This belongs to a larger set of ideas concerning improper integrals, but the main reason
for these ideas are important examples like the Gamma function or Laplace
transforms. General theory is much better understood in the context of the
Lebesgue integral. Therefore, the presentation is centered on these examples. The
Riemann integral only is defined for bounded functions which are defined on a
bounded interval. If this is not the case, then the integral has not been defined. Of
course, just because the function is bounded does not mean the integral exists as
mentioned above, but if it is not bounded or if it is defined on an infinite interval,
then no definition has been given. However, one can consider limits of Riemann
integrals. The following definition pertains to the Gamma function and Laplace
transforms.
Definition 9.2.1We say that f defined on [0,∞) is improperRiemannintegrable if it is Riemann integrable on
[δ,R ]
for each R > 1 > δ > 0 and the followinglimits exist.
∫ ∞ ∫ 1 ∫ R
0 f (t)dt ≡ lδ→i0m+ δ f (t)dt+ Rl→im∞ 1 f (t)dt
The gamma functionis defined by
∫
Γ (α) ≡ ∞e− ttα−1dt
0
whenever α > 0.
Lemma 9.2.2The limits in the above definition exists for each α > 0.
Proof: Note first that as δ → 0+, the Riemann integrals
∫
1 e−ttα−1dt
δ
increase. Thus lim_{δ→0+}∫_{δ}^{1}e^{−t}t^{α−1}dt either is +∞ or it will converge to the least upper
bound thanks to completeness of ℝ. However,
∫
1 α−1 1-
δ t dt ≤ α
so the limit of these integrals exists. Also e^{−t}t^{α−1}≤ Ce^{−}
(t∕2)
for suitable C if t > 1. This
is obvious if α − 1 < 0 and in the other case it is also clear because
e−ttα− 1 −t∕2m
0 < -e−-(t∕2)-≤ e t , where m is an integer larger than α− 1
Now apply L’Hopital’s rule to conclude that the limit of this expression is 0 as t →∞.
Thus the quotient
e−ttα−1
e−(t∕2)
is less than some constant C.
∫ ∫
R − tα−1 R −(t∕2) (−1∕2) (−R ∕2) (−1∕2)
1 e t dt ≤ 1 Ce dt ≤ 2Ce − 2Ce ≤ 2Ce
Thus these integrals also converge as R →∞ because they are increasing in R and
bounded above. Hence they converge to
{ }
∫ R −tα− 1
sup 1 e t dt : R > 1
It follows that Γ
(α)
makes sense. ■
This gamma function has some fundamental properties described in the following
proposition. In case the improper integral exists, we can obviously compute it in the
form
∫ 1∕δ
lim f (t)dt
δ→0+ δ
which is used in what follows. Thus also the usual algebraic properties of the Riemann
integral are inherited by the improper integral.
( )
−δ α −(δ−1)− α ∫ δ−1 −t α−1
= lδim→0 e δ − e δ + α δ e t dt = α Γ (α)
Now it is defined that 0! = 1 and so Γ
(1)
= 0!. Suppose that Γ
(n+ 1)
= n!, what of
Γ
(n + 2)
? Is it
(n + 1)
!? if so, then by induction, the proposition is established. From
what was just shown,
Γ (n +2) = Γ (n+ 1)(n + 1) = n!(n+ 1) = (n +1)!
and so this proves the proposition. ■
The properties of the gamma function also allow for a fairly easy proof about
differentiating under the integral in a Laplace transform. First is a definition.
Definition 9.2.4A function ϕ hasexponential growth on [0,∞) if thereare positive constants λ,C such that
|ϕ(t)|
≤ Ce^{λt}for all t ≥ 0.
Theorem 9.2.5Let f
(s)
= ∫_{0}^{∞}e^{−st}ϕ
(t)
dt where t → ϕ
(t)
e^{−st}isimproper Riemann integrable for all s large enough and ϕ has exponential growth.Then for s large enough, f^{(k)
}
(s)
exists and equals∫_{0}^{∞}
(− t)
^{k}e^{−st}ϕ
(t)
dt.
Proof:Suppose true for some k ≥ 0. By definition it is so for k = 0. Then always
assuming s > λ,
|h|
< s − λ, where
|ϕ(t)|
≤ Ce^{λt},λ ≥ 0,
f (k)(s+ h)− f(k)(s) ∫ ∞ e−(s+h)t − e−st
------------------ = (− t)k -------------ϕ(t) dt
h 0 h