The integral can also be calculated using a lower semicircle including the simple
pole located at z ¼ e
Àiπ=3
¼
1À
ffiffi
3
p
i
2
. That gives the same answer as (6.181). The
calculations are left for readers as an exercise.
Example 6.7 [9, 10] It will be interesting to compare the result of Example 6.6 with
that of the following real definite integral:
I ¼
Z 1
0
dx
1 þ x 3 :
ð6:183Þ
To evaluate (6.183), it is convenient to use a sector of circle for contour (see
Fig. 6.21a). In this case, the contour integral I C estimated along a sector is described
by
I C ¼
Z R
0
dz
1 þ z 3 þ
Z
Γ R
dz
1 þ z 3 þ
Z
L
dz
1 þ z 3 :
ð6:184Þ
In the third integral of (6.184), we take argz ¼ θ (constant) so that with z ¼ re
iθ
we may have dz ¼ dre
iθ . Then, that integral is given by
Z
L
dz
1 þ z 3 ¼
Z
L
dr
1 þ r 3 e 3iθ :
ð6:185Þ
Setting 3iθ ¼ 2iπ, namely θ ¼ 2π/3, we get
Z
L
dz
1 þ r 3 e 3iθ ¼ e
2πi=3
Z 0
R
dr
1 þ r 3 ¼ Àe
2πi=3
Z R
0
dr
1 þ r 3 :
Thus, taking R ! 1 in (6.184) we have
i
−1
0
Γ
/
/
(a)
i
−1 0
−
Γ
Γ
/
/
(b)
Fig. 6.21 Sector of circle for contour integration of
1
1þz 3 that appears in Example 6.7. (a) The
integration range is [0, 1). (b) The integration range is (À1, 0]
6.8 Examples of Real Definite Integrals
239