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