traced viewing the point z ¼ 1 on the right, the contour C
0 is traced viewing the
corresponding point w ¼ 0 on the right. The above clockwise cycle results from such
a situation. The integrand of (6.265) has a simple pole at w ¼ 0.
Regarding other singularities of the integrand, there are two branch points at
w ¼ 1/a and w ¼ 1/b. Then, we appropriately choose R ) 1 (or 1/R ( 1) so that C
0
does not cut across the branch cut. To meet this condition, we have two choices.
(i) The branch cut connects w ¼ 1/a and w ¼ 1/b. (ii) Another choice of the branch
cut is that it connects w ¼ 1/a and w ¼ À 1 along with w ¼ 1/b and w ¼ 1 (see
Fig. 6.31). If we have 0 < a < b, we can choose a branch cut as that depicted in
Fig. 6.31a corresponding to Fig. 6.28a. In case a < 0 < b, the branch cut can be that
shown in Fig. 6.31b corresponding to Fig. 6.28b. (When a < b < 0, the branch cut
geometry may be that similar to Fig. 6.31a.)
Consequently, in both the above cases of (i) and (ii) there is only a simple pole at
w ¼ 0 without any other singularities inside the contour C
0 . Therefore, according to
(6.148) we have a contour integration described by
I
C
0
1
w
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1 À aw
ð
Þ1 À bw
ð
Þ
p
dw ¼ À2πi Res
1
w
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1 À aw
ð
Þ1 À bw
ð
Þ
p
w¼0
¼ À2πi Á
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1 À aw
ð
Þ1 À bw
ð
Þ
p
w¼0
¼ À2πi,
ð6:266Þ
where the above minus sign is due to the clockwise rotation of the contour integration. Hence, from (6.265) we get
I C ¼
I
C
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
z À a
ð
Þ z À b
ð
Þ
p
dz ¼ 2πi:
ð6:267Þ
Equating (6.262) and (6.267), we finally obtain the answer described by
0
1/
0 < <
(a)
0
1/
< 0 <
(b)
Fig. 6.31 Branch cuts (shown with doubled broken lines) and contour in the w-plane for the
integration of
1
w
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1Àaw
ð
Þ1Àbw ð
Þ
p
. (a) Case of 0 < a < b. (b) Case of a < 0 < b. Notice that in both cases
the contour C
0 is traced clockwise (see text)
262
6 Theory of Analytic Functions
0 is traced viewing the
corresponding point w ¼ 0 on the right. The above clockwise cycle results from such
a situation. The integrand of (6.265) has a simple pole at w ¼ 0.
Regarding other singularities of the integrand, there are two branch points at
w ¼ 1/a and w ¼ 1/b. Then, we appropriately choose R ) 1 (or 1/R ( 1) so that C
0
does not cut across the branch cut. To meet this condition, we have two choices.
(i) The branch cut connects w ¼ 1/a and w ¼ 1/b. (ii) Another choice of the branch
cut is that it connects w ¼ 1/a and w ¼ À 1 along with w ¼ 1/b and w ¼ 1 (see
Fig. 6.31). If we have 0 < a < b, we can choose a branch cut as that depicted in
Fig. 6.31a corresponding to Fig. 6.28a. In case a < 0 < b, the branch cut can be that
shown in Fig. 6.31b corresponding to Fig. 6.28b. (When a < b < 0, the branch cut
geometry may be that similar to Fig. 6.31a.)
Consequently, in both the above cases of (i) and (ii) there is only a simple pole at
w ¼ 0 without any other singularities inside the contour C
0 . Therefore, according to
(6.148) we have a contour integration described by
I
C
0
1
w
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1 À aw
ð
Þ1 À bw
ð
Þ
p
dw ¼ À2πi Res
1
w
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1 À aw
ð
Þ1 À bw
ð
Þ
p
w¼0
¼ À2πi Á
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi
1 À aw
ð
Þ1 À bw
ð
Þ
p
w¼0
¼ À2πi,
ð6:266Þ
where the above minus sign is due to the clockwise rotation of the contour integration. Hence, from (6.265) we get
I C ¼
I
C
1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
z À a
ð
Þ z À b
ð
Þ
p
dz ¼ 2πi:
ð6:267Þ
Equating (6.262) and (6.267), we finally obtain the answer described by
0
1/
0 < <
(a)
0
1/
< 0 <
(b)
Fig. 6.31 Branch cuts (shown with doubled broken lines) and contour in the w-plane for the
integration of
1
w
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1Àaw
ð
Þ1Àbw ð
Þ
p
. (a) Case of 0 < a < b. (b) Case of a < 0 < b. Notice that in both cases
the contour C
0 is traced clockwise (see text)
262
6 Theory of Analytic Functions