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 or if it is on an infinite interval, 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 9.6.1We say that f defined on [0,∞) is improperRiemann integrable ifit is Riemann integrable on
[δ,R ]
for each R > 1 > δ > 0 and the following limitsexist.
∫ ∞ ∫ 1 ∫ R
f (t)dt ≡ lim f (t)dt+ lim f (t)dt
0 δ→0+ δ R→ ∞ 1
The gamma functionis defined by
∫ ∞
Γ (α) ≡ e−ttα−1dt
0
whenever α > 0.
Lemma 9.6.2The 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 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→im0+ f (t)dt
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 −tα− 1
= lδim→0 e δ − e δ + α δ e t dt = αΓ (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 9.6.4A function ϕ hasexponential growth on [0,∞) if there arepositive constants λ,C such that
|ϕ(t)|
≤ Ce^{λt}for all t ≥ 0.
Theorem 9.6.5Let f
(s)
= ∫_{0}^{∞}e^{−st}ϕ
(t)
dt where t → ϕ
(t)
e^{−st}is improperRiemann integrable for all s large enough and ϕ has exponential growth. Then for slarge 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) ∞ k e−-(s+h)t −-e−-st
h = 0 (− t) h ϕ (t)dt
∫ ∞ k ( e−ht − 1) ∫ ∞ k ( )
= (− t) e− st ---h--- ϕ(t)dt = (− t) e−st (− t)eθ(h,t) ϕ(t)dt
0 0
where θ
(h,t)
is between −ht and 0, this by the mean value theorem. Thus by mean value
theorem again,