hence, can be expanded in the Taylor’s series around any complex number.
Expanding it around the origin we have
e
iz
¼
X 1
n¼0
iz
ð Þ
n
n!
¼ 1 þ
X 1
n¼1
iz
ð Þ
n
n!
:
ð6:195Þ
As already mentioned in (6.173) through (6.178), among terms of RHS in (6.195)
only the first term, i.e., 1, contributes to the integral of the second term of RHS in
(6.194). Thus, as in (6.178) we get
Z
Γ 0
e
iz
z
dz ¼ À1 Â iπ ¼ Àiπ:
ð6:196Þ
Notice that in (6.196) the argument is decreasing from π ! 0 during the contour
integration. Within the contour C, there is no singular point, and so I C ¼ 0 due to
Theorem 6.10 (Cauchy’s integral theorem). Thus, from (6.194) we obtain
P
Z 1
À1
e
iz
z
dz ¼ À
Z
Γ 0
e
iz
z
dz ¼ iπ:
ð6:197Þ
Exchanging the variable z ! À z in (6.197), we have
P À
Z À1
1
e
Àiz
Àz
dz
!
¼ P
Z 1
À1
e
Àiz
Àz
dz
!
¼ iπ:
ð6:198Þ
Summing both sides of (6.197) and (6.198), we get
P
Z 1
À1
e
iz
z
dz þ P
Z 1
À1
e
Àiz
Àz
dz ¼ 2iπ:
ð6:199Þ
Rewriting (6.199), we have
0
−
Γ
Γ
Fig. 6.22 Contour for the
integration of
sin z
z that
appears in Example 6.8
6.8 Examples of Real Definite Integrals
243