Solutions to Complex Integral Additional Problems

1) ∫ _-¥  ^¥    sin x dx / (x²   + 2x + 4)

In terms of f(z) we can write:

f(z)=  exp(iz)   dz / (z²  - 2z + 4)

so the integral becomes:

∫ _-¥  ^¥    exp(iz)   dz / (z²  - 2z + 4)

Factor f(z) to get:

exp(iz)   dz / (z²  - 2z + 4) =    exp(iz)/ (z + 1 - iÖ3) (z + 1 + iÖ3)

 

Then singularities (poles ) occur at:

z= -1 + -iÖ3   and z = -1  -iÖ3

Therefore:

Res f(z) =  lim _{z ® 1+iÖ3}       [exp(iz)/ (z + 1 + iÖ3)]

=  exp i(- 1+ -iÖ3)/  2iÖ3   =   exp(-i) exp(-Ö3) / 2i Ö3   exp (-Ö3)

But recall exp (-i) = sin(-1) =  -sin (1)  so that:

Res f(z) = - sin(1)/ 2i Ö3   exp (-Ö3)

Then:

∫ _-¥  ^¥    exp(iz)   dz / (z²  - 2z + 4)  =

2 pi [- sin(1)/ 2i Ö3   exp (-Ö3)]  =    -pi [- sin(1)/ Ö3   e^-Ö3  ]

=  ∫ _-¥  ^¥    sin x dx / (x²   + 2x + 4)

Note: this soln. is for  the upper half plane. To modify the development for the case where the singularity is in the lower half plane, or m < 0 (i.e. m = -i) then we have:

∫ _-¥  ^¥    exp(im x)  f(x) dx  = - 2 pi å (Res)

(See e.g. the accompanying graphic. Note that  in the lower graphic, that should be - i Ö3   along the Im -axis)

2)  ∫ _-¥  ^¥    cos x dx / (x⁴     +  1)

To obtain singularities:

x⁴     +  1 = 0   or  x⁴     =   -1

Then: x =   (-1)^1/4 = (-1 + 0i) ^1/4

=   ⁴Ö1 [cos (p + 2kp/ 4) + isin((p + 2kp/ 4)]

For k= 0: 

(-1)^1/4 =   cis(p/ 4) =  exp (ip/ 4)

For k=1:

(-1)^1/4 =   cis(3p/ 4) =  exp (i3p/ 4)

Both poles are in the upper half plane (see e,g, diagram for Jan. 27 post)

The above are associated with two simple poles, for which:

Res f(z) _{z = exp pi/4}       =   lim _{z ® exp pi/4}      exp(iz)  / 4 z³     =

 exp(- p/4) / 4 exp(3pi/4) =  exp(- p/4) /  2Ö2 – 2iÖ2  =

exp(- p/4) /  2Ö2 (1– i)

And:

Res f(z) _{z = exp 3pi/4}       =   lim _{z ® exp 3pi/4}      exp(iz)  / 4 z³     =

 exp(- 3p/4) / 4 [exp(3pi/4)] ³ =  exp(- 3p/4) / 4 exp(7pi/4)

=  exp(- 3p/4) / 4[ cos(9p/4) + i sin(9p/4)]

=  exp(- 3p/4) /  2Ö2 + 2iÖ2  =    exp(- 3p/4) /  2Ö2 (1 +i)

Finally:

∫ _-¥  ^¥    exp(iz) dz / (z⁴     +  1)  =

 2 pi [exp(- p/4) /  2Ö2 (1– i)  +    exp(- 3p/4) /  2Ö2 (1 +i) ]

= ∫ _-¥  ^¥    cos x dx / (x⁴     +  1)