Menu Close

calculate-xsin-2x-1-x-2-2-dx-

Tinku Tara 0 Comments
Pin




Question Number 39033 by maxmathsup by imad last updated on 01/Jul/18
calculate ∫_(−∞) ^(+∞) ((xsin(2x))/((1+x^2 )^2 ))dx
$${calculate}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{xsin}\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$
Commented by math khazana by abdo last updated on 02/Jul/18
let I = ∫_(−∞) ^(+∞) ((x sin(2x))/((1+x^2 )^2 ))dx I = Im( ∫_(−∞) ^(+∞) ((x e^(ix) )/((1+x^2 )^2 ))dx) let ϕ(z)=((z e^(iz) )/((1+z^2 )^2 )) ϕ(z) = ((z e^(iz) )/((z−i)^2 (z+i)^2 )) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i)=lim_(z→i) (1/((2−1)!)) {(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { ((z e^(iz) )/((z+i)^2 ))}^((1)) =lim_(z→i) (((e^(iz) +iz e^(iz) )(z+i)^2 −2(z+i)z e^(iz) )/((z+i)^4 )) =lim_(z→i) (((1+iz)e^(ix) (z+i) −2 ze^(iz) )/((z+i)^3 )) =((−2i e^(−1) )/((2i)^3 )) = ((−2i e^(−1) )/(−8i)) = (e^(−1) /4) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (e^(−1) /4) = ((iπe^(−1) )/2) ⇒ I = Im(∫_(−∞) ^(+∞) ϕ(z)dz ) =(π/(2e)) .
$${let}\:{I}\:\:=\:\int_{−\infty} ^{+\infty} \:\frac{{x}\:{sin}\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$${I}\:=\:{Im}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{x}\:{e}^{{ix}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\right)\:{let}\:\varphi\left({z}\right)=\frac{{z}\:{e}^{{iz}} }{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}\:{e}^{{iz}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} } \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right) \\ $$$${Res}\left(\varphi,{i}\right)={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\left\{\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\left\{\:\frac{{z}\:{e}^{{iz}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \frac{\left({e}^{{iz}} \:\:+{iz}\:{e}^{{iz}} \right)\left({z}+{i}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{i}\right){z}\:{e}^{{iz}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{\left(\mathrm{1}+{iz}\right){e}^{{ix}} \left({z}+{i}\right)\:−\mathrm{2}\:{ze}^{{iz}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{−\mathrm{2}{i}\:{e}^{−\mathrm{1}} }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:=\:\frac{−\mathrm{2}{i}\:{e}^{−\mathrm{1}} }{−\mathrm{8}{i}}\:=\:\frac{{e}^{−\mathrm{1}} }{\mathrm{4}}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{{e}^{−\mathrm{1}} }{\mathrm{4}}\:=\:\frac{{i}\pi{e}^{−\mathrm{1}} }{\mathrm{2}}\:\:\Rightarrow \\ $$$${I}\:\:=\:{Im}\left(\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:\right)\:=\frac{\pi}{\mathrm{2}{e}}\:. \\ $$
Commented by math khazana by abdo last updated on 02/Jul/18
the Q is find I = ∫_(−∞) ^(+∞) ((x sin(x))/((1+x^2 )^2 ))dx
$${the}\:{Q}\:{is}\:{find}\:\:{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{x}\:{sin}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *