10. Appendix

622 Appendix B
y
-R R
R
C
x
In this case the pole occurs at the origin
so we have to exclude the origin from C
by adding another semicircle with radius
‰. By applying the Residue Theorem we
find that:
C
1
dz 0 (4.5m)
z
To obtain the integral in (4.5l) we have
to take the limits: R ⇒ ∞ and ‰ ⇒ 0.
We note that the integrals over the two semicircles of radius R and ‰ cancel
each other since the integration over the angle ı is from 0 to apple for the larger
semicircle and from apple to 0 for the smaller semicircle.
Thus
∞
1
P
∞ x ′ E dx′ 0 (4.5n)
Putting all these integration results back into (4.5b) we obtain:
Re[f (E)] 1
i (iapple)
apple 2° (E E ′ i iapple
) i° 2° (E E ′
) i°
1
1
2° (E E ′ ) i°
1
(E E ′
) i°
1 (E E
°
′ )
(E E ′ ) 2 ° 2
(4.5o)
For readers who are familiar with complex analysis and contour integrals,
there is a faster way to obtain the same result. The first step is to make a
transformation: z ′ z E ′ so that the integral in (4.5b) becomes:
Re[f (E)] 1
apple P
∞
(1)
∞ (z ′ E ′ E)[(z ′ ) 2 ° 2 ] dz′
(4.5p)
The integrand now has three poles at E E ′ , i° and i°. To calculate the
principal value of the integral one constructs two contours A and B as shown
in the following figures.