A contour integral from class
√
√
π 2 −√2
cos 2x
dx
=
e
Problem: Prove that
2.
cos
x4 + 1
8
0
Solution: Since the integrand is even, we have
Z ∞ i2x
Z ∞
Z
Z
cos 2x
1 ∞ cos 2x
1 ∞ Re(ei2x )
1
e
dx =
dx =
dx = Re
dx ,
4
x4 + 1
2 −∞ x4 + 1
2 −∞ x4 + 1
2
0
−∞ x + 1
Z
∞
(1)
so we’ll compute the last integral.
The singular points of x41+1 is the set {ei(π/4+2πi/4k) : k = 0, 1, 2, 3}, so the only singular
points above the x-axis are z0 = eiπ/4 , z1 = e3πi/4 . In the usual notation, we have
Z R i2x
Z
e
ei2z
ei2z
ei2z
dx +
dz = 2πi Resz=z0 4
+ Resz=z1 4
.
(2)
4
4
z +1
z +1
−R x + 1
CR z + 1
We compute the residues by §76, Theorem 2, with p(z) = ei2z , q(z) = z 4 + 1. We have
iπ/4
ei2z ei2e
ei2z
= 3
= 3iπ/4 ,
Resz=z0 4
z +1
4z z=z0 4e
(provided q 0 (z0 ) 6= 0, which is certainly the case). Thus
√
√
√
√
√
ei2z
e2i( 2/2+i 2/2)
e− 2 e 2i
e− 2 i(√2−3π/4)
Resz=z0 4
=
=
=
e
.
z +1
4e3iπ/4
4e3iπ/4
4
Similarly,
(3)
√
e− 2 i(−√2−π/4)
ei2z
=
e
.
(4)
Resz=z1 4
z +1
4
By (3) and (4), the RHS of (2) equals
"
√
√
√
ei2z
ei2z
e− 2 Re 2πi Resz=z0 4
+ Resz=z1 4
= Re 2πi
cos( 2 − 3π/4) + i sin( 2 − 3π/4)
z +1
z +1
4
i
√
√
+ cos(− 2 − π/4) + i sin(− 2 − π/4)
√
√
√
√
√
e− 2 h
= −2π
(sin 2)(− 2/2) + (cos 2)(− 2/2)
4
i
√
√
√
√
+(sin(− 2))(− 2/2) + (cos(− 2))(− 2/2)
√
√
π 2 − √2
=
e
cos 2.
(5)
4
Here we used the trig addition formula sin(θ + ψ) = cos θ sin ψ + sin θ cos ψ. Combining (1),
(2), (5), we have
√
Z ∞ i2x
Z ∞
√
cos 2x
1
e
π 2 −√2
dx
=
Re
dx
=
e
cos
2,
(6)
4
x4 + 1
2
8
0
−∞ x + 1
provided
Z
lim
R→∞
CR
ei2z
dz = 0.
z4 + 1
(7)
We’ve done this type of estimate in class. We have
Z
Z
π e2iReiθ
i2z
e
iθ
= dz
·
Rie
dθ
4
4
i4θ
0 R e +1
CR z + 1
Z π 2iReiθ e
≤
4 i4θ
· |Rieiθ |dθ
R
e
+
1
0
Z π 2iR(cos θ+i sin θ)
|e
|
· R dθ
=
|R4 ei4θ + 1|
0
Z π 2iR cos θ
|e
| · |e−2R sin θ |
=
· R dθ.
|R4 ei4θ + 1|
0
Since sin θ ∈ [0, 1] for θ ∈ [0, π], e−2R sin θ ∈ [e−2R , 1]. Also, |e2iR cos θ | = 1, so
Z
Z π
ei2z
R
≤
dz
dθ.
4
4
i4θ
z +1
|R e + 1|
CR
0
Finally,
R4
R4
|R e + 1| ≥ | |R e | − |1| | = R − 1 > R −
=
,
2
2
√
> 1, i.e. as soon as R > 4 2. Thus
Z
Z π
ei2z
R
2π
0 ≤ lim dz ≤ lim
dθ = 3 = 0,
4
4
R→∞
R→∞ 0 R /2
R
CR z + 1
4 i4θ
for
R4
2
so this limit is zero.
4 i4θ
4
4