A contour integral from class - BU
A contour integral from class
Problem: Prove that
cos 2x
dx =
2
e- 2
cos
2.
0 x4 + 1
8
Solution: Since the integrand is even, we have
cos 2x
1 cos 2x
1 Re(ei2x)
1
ei2x
dx =
dx =
dx = Re
dx , (1)
0 x4 + 1
2 - x4 + 1
2 - x4 + 1
2
- x4 + 1
so we'll compute the last integral.
The
singular
points
of
1 x4+1
is
the
set
{ei(/4+2i/4k)
:
k
=
0, 1, 2, 3},
so
the
only
singular
points above the x-axis are z0 = ei/4, z1 = e3i/4. In the usual notation, we have
R ei2x
ei2z
ei2z
ei2z
-R
x4
+
dx 1
+
dz = 2i CR z4 + 1
Resz=z0 z4 + 1 + Resz=z1 z4 + 1
.
(2)
We compute the residues by ?76, Theorem 2, with p(z) = ei2z, q(z) = z4 + 1. We have
ei2z
ei2z
ei2ei/4
Resz=z0 z4 + 1
=
4z3
=
z=z0
, 4e3i/4
(provided q (z0) = 0, which is certainly the case). Thus
ei2z Resz=z0 z4 + 1
=
e2i( 2/2+i 2/2) 4e3i/4
=
e- 2e 2i 4e3i/4
=
e- 4
2
ei( 2-3/4).
(3)
Similarly,
ei2z Resz=z1 z4 + 1
=
e- 4
2
ei(- 2-/4).
(4)
By (3) and (4), the RHS of (2) equals
ei2z
ei2z
Re 2i Resz=z0 z4 + 1 + Resz=z1 z4 + 1
e- 2
= Re 2i
cos( 2 - 3/4) + i sin( 2 - 3/4)
4
+ cos(- 2 - /4) + i sin(- 2 - /4)
e- 2
= -2
(sin 2)(- 2/2) + (cos 2)(- 2/2)
4
+(sin(- 2))(- 2/2) + (cos(- 2))(- 2/2)
=
2
e- 2
cos
2.
(5)
4
Here we used the trig addition formula sin( + ) = cos sin + sin cos . Combining (1), (2), (5), we have
cos 2x
1
dx = Re
ei2x
dx =
2
e- 2
cos
2,
(6)
0 x4 + 1
2
- x4 + 1
8
provided
ei2z
lim
dz = 0.
(7)
R CR z4 + 1
We've done this type of estimate in class. We have
ei2z dz =
CR z4 + 1
e2iRei ? Rieid 0 R4ei4 + 1
e2iRei
? |Riei|d
0 R4ei4 + 1
|e2iR(cos +i sin )|
=
? R d
0 |R4ei4 + 1|
|e2iR cos | ? |e-2R sin |
=
? R d.
0
|R4ei4 + 1|
Since sin [0, 1] for [0, ], e-2R sin [e-2R, 1]. Also, |e2iR cos | = 1, so
ei2z
R
dz
d.
CR z4 + 1
0 |R4ei4 + 1|
Finally,
|R4ei4
+ 1|
|
|R4ei4| - |1|
|
=
R4 - 1
>
R4
-
R4
=
R4 ,
22
for
R4 2
>
1,
i.e.
as
soon
as
R
>
4 2.
Thus
ei2z
R
2
0 lim
dz lim
d = = 0,
R CR z4 + 1
R 0 R4/2
R3
so this limit is zero.
................
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