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744 CHAPTER 22 Singularities and the Residue Theorem

2π
In summary, to evaluate K(cos(θ),sin(θ)dθ, begin by computing the function
0
1 1 1 1 1

f (z) = K z + , z − . (22.5)
2 z 2i z iz
Then
2π

K(cos(θ), sin(θ))dθ = 2πi Res( f, z j ) (22.6)
0
|z j |<1
with this sum over all singularities z j of f (z) enclosed by the unit circle.

EXAMPLE 22.19

Evaluate
2π 2
sin (θ)
dθ.
2 + cos(θ)
0
2
Here K(x, y) = y /(2 + x), and
2
sin (θ)
K(cos(θ),sin(θ)) = .
2 + cos(θ)
Let x = cos(θ) = (z + 1/z)/2, and y = sin(θ) = (z − 1/z)/2i in K(x, y), and multiply by 1/iz to
produce the complex function of equation (22.5):

2 4 2
[(z − 1/z)/2i] 1 i z − 2z + 1
f (z) = = .
2
2 + (z + 1/z)/2 iz 2 z (z + 4z + 1)
2
√
2
f has a double pole at 0 and simple poles at zeros of z + 4z + 1, which are −2 ± 3. Only the
√
poles 0 and 2 − 3 are enclosed by γ . By equation (22.6),
2π 2 √
sin (θ)
dθ = 2πi[Res( f,0) + Res( f,−2 + 3)].
2 + cos(θ)
0
Compute these residues:
4
2
d d i z − 2z + 1
2
Res( f,0) = lim (z f (z)) = lim
2
z→0 dz z→0 dz 2 z + 4z + 1
3
2
4
5
i z + 6z + 2z − 4z − 3z − 2
= lim 2 =−2i
2
2 z→0 (z + 4z + 1) 2
and
4
2
√ i z − 2z + 1
Res( f,−2 + 3) =
2 2z(z + 4z + 1) + z (2z + 4) z=−2+ 3
2
2
√
√
i 42 − 24 3
= √ .
2 −12 + 7 3
Therefore,
√ √
sin (θ) i 42 − 24 3 90 − 52 3
2π 2
dθ = 2πi −2i + √ = √ π.
2 + cos(θ) 2 −12 + 7 3
0 12 − 7 3
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October 14, 2010 15:37 THM/NEIL Page-744 27410_22_ch22_p729-750