14—Complex Variables 3581
2
1
(a^2 +z^2 )^2
=
1
2
1
(z−ia)^2 (z+ia)^2
=
1
2
1
(z−ia)^2 (z−ia+ 2ia)^2
=
1
2
1
(z−ia)^2 (2ia)^2
[
1 + (z−ia)/ 2 ia
] 2
=
1
2
1
(z−ia)^2 (2ia)^2
[
1 − 2
(z−ia)
2 ia
+···
]
=
1
2
1
(z−ia)^2 (2ia)^2
+
1
2
(−2)
1
(z−ia)(2ia)^3
+···
The residue is the coefficient of the 1 /(z−ia)term, so the integral is
∫∞
0dx/(a^2 +x^2 )^2 = 2πi.(−1).
1
(2ia)^3
=
π
4 a^3
Is this plausible? The dimensions came out as expected, and to estimate the size of the coefficient,
π/ 4 , look back at the result Eq. (14.12). Seta= 1and compare theπthere to theπ/ 4 here. The
range of integration is half as big, so that accounts for a factor of two. The integrands are always less
than one, so in the second case, where the denominator is squared, the integrand is always less than
that of Eq. (14.12). The integral must be less, and it is. Why less by a factor of two? Dunno, but plot
a few points and sketch a graph to see if you believe it. (Or use parametric differentiation to relate the
two.)
Example 7
A trigonometric integral:
∫ 2 π0 dθ
/
(a+bcosθ). The first observation is that unless|a|>|b|then this
denominator will go to zero somewhere in the range of integration (assuming thataandbare real).
Next, the result can’t depend on the relative sign ofaandb, because the change of variablesθ′=θ+π
changes the coefficient ofbwhile the periodicity of the cosine means that you can leave the limits alone.
I may as well assume thataandbare positive. The trick now is to use Euler’s formula and express the
cosine in terms of exponentials.
Letz=eiθ, then cosθ=
1
2
[
z+
1
z
]
and dz=ieiθdθ=iz dθ
Asθgoes from 0 to 2 π, the complex variablezgoes around the unit circle. The integral is then
∫ 2 π0dθ
1
(a+bcosθ)
=
∫
Cdz
iz
1
a+b
(
z+z^1
)
/ 2
The integrand obviously has some poles, so the first task is to locate them.