1}−\mathrm{i} \\ $$$$\mathrm{and}\:\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{1}} \right)\:\:\mathrm{but}\:\varphi\left(\mathrm{z}\right)=\frac{\mathrm{ze}^{\mathrm{iz}} }{\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{z}_{\mathrm{1}} \right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{z}_{\mathrm{1}} } \:\:\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)\varphi\left(\mathrm{z}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{z}_{\mathrm{1}} } \:\:\:\frac{\mathrm{ze}^{\mathrm{iz}} }{\mathrm{z}−\mathrm{z}_{\mathrm{2}} }\:\:\:=\frac{\mathrm{z}_{\mathrm{1}} \mathrm{e}^{\mathrm{iz}_{\mathrm{1}} } }{\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} } \\ $$$$=\frac{\left(−\mathrm{1}+\mathrm{i}\right)\mathrm{e}^{\mathrm{i}\left(−\mathrm{1}+\mathrm{i}\right)} }{\mathrm{2i}}\:=\frac{\left(\left(−\mathrm{1}+\mathrm{i}\right)\mathrm{e}^{−\mathrm{i}−\mathrm{1}} \right.}{\mathrm{2i}}\:=\frac{\mathrm{e}^{−\mathrm{1}} \left(−\mathrm{1}+\mathrm{i}\right)\left(\mathrm{cos1}−\mathrm{isin1}\right)}{\mathrm{2i}} \\ $$$$=\frac{\mathrm{e}^{−\mathrm{1}} }{\mathrm{2i}}\left\{−\mathrm{cos1}+\mathrm{isin1}\:+\mathrm{icos1}+\mathrm{sin1}\right\}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi.\frac{\mathrm{e}^{−\mathrm{1}} }{\mathrm{2i}}\left\{\mathrm{sin1}−\mathrm{cos1}\:+\mathrm{i}\left(\mathrm{cos1}+\mathrm{sin1}\right)\right\}\:\Rightarrow \\ $$$$\Phi=\frac{\pi}{\mathrm{e}}\left(\mathrm{cos}\left(\mathrm{1}\right)+\mathrm{sin}\left(\mathrm{1}\right)\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com