lim
R!1
I e C
¼
Z À1
1
dx
1 þ x 2 þ lim
R!1
Z
e
Γ R
dz
1 þ z 2 ¼
Z À1
1
dx
1 þ x 2
¼ 2πi Res f Ài
ð Þ ¼ Àπ:
ð6:157Þ
Note that in the above equation the contour integral is taken in the counterclockwise direction and, hence, that the real definite integral has been taken from 1 to
À1. Notice also that
Res f Ài
ð Þ ¼ f z
ð Þ z þ i
ð
Þj z¼Ài ¼ À
1
2i
,
because f (z) has a simple pole at z ¼ À i in the lower semicircle (Fig. 6.18) for which
the residue has been taken. From (6.157), we get the same result as before such that
Z 1
À1
dx
1 þ x 2 ¼ π:
Alternatively, we can use the method of partial fraction decomposition. In that
case, as the integrand we have
1
1 þ z 2 ¼
1
2i
1
z À i
À
1
z þ i
:
ð6:158Þ
The residue of
1
1þz 2 at z ¼ i is
1
2i [i.e., the coefficient of
1
zÀi in (6.158)], giving the
same result as (6.153).
Example 6.5 Evaluate the following real definite integral:
I ¼
Z 1
À1
dx
1 þ x 2
ð
Þ
3
:
ð6:159Þ
As in the case of Example 6.4, we evaluate a contour integration described by
I C ¼
I
C
1
1 þ z 2
ð
Þ
3
dz ¼
Z R
ÀR
dz
1 þ z 2
ð
Þ
3
þ
Z
Γ R
dz
1 þ z 2
ð
Þ
3
,
ð6:160Þ
where C stands for the closed curve comprising the interval [ÀR, R] and the upper
semicircle Γ R (see Fig. 6.18).
The function f z
ð Þ
1
1þz 2
ð
Þ
3 has two isolated poles at z ¼ Æ i. In the present case,
however, the poles are of order of 3. Hence, we use (6.147) to estimate the residue.
The inside of the upper semicircle contains the pole of order 3 at z ¼ i. Therefore, we
have
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6 Theory of Analytic Functions
R!1
I e C
¼
Z À1
1
dx
1 þ x 2 þ lim
R!1
Z
e
Γ R
dz
1 þ z 2 ¼
Z À1
1
dx
1 þ x 2
¼ 2πi Res f Ài
ð Þ ¼ Àπ:
ð6:157Þ
Note that in the above equation the contour integral is taken in the counterclockwise direction and, hence, that the real definite integral has been taken from 1 to
À1. Notice also that
Res f Ài
ð Þ ¼ f z
ð Þ z þ i
ð
Þj z¼Ài ¼ À
1
2i
,
because f (z) has a simple pole at z ¼ À i in the lower semicircle (Fig. 6.18) for which
the residue has been taken. From (6.157), we get the same result as before such that
Z 1
À1
dx
1 þ x 2 ¼ π:
Alternatively, we can use the method of partial fraction decomposition. In that
case, as the integrand we have
1
1 þ z 2 ¼
1
2i
1
z À i
À
1
z þ i
:
ð6:158Þ
The residue of
1
1þz 2 at z ¼ i is
1
2i [i.e., the coefficient of
1
zÀi in (6.158)], giving the
same result as (6.153).
Example 6.5 Evaluate the following real definite integral:
I ¼
Z 1
À1
dx
1 þ x 2
ð
Þ
3
:
ð6:159Þ
As in the case of Example 6.4, we evaluate a contour integration described by
I C ¼
I
C
1
1 þ z 2
ð
Þ
3
dz ¼
Z R
ÀR
dz
1 þ z 2
ð
Þ
3
þ
Z
Γ R
dz
1 þ z 2
ð
Þ
3
,
ð6:160Þ
where C stands for the closed curve comprising the interval [ÀR, R] and the upper
semicircle Γ R (see Fig. 6.18).
The function f z
ð Þ
1
1þz 2
ð
Þ
3 has two isolated poles at z ¼ Æ i. In the present case,
however, the poles are of order of 3. Hence, we use (6.147) to estimate the residue.
The inside of the upper semicircle contains the pole of order 3 at z ¼ i. Therefore, we
have
232
6 Theory of Analytic Functions