10. Appendix

C
dz ′
z ′ i°
Solution to Problem 4.5 621
(4.5f)
In order to avoid the pole at z ′ i°, we construct the contour C with the
half-circle shown in the following figure:
C
y
-R R
R
C
dz ′
z ′ i°
R
dx
R x i°
apple
0
x
C consists of two parts: (1) a straight
line along the x-axis connecting (R,0)
and (R,0) and (2) a semi-circle in the
lower half of the y-plane with radius R
and centered at the origin. Since C does
not enclose any pole we obtain from the
Residue Theorem
C
dz ′
z ′ i°
0 (4.5g)
We now explicitly decompose the integration
over C into one integral over the horizontal
axis and one over the semi-circle:
Rieiı dı
Re iı i°
(4.5h)
Next we take the limit R ⇒ ∞. The first integral of (4.5h) becomes
∞
dx
P
which is just what we want to calculate.
∞ x i°
In the limit RÊ ≫ °, the second integral becomes
apple
0
Rieiı dı
Re iı i° ⇒
apple
0
Rie iı dı
i(apple) (4.5i)
iı
Re
Substituting these results back into (4.5g) we get:
∞
dx
P
iapple (4.5j)
∞ x i°
Similarly we can obtain:
∞
dx
P
iapple (4.5k)
∞ x i°
Finally, to evaluate the integral:
∞
P
∞
1
x ′ E dx′ ∞
P
∞
1
dx (4.5l)
x
we have to use a different contour C as shown in the following figure.