10.1. THE DIRICHLET INTEGRAL 231
You just note∫ b
a f (t)dt ∈ C makes sense if f ∈ L1 and then such a θ exists. Thensome short computations with intervals [an,bn] ⊆ (a,b) show that the left side is equalto∫ b
a θ f (t)dt. Since this is real,∣∣∣∣∫ b
af (t)dt
∣∣∣∣= ∫ b
aθ f (t)dt =
∫ b
aRe(θ f (t))dt ≤
∫ b
a| f (t)|dt.
10.1 The Dirichlet IntegralThere is a very important improper integral involving sin(x)/x. You can show with a littleestimating that x→ sin(x)/x is not in L1 (0,∞) . However, one can show that this functionis improper Riemann integrable. The following lemma is on a very important improperintegral known as the Dirichlet integral after Dirichlet who first used it. See Problem 34on Page 220 or else Problem 5 on 245 for more hints on how to show this. Here the actualvalue of this integral is obtained along with its existence. First note that for x > 0∫
∞
re−txdt = lim
R→∞
∫ R
re−txdt = lim
R→∞
−e−tx
x|Rr = lim
R→∞
(−e−tR
x+
e−rx
x
)=
e−rx
x
Lemma 10.1.1 π
2 =∫
∞
0sin(x)
x dx. Here x→ sinxx is improper Riemann integrable.
Proof: By limx→0sinx
x = 1, we can assume sinxx is continuous on [0,1] . Now let
[an,bn]⊆ (0,∞) . Then
∫ bn
an
sinxx
dx =∫ bn
an
sin(x)
=1/x︷ ︸︸ ︷∫∞
0e−txdtdx =
∫ bn
0sin(x)
=1/x︷ ︸︸ ︷∫∞
0e−txdtdx+ e(n)
where limn→∞ e(n)≡ limn→∞
(−∫ an
0sinx
x dx)= 0. Thus∫ bn
an
sinxx
dx = e(n)+∫ bn
0sin(x)
∫ bn
0e−txdtdx+
∫ bn
0sin(x)
∫∞
bn
e−txdtdx
= e(n)+∫ bn
0sin(x)
∫ bn
0e−txdtdx+E (n) (10.1)
where |E (n)| ≤∫ bn
0 |sin(x)| e−bnx
x dx. Now
|E (n)| ≤∫ bn
0
<1|sinx|
xe−bnxdx <
1bn
.
Now interchange the order of integration in 10.1 using the Fubini theorem presented earlier,Theorem 9.9.4. ∫ bn
an
sinxx
dx = e(n)+E (n)+∫ bn
0
∫ bn
0e−tx sin(x)dxdt
Some tedious integration by parts on the inside integral on the right gives∫ bn
an
sinxx
dx = e(n)+E (n)
+∫ bn
0
[1
t2 +1
((1− (cosbn)e−bnt + t (sinbn)e−bnt
))]dt