Menu Close

find-the-value-of-0-x-2-1-x-2-1-sin-x-x-dx-

Tinku Tara 0 Comments
Pin




Question Number 36204 by prof Abdo imad last updated on 30/May/18
find the value of ∫_0 ^∞ ((x^2 −1)/(x^2 +1)) ((sin(x))/x)dx
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\frac{{sin}\left({x}\right)}{{x}}{dx}\: \\ $$
Commented by math khazana by abdo last updated on 17/Aug/18
let I = ∫_0 ^∞ ((x^2 −1)/(x^2 +1)) ((sinx)/x) dx we hsve2I = ∫_(−∞) ^(+∞) ((x^2 −1)/(x^2 +1)) ((sinx)/x)?dx =Im( ∫_(−∞) ^(+∞) ((x^2 −1)/(x(x^2 +1))) e^(ix) dx) let?consider the complex function ϕ(z) = ((z^2 −1)/(z(z^2 +1))) e^(iz) the poles of ϕ are 0,i and −i ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ,0)+Res(ϕ,i)} but ϕ(z) = (((z^2 −1)e^(iz) )/(z(z−i)(z+i))) ⇒ Res(ϕ,0) =lim_(z→0) zϕ(z) = (1/1) =1 Res(ϕ,i) = lim_(z→i) (z−i)ϕ(z) =((−2 e^(−1) )/(i(2i))) = e^(−1) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{1+e^(−1) } ⇒ 2I =Im( ∫_(−∞) ^(+∞) ϕ(z)dz) =2π{1+e^(−1) } ⇒ I = π +(π/e) .
$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\frac{{sinx}}{{x}}\:{dx} \\ $$$${we}\:{hsve}\mathrm{2}{I}\:=\:\int_{−\infty} ^{+\infty} \:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:\frac{{sinx}}{{x}}?{dx} \\ $$$$={Im}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}\left({x}^{\mathrm{2}} +\mathrm{1}\right)}\:{e}^{{ix}} {dx}\right)\:{let}?{consider}\:{the}\: \\ $$$${complex}\:{function}\:\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} −\mathrm{1}}{{z}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}\:{e}^{{iz}} \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\mathrm{0},{i}\:{and}\:−{i} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,\mathrm{0}\right)+{Res}\left(\varphi,{i}\right)\right\} \\ $$$${but}\:\varphi\left({z}\right)\:=\:\frac{\left({z}^{\mathrm{2}} −\mathrm{1}\right){e}^{{iz}} }{{z}\left({z}−{i}\right)\left({z}+{i}\right)}\:\Rightarrow \\ $$$${Res}\left(\varphi,\mathrm{0}\right)\:={lim}_{{z}\rightarrow\mathrm{0}} {z}\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}}\:=\mathrm{1} \\ $$$${Res}\left(\varphi,{i}\right)\:=\:{lim}_{{z}\rightarrow{i}} \left({z}−{i}\right)\varphi\left({z}\right)\:=\frac{−\mathrm{2}\:{e}^{−\mathrm{1}} }{{i}\left(\mathrm{2}{i}\right)} \\ $$$$=\:{e}^{−\mathrm{1}} \:\Rightarrow\:\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\mathrm{1}+{e}^{−\mathrm{1}} \right\}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:={Im}\left(\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\right)\:=\mathrm{2}\pi\left\{\mathrm{1}+{e}^{−\mathrm{1}} \right\}\:\Rightarrow \\ $$$${I}\:\:=\:\pi\:\:+\frac{\pi}{{e}}\:\:. \\ $$

Pin

Leave a Reply

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