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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