P
Z 1
À1
e
iz
À e
Àiz
ð
Þ
z
dz ¼ 2iP
Z 1
À1
sin z
z
dz ¼ 2iπ:
ð6:200Þ
That is, the answer is
P
Z 1
À1
sin z
z
dz ¼ P
Z 1
À1
sin x
x
dx ¼ I ¼ π:
ð6:201Þ
Notice that as mentioned in (6.137) z ¼ 0 is a removable singularity of
sin z
z , and so
the symbol P is superfluous in (6.201).
Example 6.9 Evaluate the following real definite integral:
I ¼
Z 1
À1
sin
2 x
x 2 dx:
ð6:202Þ
From the trigonometric theorem, we have
sin
2 x ¼
1
2
1 À cos 2x
ð
Þ :
ð6:203Þ
Then, the integrand of (6.202) with the variable changed is rewritten using (6.37)
as
sin
2 z
z 2 ¼
1
4
1 À e
2iz
z 2 þ
1 À e
À2iz
z 2
!
:
ð6:204Þ
Expanding e
2iz as before, we have
e
2iz
¼
X 1
n¼0
2iz
ð Þ
n
n!
¼ 1 þ 2iz þ
X 1
n¼2
2iz
ð Þ
n
n!
:
ð6:205Þ
Then, we get
1 À e
2iz
¼ À2iz À
X 1
n¼2
2iz
ð Þ
n
n!
,
1 À e
2iz
z 2 ¼ À
2i
z
À
X 1
n¼2
2i
ð Þ
n z
nÀ2
n!
:
ð6:206Þ
As in Example 6.8, we wish to use (6.206) to estimate the second term of the
following contour integral I C :
244
6 Theory of Analytic Functions
Z 1
À1
e
iz
À e
Àiz
ð
Þ
z
dz ¼ 2iP
Z 1
À1
sin z
z
dz ¼ 2iπ:
ð6:200Þ
That is, the answer is
P
Z 1
À1
sin z
z
dz ¼ P
Z 1
À1
sin x
x
dx ¼ I ¼ π:
ð6:201Þ
Notice that as mentioned in (6.137) z ¼ 0 is a removable singularity of
sin z
z , and so
the symbol P is superfluous in (6.201).
Example 6.9 Evaluate the following real definite integral:
I ¼
Z 1
À1
sin
2 x
x 2 dx:
ð6:202Þ
From the trigonometric theorem, we have
sin
2 x ¼
1
2
1 À cos 2x
ð
Þ :
ð6:203Þ
Then, the integrand of (6.202) with the variable changed is rewritten using (6.37)
as
sin
2 z
z 2 ¼
1
4
1 À e
2iz
z 2 þ
1 À e
À2iz
z 2
!
:
ð6:204Þ
Expanding e
2iz as before, we have
e
2iz
¼
X 1
n¼0
2iz
ð Þ
n
n!
¼ 1 þ 2iz þ
X 1
n¼2
2iz
ð Þ
n
n!
:
ð6:205Þ
Then, we get
1 À e
2iz
¼ À2iz À
X 1
n¼2
2iz
ð Þ
n
n!
,
1 À e
2iz
z 2 ¼ À
2i
z
À
X 1
n¼2
2i
ð Þ
n z
nÀ2
n!
:
ð6:206Þ
As in Example 6.8, we wish to use (6.206) to estimate the second term of the
following contour integral I C :
244
6 Theory of Analytic Functions