11.3. FOURIER TRANSFORM 269
With this convention, there is a very important improper integral involving sin(x)/x.You can show with a little estimating that x→ sin(x)/x is not in L1 (0,∞) . Nevertheless, alot can be said about improper integrals involving this function.
Theorem 11.3.3 The following hold
1.∫
∞
0sinu
u du = π
2
2. limr→∞
∫∞
δ
sin(ru)u du = 0 whenever δ > 0.
3. If f ∈ L1 (R) , then limr→∞
∫R sin(ru) f (u)du = 0. This is called the Riemann Leb-
esgue lemma.
Proof: You know 1u =
∫∞
0 e−utdt. Therefore, using Fubini’s theorem,∫ r
0
sinuu
du =∫ r
0sin(u)
∫∞
0e−utdtdu =
∫∞
0
∫ r
0e−ut sin(u)dudt
Now you integrate that inside integral by parts to obtain∫∞
0
(1
t2 +1− e−tr cos(r)+ t sin(r)
1+ t2
)dt.
This integrand converges to 1t2+1 as r→∞ for each t > 0. I would like to use the dominated
convergence theorem. That second term is of the form
e−tr
√1+ t2 cos(r−φ (t,r))
1+ t2 ≤ 1
(1+ t2)1/2 e−tr.
For r > 1, this is no larger than 1
(1+t2)1/2 e−t which is obviously in L1 and so one can apply
the dominated convergence theorem and conclude that
limr→∞
∫ r
0
sinuu
du =∫
∞
0
11+ t2 dt =
π
2.
This shows part 1.Now consider
∫∞
δ
sin(ru)u du. It equals
∫∞
0sin(ru)
u du−∫
δ
0sin(ru)
u du which can be seen fromthe definition of what the improper integral means. Let ru = t so rdu = dt and∫
∞
δ
sin(ru)u
du =∫
∞
0
sin(t)t
r1r
dt−∫ rδ
0
sin(t)t
dt =π
2−∫ rδ
0
sin(t)t
dt
so limr→∞
∫∞
δ
sin(ru)u du = limr→∞
(π
2 −∫ rδ
0sin(t)
t dt)= 0 from the first part.
Now consider the Riemann Lebesgue lemma. Let h∈C∞c (R) such that
∫R | f −h|dmp <
ε. Then∣∣∣∣∫R sin(ru) f (u)du∣∣∣∣ ≤ ∣∣∣∣∫R sin(ru)( f (u)−h(u))du
∣∣∣∣+ ∣∣∣∣∫R sin(ru)h(u)∣∣∣∣
≤ ε +
∣∣∣∣∫R sin(ru)h(u)∣∣∣∣