This belongs to a larger set of ideas concerning improper integrals. I will just give enough of an
introduction to this to present the very important gamma function. 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, then there is no hope for it at all. However, one can consider
limits of Riemann integrals. The following definition is sufficient to deal with the gamma function in the
generality needed in this book.
Definition A.1.1We say that f defined on [0,∞) is improper Riemann integrable if it is Riemannintegrable on
[δ,R]
for each R > 1 > δ > 0 and the following limits exist.
∫ ∫ ∫
∞ 1 R
0 f (t)dt ≡ lδ→im0+ δ f (t)dt+ Rli→m∞ 1 f (t)dt
The gamma functionis defined by
∫ ∞
Γ (α) ≡ e−ttα−1dt
0
whenever α > 0.
Lemma A.1.2The 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 1
tα− 1dt ≤--
δ 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 exponential growth exceeds polynomial growth.
Thus
∫ R ∫ R
e−ttα−1dt ≤ Ce −(t∕2)dt ≤ 2Ce(−1∕2) − 2Ce(−R∕2) ≤ 2Ce (−1∕2)
1 1
Thus these integrals also converge as R →∞. 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
lδ→i0m+ δ f (t)dt
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 A.1.4A function ϕ has exponential growth on [0,∞) if there are positive constantsλ,C such that
|ϕ (t)|
≤ Ce^{λt}for all t.
Theorem A.1.5Let f
(s)
= ∫_{0}^{∞}e^{−st}ϕ
(t)
dt where t → ϕ
(t)
e^{−st}is improper Riemann integrablefor all s large enough and ϕ has exponential growth. Then for s large enough, f^{(k)
}
(s)
exists andequals∫_{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,
(k) (k) ∫ ∞ −(s+h)t −st
f---(s+-h)−-f---(s) = (− t)k e------−-e--ϕ (t)dt
h 0 h