14—Complex Variables 356zero. It goes to zero faster than any inverse power ofy, so even with the length of the contour going
asπR, the combination vanishes.
C 2 C 3 C 4
As before, when you pushC 1 up toC 2 and toC 3 , nothing has changed, because the contour
has crossed no singularities. The transition toC 4 happens because the pairs of straight line segments
cancel when they are pushed together and made to coincide. The large contour is pushed to+i∞
where the negative exponential kills it. All that’s left is the sum over the two residues ataeiπ/^4 and
ae^3 iπ/^4. ∫
C 1=
∫
C 4= 2πi
∑
Reseikz
a^4 +z^4
The denominator factors as
a^4 +z^4 = (z−aeiπ/^4 )(z−ae^3 iπ/^4 )(z−ae^5 iπ/^4 )(z−ae^7 iπ/^4 )
The residue ataeiπ/^4 =a(1 +i)/
√
2 is the coefficient of 1 /(z−aeiπ/^4 ), so it is
eika(1+i)/
√ 2(aeiπ/^4 −ae^3 iπ/^4 )(aeiπ/^4 −ae^5 iπ/^4 )(aeiπ/^4 −ae^7 iπ/^4 )
1
2 3
Do you have to do a lot of algebra to evaluate this denominator? Maybe you will prefer that to the
alternative:draw a picture. The distance from the center to a corner of the square isa, so each side
has lengtha
√
2. The first factor in the denominator of the residue is the line labeled “1” in the figure,so it isa
√
2. Similarly the second and third factors (labeled in the diagram) are 2 a(1 +i)/
√
2 andia
√
2. This residue is thenRes
eiπ/^4=
eika(1+i)/
√
2
(a
√
2
)
(2a(1 +i)/
√
2)(ia
√
2)
=
eika(1+i)/
√
2a^32
√
2(−1 +i)
(14.13)
For the other pole, ate^3 iπ/^4 , the result is
Res
e^3 iπ/^4=
eika(−1+i)/
√
2(−a
√
2)(2a(−1 +i)/
√
2)(ia
√
2)
=
eika(−1+i)/
√
2a^32
√
2(1 +i)
(14.14)
The final result for the integral Eq. (14.10) is then the sum of these (× 2 πi)
∫+∞
−∞eikxdx
a^4 +x^4
= 2πi
[
( 14. 13 ) + ( 14. 14 )
]
=
πe−ka/
√
2a^3
cos[(ka/
√
2)−π/4] (14.15)
This would be a challenge to do by other means, without using contour integration. It is probably
possible, but would be much harder. Does the result make any sense? The dimensions work, because