mathsec

Evaluation of improper integrals(residues)

int{0}{infty}{{dx/{x^3+1}}}={{2 pi}/{3sqrt{3}}}

To establsih the integration formula we use the closed contour shown in the figure, where R>1

int{C_1}{}{{dz/{z^3+1}}}+int{C_2}{}{{dz/{z^3+1}}}+int{C_R}{}{{dz/{z^3+1}}}=

{=}2 pi i Res(z=z_n){1/{z^3+1}}

Where the legs of the closed contour are as indicated in the figure. Since C_1 has parametric representationz=r (0<=r<=R),

int{C_1}{}{{dz/{z^3+1}}}=int{0}{R}{{dr/{r^3+1}}}

and C_2 can be represented by z=re^{2 pi i/3} (0<=r<=R)

dz=e^{2 pi i/3}dr

int{C_2}{}{{dz/{z^3+1}}}=int{-R}{0}{{{e^{2 pi i/3}}/{{r^3}{e^{2 pi i}}+1}}dr}={-}int{0}{R}{{{e^{2 pi i/3}}/{{r^3}{e^{2 pi i}}+1}}dr}

{=}{-}{e^{2 pi i/3}}int{0}{R}{{{dr}/{{r^3}{e^{2 pi i}}+1}}}

e^{2 pi i} = cos 2 pi + i sin 2 pi = 1

int{C_2}{}{{dz/{z^3+1}}}={-}{e^{2 pi i/3}}int{0}{R}{{{dr}/{{r^3}+1}}}

we find the singularities

z^3+1=0

z=e^({i/3}(pi + 2 pi n))  (n=0,1,2)

z_0=e^(i pi /3) z_1=e^(i pi) z_2=e^(i 5 pi /3)

z_0 lies inside the closed contour

p(z)=1 q(z)=z^3+1

p(z_0)=1<>0 q(z_0)=0, q prime (z) = 3 z^2

2 pi i Res(z=z_0){1/{z^3+1}}=2 pi i{1/{3{z_0}^2}}= 2 pi i{{z_0}/{3{z_0}^3}}

{z_0}^3=-1

2 pi i Res(z=z_0){1/{z^3+1}}= {-} 2 pi i{{z_0}/{3}}=  {-} 2 pi i{{e^(i pi /3)}/{3}}

consequently,

int{C_1}{}{{dz/{z^3+1}}}+int{C_2}{}{{dz/{z^3+1}}}+int{C_R}{}{{dz/{z^3+1}}}=

{=}(1-{e^{2 pi i/3}})int{0}{R}{{dr/{r^3+1}}}+int{C_R}{}{{dz/{z^3+1}}}={-} 2 pi i{{e^(i pi /3)}/{3}}

furthermore,

M_R={1/{R^3-1}} Perimeter_R={2 pi R/3}

delim{|}{int{C_R}{}{{dz/{z^3+1}}}}{|} <={1/{{R^3}-1}}{2 pi R/3} right 0 as R right infty

int{0}{infty}{{dr/{r^3+1}}}={-} 2 pi i{{e^(i pi /3)}/{3(1-{e^{2 pi i/3}})}}={-} 2 pi i {(0.5+i{sqrt{3}}/2)/(1.5-i{sqrt{3}}/2)}

int{0}{infty}{{dr/{r^3+1}}}={-} 2 pi i {(0.5+i{sqrt{3}}/2)/(1.5-i{sqrt{3}}/2)}{(1.5+i{sqrt{3}}/2)/(1.5+i{sqrt{3}}/2)}

int{0}{infty}{{dr/{r^3+1}}}={{2 pi}/{3sqrt{3}}}