Mathematical Tools for Physics

14—Complex Variables 433

if I extend the integration limits to the whole axis (times^1 / 2 ).

1

2

∫

C 1

dz

(a^2 +z^2 )^2

C 1

As with Eq. ( 8 ), push the contour up and it is caught on the pole atz=ia. That’s curveC 5 following that
equation. This time however, the pole is second order, so it take a (little) more work to evaluate the residue.

1

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

∫∞

0

dx/(a^2 +x^2 )^2 = 2πi.(−1).

1

(2ia)^3

=

π

4 a^3

Is this plausible? The dimensions came out as I expected, and to estimate the size of the coefficient,π/ 4 , look
back at the result Eq. ( 9 ). 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. ( 9 ). 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.

Example 7

A trigonometric integral:

∫ 2 π

0 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