Mathematical Tools for Physics

14—Complex Variables 429

move across a singularity, you can distort a contour at will. I will push the contourC 1 up, but I have to leave
the endpoints where they are and I can’t let it cross the pole atia. Those are my only constraints.

C 2 C 3 C 4 C 5

As I push the contour fromC 1 up toC 2 , nothing has changed, and the same applies toC 3. The next two
steps however, requires some comment. InC 3 the two straight-line segments that parallel they-axis are going
in opposite directions, and as they are squeezed together, they cancel each other; they are integrals of the same
function in reverse directions. In the final step, toC 5 , I pushed the contour all the way to+i∞and eliminated
it. How does that happen? On a big circle of radiusR, the function 1 /(a^2 +z^2 )has a magnitude approximately
1 /R^2. As you push the top curve inC 4 out, forming a big circle, its length isπR. The product of these isπ/R,
and that approaches zero asR→∞. All that is left is the single closed loop inC 5 , and I evaluate that with the
residue theorem. ∫

C 1

=

∫

C 5

= 2πiRes

z=ia

1

a^2 +z^2

Compute this residue by examining the behavior near the pole atia.

1

a^2 +z^2

=

1

(z−ia)(z+ia)

≈

1

(z−ia)(2ia)

Near the pointz=iathe value ofz+iais nearly 2 ia, so the coefficient of 1 /(z−ia)is 1 /(2ia), and that is the
residue. The integral is 2 πitimes this residue, so
∫∞

−∞

dx

1

a^2 +x^2

= 2πi.

1

2 ia

=

π

a

(9)

The most obvious check on this result is that it has the correct dimensions. [dz/z^2 ] =L/L^2 = 1/L, a reciprocal
length (assumingais a length). What happens if you push the contourdowninstead of up? See problem 10

Example 4
How about the more complicated integral, Eq. ( 7 )? There are more poles, so that’s where to start. The
denominator vanishes wherez^4 =−a^4 , or at

z=a

(

eiπ+2inπ

) 1 / 4

=aeiπ/^4 einπ/^2