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Tentauppgift, komplexanalys
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
∞
−∞
< e2ix
dx
x2 + 2
along the curve ΓR
The function f has the single pole
res{f,
Z
√
√
2i inside ΓR (given R >
√
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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