Menu Close

calculate-0-x-2-cos-pix-x-2-4-2-dx-

Tinku Tara 0 Comments
Pin




Question Number 38724 by maxmathsup by imad last updated on 28/Jun/18
calculate ∫_0 ^∞ ((x^2 cos(πx))/((x^2 +4)^2 ))dx
$${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}} {cos}\left(\pi{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by maxmathsup by imad last updated on 30/Jun/18
let I =∫_0 ^∞ ((x^2 cos(πx))/((x^2 +4)^2 ))dx 2I = ∫_(−∞) ^(+∞) ((x^2 cos(πx))/((x^2 +4)^2 ))dx =Re( ∫_(−∞) ^(+∞) ((x^2 e^(iπx) )/((x^2 +4)^2 ))) let ϕ(z)=((z^2 e^(iπz) )/((z^2 +4)^2 )) we have ϕ(z) =((z^2 e^(iπz) )/((z−2i)^2 (z+2i)^2 )) so the poles of ϕ are 2i and −2i ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,2i) but Res(ϕ,2i) =lim_(z→2i) (1/((2−1)!)) { (z−2i)^2 ϕ(z)}^((1)) =lim_(z→2i) { ((z^2 e^(iπz) )/((z+2i)^2 ))}^((1)) =lim_(z→2i) { (((2z e^(iπz) +iπz^2 e^(iπz) )(z+2i)^2 −2(z+2i)z^2 e^(iπz) )/((z+2i)^4 ))} lim_(z→2i) (((2z+iπz^2 )e^(iπz) (z+2i)−2z^2 e^(iπz) )/((z+2i)^3 )) = (((4i −4iπ)e^(−2π) (4i) +8 e^(−2π) )/((4i)^3 )) =(((−16 +16 π +8)e^(−2π) )/(−64i)) =(((8−16π)e^(−2π) )/(64i)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (((8−16π)e^(−2π) )/(64i)) =π (((8−16π)e^(−2π) )/(32)) =(π/(32)) 8(1−2π)e^(−2π) =(π/4)(1−2π)e^(−2π) ⇒
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}^{\mathrm{2}} {cos}\left(\pi{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }{dx} \\ $$$$\mathrm{2}{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} {cos}\left(\pi{x}\right)}{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }{dx}\:\:={Re}\left(\:\int_{−\infty} ^{+\infty} \:\frac{{x}^{\mathrm{2}} \:{e}^{{i}\pi{x}} }{\left({x}^{\mathrm{2}} \:+\mathrm{4}\right)^{\mathrm{2}} }\right)\:{let} \\ $$$$\varphi\left({z}\right)=\frac{{z}^{\mathrm{2}} \:{e}^{{i}\pi{z}} }{\left({z}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\:\:{we}\:{have}\:\varphi\left({z}\right)\:=\frac{{z}^{\mathrm{2}} \:{e}^{{i}\pi{z}} }{\left({z}−\mathrm{2}{i}\right)^{\mathrm{2}} \left({z}+\mathrm{2}{i}\right)^{\mathrm{2}} }\:{so}\:{the}\:{poles}\:{of}\:\varphi\:{are}\:\mathrm{2}{i}\:{and}\:−\mathrm{2}{i} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\mathrm{2}{i}\right)\:{but} \\ $$$${Res}\left(\varphi,\mathrm{2}{i}\right)\:={lim}_{{z}\rightarrow\mathrm{2}{i}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\:\left\{\:\left({z}−\mathrm{2}{i}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow\mathrm{2}{i}} \:\:\left\{\:\frac{{z}^{\mathrm{2}} {e}^{{i}\pi{z}} }{\left({z}+\mathrm{2}{i}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} ={lim}_{{z}\rightarrow\mathrm{2}{i}} \:\left\{\:\frac{\left(\mathrm{2}{z}\:{e}^{{i}\pi{z}} \:+{i}\pi{z}^{\mathrm{2}} \:{e}^{{i}\pi{z}} \right)\left({z}+\mathrm{2}{i}\right)^{\mathrm{2}} \:−\mathrm{2}\left({z}+\mathrm{2}{i}\right){z}^{\mathrm{2}} {e}^{{i}\pi{z}} }{\left({z}+\mathrm{2}{i}\right)^{\mathrm{4}} }\right\} \\ $$$${lim}_{{z}\rightarrow\mathrm{2}{i}} \:\:\frac{\left(\mathrm{2}{z}+{i}\pi{z}^{\mathrm{2}} \right){e}^{{i}\pi{z}} \left({z}+\mathrm{2}{i}\right)−\mathrm{2}{z}^{\mathrm{2}} \:{e}^{{i}\pi{z}} }{\left({z}+\mathrm{2}{i}\right)^{\mathrm{3}} } \\ $$$$=\:\frac{\left(\mathrm{4}{i}\:−\mathrm{4}{i}\pi\right){e}^{−\mathrm{2}\pi} \left(\mathrm{4}{i}\right)\:+\mathrm{8}\:{e}^{−\mathrm{2}\pi} }{\left(\mathrm{4}{i}\right)^{\mathrm{3}} }\:=\frac{\left(−\mathrm{16}\:+\mathrm{16}\:\pi\:+\mathrm{8}\right){e}^{−\mathrm{2}\pi} }{−\mathrm{64}{i}}\:=\frac{\left(\mathrm{8}−\mathrm{16}\pi\right){e}^{−\mathrm{2}\pi} }{\mathrm{64}{i}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\left(\mathrm{8}−\mathrm{16}\pi\right){e}^{−\mathrm{2}\pi} }{\mathrm{64}{i}}\:=\pi\:\frac{\left(\mathrm{8}−\mathrm{16}\pi\right){e}^{−\mathrm{2}\pi} }{\mathrm{32}} \\ $$$$=\frac{\pi}{\mathrm{32}}\:\mathrm{8}\left(\mathrm{1}−\mathrm{2}\pi\right){e}^{−\mathrm{2}\pi} \:=\frac{\pi}{\mathrm{4}}\left(\mathrm{1}−\mathrm{2}\pi\right){e}^{−\mathrm{2}\pi} \:\:\Rightarrow \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 30/Jun/18
2I = (π/4)(1−2π)e^(−2π) ⇒ I =(π/8)(1−2π)e^(−2π) . π
$$\mathrm{2}{I}\:=\:\frac{\pi}{\mathrm{4}}\left(\mathrm{1}−\mathrm{2}\pi\right){e}^{−\mathrm{2}\pi} \:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{8}}\left(\mathrm{1}−\mathrm{2}\pi\right){e}^{−\mathrm{2}\pi} . \\ $$$$\pi \\ $$

Pin

Leave a Reply

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