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14—Complex Variables 355

as long as it doesn’t move across a singularity, you can distort a contour at will. I will push the contour
C 1 up, but I have to leave the endpoints where they are in order not let the contour cross the pole at

ia. Those are my sole 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 the

y-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

(14.12)

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 problem14.10

Example 4
How about the more complicated integral, Eq. (14.10)? 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

∫

C 1

eikzdz

a^4 +z^4

C 1

I’m going to use the same method as before, pushing the contour past some poles, but I have to be a

bit more careful this time. The exponential, not the 1 /z^4 , will play the dominant role in the behavior

at infinity. Ifkis positive then ifz=iy, the exponentialei

(^2) ky

=e−ky→ 0 asy→+∞. It will blow

up in the−i∞direction. Of course ifkis negative the reverse holds.

Assumek > 0 , then in order to push the contour into a region where I can determine that the

integral along it is zero, I have to push it toward+i∞. That’s where the exponential drops rapidly to