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Calculate the integral

Z
0

cos(2x)
dx
x2 + 2

Solution: First note that the integrand is even, hence
Z

0

Integrate f (z) =

e2iz
z 2 +2

1
cos(2x)
dx =
x2 + 2
2

−∞

cos(2x)
1
dx =
x2 + 2
2

Z

−∞


&lt; e2ix
dx
x2 + 2

along the curve ΓR

The function f has the single pole

res{f,

Z

2i inside ΓR (given R &gt;

2), and

√ 
e−2 2
e2iz 
e2iz

2i} = lim
z

2i
=
lim
=

2
2i
2 2i
z→ 2i z + 2
z→ 2i z +

hence

πe−2
f dz = √
2
ΓR

Z

2



By Jordan, the integral along ΓR \[−R, R] that is, the integral along Reiθ θ ∈ (0, π) equals
0, and since e2iz = cos(2z) + i sin(2z) where sin(2z) is odd, we see that

Z
[−R,R]

cos(2z)
πe−2

dz
=
z2 + 2
2

2

Z
−i
[−R,R]

|
That is,
Z
0

=0

πe−2 2
cos(2x)

dx
=
x2 + 2
2 2

1

sin(2z)
dz
z2 + 2
{z
}

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