Page 763 - Advanced_Engineering_Mathematics o'neil

22.3 Evaluation of Real Integrals 743

Then
cos(cx) sin(cx)
∞ ∞

dx + i dx
2
2
2
2
2
2
2
−∞ (x + α )(x + β ) −∞ (x + α )(x + β )
2
e e
−cα −cβ
= 2πi +
2αi(β − α ) 2βi(α − β )
2
2
2
2
π e −cα e −cβ
= − .
β − α 2 α β
2
Separate real and imaginary parts to obtain
∞
cos(cx) π e e
−cα −cβ
dx = −
2
2
2
2
2
−∞ (x + α )(x + β ) β − α 2 α β
and

∞
sin(cx)
dx = 0.
2
−∞ (x + α )(x + β )
2
2
2
The last integral is obvious because the integrand is odd.
22.3.3 Rational Functions of Cosine and Sine
Let K(x, y) be a quotient of polynomials in x and y. For example,
3
2
x y − 2xy + x − 2y
K(x, y) = .
4
x + xy − 8
4
Such a function is called a rational function of x and y. If we replace x = cos(θ) and y = sin(θ),
we obtain a rational function of cosine and sine. We want a way to evaluate the integral of such
a function over [0,2π]. This will be an integral of the form
2π
K(cos(θ), sin(θ))dθ.
0
Again, the idea is to express this real integral as a complex integral, which is then evaluated
using the residue theorem.
iθ
iθ
Let γ be the unit circle about the origin γ(θ) = e for 0 ≤ θ ≤ 2π. On this curve, z = e and
z = e −iθ = 1/z,so
1 1 1 1
cos(θ) = z + and sin(θ) = z − .
2 z 2i z
Furthermore, on γ ,
iθ
dz = ie dθ = izdθ,
so
1
dθ = dz.
iz
Therefore,
1 1 1 1 1 1
2π
iθ
K z + , z − dz = K(cos(θ),sin(θ)) ie dθ
2 z 2i z iz ie iθ
γ 0
2π
= K(cos(θ),sin(θ))dθ.
0
Evaluate the integral on the left using the residue theorem, yielding the integral we want.
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October 14, 2010 15:37 THM/NEIL Page-743 27410_22_ch22_p729-750