Menu Close

let-f-t-0-cos-tx-2-x-2-2-dx-1-find-a-simple-form-of-f-t-2-calculate-0-cos-3x-2-x-2-2-dx-




Question Number 36203 by prof Abdo imad last updated on 30/May/18
let f(t) = ∫_0 ^∞    ((cos(tx))/((2+x^2 )^2 ))dx  1) find a simple form  of f(t)  2) calculate ∫_0 ^∞     ((cos(3x))/((2+x^2 )^2 ))dx
$${let}\:{f}\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left({tx}\right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:\:{of}\:{f}\left({t}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$
Commented by prof Abdo imad last updated on 01/Jun/18
we have  2f(t) = ∫_(−∞) ^(+∞)   ((cos(tx))/((2+x^2 )^2 ))dx  =Re( ∫_(−∞) ^(+∞)    (e^(itx) /((2+x^2 )^2 ))dx) let introduce the  complex?function ϕ(z) = (e^(itz) /((2+z^2 )^2 ))  ϕ(z) = (e^(itz) /((z−i(√2))^2 (z +i(√2))^2 )) so the poles of ϕ are  i(√2)  and −i(√2) (doubles)  ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ Res( ϕ,i(√2))  Res(ϕ,i(√2)) =lim_(z→i(√2)) { (z−i(√2))^2 ϕ(z)}^′   =lim_(z→i(√2))   {  (e^(itz) /((z+i(√2))^2 ))}^′   =lim_(z→i(√2))   { ((it e^(itz) (z+i(√2))^2   −2(z+i(√2))e^(itz) )/((z +i(√2))^4 ))}  =lim_(z→i(√2))    e^(itz)  ((it(z +i(√2)) −2)/((z +i(√2))^3 ))  =e^(it(i(√2)))  ((it(2i(√2)) −2)/((2i(√2))^3 ))  = ((−2t(√2)−2)/(−8i(2(√2)))) e^(−t(√2))   =  ((t(√2) +1)/(8i(√2))) e^(−t(√2))   ∫_(−∞) ^(+∞)  ϕ(z)dz =2iπ  ((1+t(√2))/(8i(√2))) e^(−t(√2))   = (π/(4(√2)))(1+t(√2)) e^(−t(√2))   but we have  2f(t)= Re( ∫_(−∞) ^(+∞) ϕ(z)dz) ⇒  ★ f(t) = (π/(8(√2))) (1+t(√2))e^(−t(√2))  ★  2) we have ∫_0 ^∞    ((cos(3x))/((2+x^2 )^2 ))dx =f(3) ⇒  ∫_0 ^∞     ((cos(3x))/((2+x^2 )^2 ))dx = (π/(8(√2)))( 1+3(√2))e^(−3(√2))  .
$${we}\:{have}\:\:\mathrm{2}{f}\left({t}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{cos}\left({tx}\right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx} \\ $$$$={Re}\left(\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{e}^{{itx}} }{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\right)\:{let}\:{introduce}\:{the} \\ $$$${complex}?{function}\:\varphi\left({z}\right)\:=\:\frac{{e}^{{itz}} }{\left(\mathrm{2}+{z}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\varphi\left({z}\right)\:=\:\frac{{e}^{{itz}} }{\left({z}−{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \left({z}\:+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$${i}\sqrt{\mathrm{2}}\:\:{and}\:−{i}\sqrt{\mathrm{2}}\:\left({doubles}\right) \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\:\varphi,{i}\sqrt{\mathrm{2}}\right) \\ $$$${Res}\left(\varphi,{i}\sqrt{\mathrm{2}}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \left\{\:\left({z}−{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{'} \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \:\:\left\{\:\:\frac{{e}^{{itz}} }{\left({z}+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\right\}^{'} \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \:\:\left\{\:\frac{{it}\:{e}^{{itz}} \left({z}+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:\:−\mathrm{2}\left({z}+{i}\sqrt{\mathrm{2}}\right){e}^{{itz}} }{\left({z}\:+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{4}} }\right\} \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{2}}} \:\:\:{e}^{{itz}} \:\frac{{it}\left({z}\:+{i}\sqrt{\mathrm{2}}\right)\:−\mathrm{2}}{\left({z}\:+{i}\sqrt{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$={e}^{{it}\left({i}\sqrt{\mathrm{2}}\right)} \:\frac{{it}\left(\mathrm{2}{i}\sqrt{\mathrm{2}}\right)\:−\mathrm{2}}{\left(\mathrm{2}{i}\sqrt{\mathrm{2}}\right)^{\mathrm{3}} } \\ $$$$=\:\frac{−\mathrm{2}{t}\sqrt{\mathrm{2}}−\mathrm{2}}{−\mathrm{8}{i}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)}\:{e}^{−{t}\sqrt{\mathrm{2}}} \:\:=\:\:\frac{{t}\sqrt{\mathrm{2}}\:+\mathrm{1}}{\mathrm{8}{i}\sqrt{\mathrm{2}}}\:{e}^{−{t}\sqrt{\mathrm{2}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:\frac{\mathrm{1}+{t}\sqrt{\mathrm{2}}}{\mathrm{8}{i}\sqrt{\mathrm{2}}}\:{e}^{−{t}\sqrt{\mathrm{2}}} \\ $$$$=\:\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\left(\mathrm{1}+{t}\sqrt{\mathrm{2}}\right)\:{e}^{−{t}\sqrt{\mathrm{2}}} \:\:{but}\:{we}\:{have} \\ $$$$\mathrm{2}{f}\left({t}\right)=\:{Re}\left(\:\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\right)\:\Rightarrow \\ $$$$\bigstar\:{f}\left({t}\right)\:=\:\frac{\pi}{\mathrm{8}\sqrt{\mathrm{2}}}\:\left(\mathrm{1}+{t}\sqrt{\mathrm{2}}\right){e}^{−{t}\sqrt{\mathrm{2}}} \:\bigstar \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:={f}\left(\mathrm{3}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{cos}\left(\mathrm{3}{x}\right)}{\left(\mathrm{2}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:=\:\frac{\pi}{\mathrm{8}\sqrt{\mathrm{2}}}\left(\:\mathrm{1}+\mathrm{3}\sqrt{\mathrm{2}}\right){e}^{−\mathrm{3}\sqrt{\mathrm{2}}} \:. \\ $$

Leave a Reply

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