Menu Close

let-f-xsin-x-1-x-2-2-dx-calculate-f-and-f-from-R-

Tinku Tara 0 Comments
Pin




Question Number 48040 by maxmathsup by imad last updated on 18/Nov/18
let f(α)=∫_(−∞) ^(+∞) ((xsin(αx))/((1+x^2 )^2 ))dx calculate f(α) and f^′ (α).(α from R) .
$${let}\:{f}\left(\alpha\right)=\int_{−\infty} ^{+\infty} \:\:\frac{{xsin}\left(\alpha{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$${calculate}\:{f}\left(\alpha\right)\:{and}\:{f}^{'} \left(\alpha\right).\left(\alpha\:{from}\:{R}\right)\:. \\ $$
Commented by Abdo msup. last updated on 20/Nov/18
1) we have?f(α)=Im (∫_(−∞) ^(+∞) ((x^ e^(iαx) )/((1+x^2 )^2 ))dx) let ϕ(z)=((z e^(iαz) )/((z^2 +1)^2 )) ⇒ϕ(z)=((z e^(iαz) )/((z−i)^2 (z+i)^2 )) so the poles of ϕ are iand −i(doubles) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) but Res(ϕ,i)=lim_(z→i) (1/((2−1!)){(z−i)^2 ϕ(z)}^((1)) =lim_(z→i) { ((z e^(iαz) )/((z+i)^2 ))}^((1)) =lim_(z→i) (((e^(iαz) +iαz e^(iαz) )(z+i)^2 −2(z+i)z e^(iαz) )/((z+i)^4 )) =lim_(z→i) (((1+iαz)e^(iαz) (z+i)−2z e^(iαz) )/((z+i)^3 )) =(((1−α)(2i)e^(−α) −2i e^(−α) )/((2i)^3 )) =(((2i−2iα−2i)e^(−α) )/(−8i)) =(α/4) e^(−α) ⇒∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (α/4) e^(−α) =i((απ)/2) e^(−α) ⇒ f(α) =((πα)/2) e^(−α) 2) f^′ (α) =(π/2){e^(−α) −α e^(−α) }=(π/2){1−α} e^(−α) but f^′ (α)= ∫_(−∞) ^(+∞) (∂/∂α)(((xsin(αx))/((1+x^2 )^2 )))dx =∫_(−∞) ^(+∞) ((x^2 cos(αx))/((1+x^2 )^2 ))dx ⇒ ∫_(−∞) ^(+∞) ((x^2 cos(αx))/((1+x^2 )^2 ))dx =(π/2)(1−α)e^(−α) .
$$\left.\mathrm{1}\right)\:{we}\:{have}?{f}\left(\alpha\right)={Im}\:\left(\int_{−\infty} ^{+\infty} \:\frac{{x}^{} {e}^{{i}\alpha{x}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\right)\:{let} \\ $$$$\varphi\left({z}\right)=\frac{{z}\:{e}^{{i}\alpha{z}} }{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }\:\:\Rightarrow\varphi\left({z}\right)=\frac{{z}\:{e}^{{i}\alpha{z}} }{\left({z}−{i}\right)^{\mathrm{2}} \left({z}+{i}\right)^{\mathrm{2}} }\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$${iand}\:−{i}\left({doubles}\right)\:\Rightarrow\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\right)\:{but} \\ $$$${Res}\left(\varphi,{i}\right)={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}!\right.}\left\{\left({z}−{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\left\{\:\frac{{z}\:{e}^{{i}\alpha{z}} }{\left({z}+{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{\left({e}^{{i}\alpha{z}} \:+{i}\alpha{z}\:{e}^{{i}\alpha{z}} \right)\left({z}+{i}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+{i}\right){z}\:{e}^{{i}\alpha{z}} }{\left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{i}} \:\frac{\left(\mathrm{1}+{i}\alpha{z}\right){e}^{{i}\alpha{z}} \left({z}+{i}\right)−\mathrm{2}{z}\:{e}^{{i}\alpha{z}} }{\left({z}+{i}\right)^{\mathrm{3}} } \\ $$$$=\frac{\left(\mathrm{1}−\alpha\right)\left(\mathrm{2}{i}\right){e}^{−\alpha} \:−\mathrm{2}{i}\:{e}^{−\alpha} }{\left(\mathrm{2}{i}\right)^{\mathrm{3}} }\:=\frac{\left(\mathrm{2}{i}−\mathrm{2}{i}\alpha−\mathrm{2}{i}\right){e}^{−\alpha} }{−\mathrm{8}{i}} \\ $$$$=\frac{\alpha}{\mathrm{4}}\:{e}^{−\alpha} \:\Rightarrow\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\alpha}{\mathrm{4}}\:{e}^{−\alpha} \:\:={i}\frac{\alpha\pi}{\mathrm{2}}\:{e}^{−\alpha} \:\Rightarrow \\ $$$${f}\left(\alpha\right)\:=\frac{\pi\alpha}{\mathrm{2}}\:{e}^{−\alpha} \\ $$$$\left.\mathrm{2}\right)\:{f}^{'} \left(\alpha\right)\:=\frac{\pi}{\mathrm{2}}\left\{{e}^{−\alpha} \:−\alpha\:{e}^{−\alpha} \right\}=\frac{\pi}{\mathrm{2}}\left\{\mathrm{1}−\alpha\right\}\:{e}^{−\alpha} \:{but} \\ $$$${f}^{'} \left(\alpha\right)=\:\int_{−\infty} ^{+\infty} \:\frac{\partial}{\partial\alpha}\left(\frac{{xsin}\left(\alpha{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\right){dx} \\ $$$$=\int_{−\infty} ^{+\infty} \:\frac{{x}^{\mathrm{2}} \:{cos}\left(\alpha{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} {cos}\left(\alpha{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\alpha\right){e}^{−\alpha} \:. \\ $$

Pin

Leave a Reply

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