Saturday, February 1, 2014

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)

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