Example Jordan’s lemma

The integral to be evaluated is:

int{-infty}{infty}{{x^3sin ax}/{x^4+4}}

We define the function


and, by computing the fourth roots of -4,

z=4^{1/4}e^{(1/4)(i pi+2ni pi)} n=(0,1,2,3)

we find that singularities

z_1=2^{1/2}e^{{i pi}/{4}} , z_2=2^{1/2}e^{{i pi}/{4}}e^{{i pi}/{2}}

z_1=2^{1/2}(cos (pi/4) + i sin (pi/4))=1+i 

z_2=2^{1/2}(cos (pi/4) + i sin (pi/4))(cos(pi/2)+i sin(pi/2))


both lie inside the simple closed contour shown above, where R>{1/4}. The other two singularities lie below the real axis. The theorem for finding rsidues at simple poles tells us

int{-R}{R}{{x^3 e^{iax}}/{x^4+4}}+int{C_r}{}{f(z)e^{iaz}dz}=2 pi i(B_1+B_2)

Note: Let two functions p and q be analytic at a point z_0. If

P(z_0)<>0, Q(z_0)=0 and Q prime(z_0)<>0

then z_0 is a simple pole of the quotient {p(z)}/{q(z)} and

Res(z=z_0){p(z)}/{q(z)}={p(z_0)}/{q prime (z_0)}

In our case:

B_1=Res(z=z_1){z^3 e^{iaz}}/{x^4+4}={{z_1}^3 e^{iaz_1}}/{4{z_1}^3}={e^{ia(1+i)}}/4


B_2=Res(z=z_2){z^3 e^{iaz}}/{x^4+4}={{z_2}^3 e^{iaz_2}}/{4{z_2}^3}={e^{ia(-1+i)}}/4



2 pi i (B_1+B_2)= pi i e^{-a} ({e^{ia}+e^{-ia}})/2

=pi i e^{-a} ( cos a + i sin a + cos a - i sin a )/2 = i pi e^{-a} cos a

we are now able to write

int{-R}{R}{{x^3 e^{iax}}/{x^4+4}}+int{C_r}{}{f(z)e^{iaz}dz}=2 pi i(B_1+B_2)

int{-R}{R}{{x^3 (cos ax + i sin ax)}/{x^4+4}}+int{C_r}{}{f(z)e^{iaz}dz}= i pi e^{-a} cos a

int{-R}{R}{{x^3 (sin ax)}/{x^4+4}}+Im int{C_r}{}{f(z)e^{iaz}dz}= i pi e^{-a} cos a

Furthermore, if z is a point in C_r, then

delim{|}{f(z)}{|}<=M_R where M_R={R^3}/{R^4 - 4}right 0 as R right infty

and this means that

delim{|}{Im int{C_r}{}{f(z)e^{iaz}dz}}{|}

Finally then

int{-R}{R}{{x^3 (sin ax)}/{x^4+4}}= i pi e^{-a} cos a