SOLUTIONS TO SELECTED QUESTIONS IN HOMEWORK 14
MATH 241
15.3.4
Proof.
A(α) =
Z
π
0
1
sin x cos αxdx =
2
π
Z
0
1 cos(1 + α)x cos(1 − α)x π
sin(1 + α)x + sin(1 − α)xdx = [−
−
]|0
2
1+α
1−α
1 1 − cos(1 + α)π 1 − cos(1 − α)π
+
]
= [
2
1+α
1−α
B(α) =
π
Z
0
1
sin x sin αxdx =
2
π
Z
0
1 sin(1 − α)x sin(1 + α)x π
−
]|0
cos(1 − α)x − cos(1 + α)xdx = [
2
1−α
1+α
1 sin(1 − α)π sin(1 + α)π
= [
−
]
2
1−α
1+α
Therefore
Z
1 ∞ 1 1 − cos(1 + α)π 1 − cos(1 − α)π
1 sin(1 − α)π sin(1 + α)π
f (x) =
[
+
] cos αx + [
−
] sin αxdα
π 0 2
1+α
1−α
2
1−α
1+α
15.3.17
Proof. For an f defined on (0, ∞), then the integral in the problem is the coefficient of Fourier cosine integral,
so A(α) = e−α , therefore the Fourier integral becomes
Z
Z
Z
2 ∞ −α
2 −∞ α
2 0 α
f (x) =
e cos αxdα =
e cos αx(−dα) =
e cos αxdα
π 0
π 0
π −∞
Then by the formula
Z
et cos λtdt =
et (λ sin λt + cos λt)
λ2 + 1
In this situation λ = x, t = α, so we get
2 eα (x sin αx + cos αx) 0
|−∞
π
x2 + 1
Note that here we are evaluating with respect to α, not x! When α → −∞, eα → 0 but (x sin αx + cos αx) is
bounded (again, notice that x is not changing! That’s why it is fixed, otherwise if x is allowed to go to infinity,
this may no longer be true), so the evaluation at −∞ is zero. Therefore the answer is f (x) =
1
2 1
π x2 +1 .
2
MATH 241
Fall 10, #11
Proof.
F { f }(α) =
Z
1
2eiαx dx =
−1
g00
2 iαx 1
2(eiα − e−iα ) 4i sin α 4 sin α
e |−1 =
=
=
iα
iα
iα
α
= f is routine to check. Therefore
F { f }(α) = F {g00 }(α) = (−iα)2 F {g}(α)
α
Therefore F {g}(α) = − α12 F { f }(α) = − 4 sin
.
α3