Chapter 11
Fundamental Transforms11.1 Gamma Function
With the Lebesgue integral, it becomes easy to consider the Gamma function and the the-ory of Laplace transforms. I will use the standard notation for the integral used in calculus,but remember that all integrals will be Lebesgue integrals taken with respect to one di-mensional Lebesgue measure. First is a very important function defined in terms of anintegral. Problem 13 on Page 183 shows that in the case of a continuous function, theRiemann integral and the Lebesgue integral are exactly the same. Thus all the standard cal-culus manipulations are valid for the Lebesgue integral provided the functions integratedare continuous. This also implies immediately that the two integrals coincide whenever thefunction is piecewise continuous on a finite interval. Recall that the value of the Riemannintegral does not depend on the value of the function at single points and the same is trueof the Lebesgue integral because single points have zero measure.
Definition 11.1.1 The gamma function α → Γ(α) is defined as
Γ(α)≡∫
∞
0e−ttα−1 dt
whenever α > 0.
Lemma 11.1.2 The integral is finite for each α > 0.
Proof: By the monotone convergence theorem, for n ∈ N
Γ(α) = limn→∞
∫ n
1/ne−ttα−1 ≤ lim sup
n→∞
(∫ 1
1/ntα−1dt +
∫ n
1Ce−t/2
)≤ 1
α+ lim
n→∞
(−2Ce−
12 n +2Ce−
12
)< ∞
The explanation for the constant is as follows. Letting m be a positive integer larger thanα−1, for t ≥ 1,e−ttα−1 < e−ttm ≤Ce−t/2 for suitable C. ■
Proposition 11.1.3 For n a positive integer, n! = Γ(n+1). In general,the followingfundamental identity holds. Γ(1) = 1,Γ(α +1) = αΓ(α)
Proof: First of all, Γ(1) = limδ→0∫
δ−1
δe−tdt = limδ→0
(e−δ − e−(δ
−1))= 1. Next,
for α > 0,
Γ(α +1) = limδ→0
∫δ−1
δ
e−ttα dt = limδ→0
[−e−ttα |δ
−1
δ+α
∫δ−1
δ
e−ttα−1dt
]
= limδ→0
(e−δ
δα − 1
δα e1/δ
+α
∫δ−1
δ
e−ttα−1dt
)= αΓ(α)
Note that limδ→0+ ln(
1δ
α e1/δ
)= limδ→0+
(α lnδ + 1
δ
)=−∞ so limδ→0+
1δ
α e1/δ= 0. 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 justshown, Γ(n+2) = Γ(n+1)(n+1) = n!(n+1) = (n+1)! and so this proves the proposi-tion. ■
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