The Riemann Integral

48 1. The Riemann Integral
Principal value integrals arise frequently in complex analysis, harmonic analysis,
and a variety of applications.
Example 1.83. Consider the exponential integral function Ei given in Example
1.60,
∫ x
e t
Ei(x) =
−∞ t dt.
Ifx < 0, the integrandis continuous for−∞ < t ≤ x, and the integralis interpreted
as an improper integral,
∫ x
e t ∫ x
e t
dt = lim
t r→∞
−r t dt.
∞
This improper integral converges absolutely by comparison with e t , since
e t
∣ t ∣ ≤ et for −∞ < t ≤ −1,
and
∫ −1
−∞
e t dt = lim
r→∞
∫ −1
−r
e t dt = 1 e .
If x > 0, then the integrand has a non-integrable singularity at t = 0, and we
interpret it as a principal value integral. We write
∫ x
−∞
e t
t dt = ∫ −1
−∞
e t
t dt+ ∫ x
−1
e t
t dt.
The first integral is interpreted as an improper integral as before. The second
integral is interpreted as a principal value integral
∫ x
e t (∫ −ǫ
e t ∫ x
p.v. dt = lim
−1 t ǫ→0 + −1 t dt+ e t )
ǫ t dt .
This principal value integral converges, since
p.v.
∫ x
−1
e t
t dt = ∫ x
−1
e t −1
t
dt+p.v.
∫ x
−1
∫
1 x
t dt =
−1
e t −1
t
dt+lnx.
The first integral makes sense as a Riemann integral since the integrand has a
removable singularity at t = 0, with
( e t )
−1
lim = 1,
t→0 t
so it extends to a continuous function on [−1,x].
Finally, if x = 0, then the integrand is unbounded at the left endpoint t =
0. The corresponding improper or principal value integral diverges, and Ei(0) is
undefined.
Example 1.84. Let f : R → R and assume, for simplicity, that f has compact
support, meaning that f = 0 outside a compact interval [−r,r]. If f is integrable,
we define the Hilbert transform Hf : R → R of f by the principal value integral
Hf(x) = 1 π p.v. ∫ ∞
−∞
f(t)
x−t dt = 1 (∫ x−ǫ
π lim
ǫ→0 + −∞
∫
f(t) ∞
x−t dt+
x+ǫ
f(t)
x−t dt )
.