Lecture Notes in Differential Equations - Bruce E. Shapiro

L [ t x−1] [ ] 1 d
= L
x dt tx = 1 [ ] d
x L dt tx
339
(32.51)
Let f(t) = t x . Then
[ ] d
L
dt tx = L[f ′ (t)] = sF (s) − f(0) = sF (s) (32.52)
since f(0) = 0, where F (s) is given by the right hand side of (32.57)
Substituting (32.52) into (32.51) gives
L [ t x−1] = 1 [ ] d
x L dt tx = s Γ(x + 1)
x s x+1 (32.53)
But from (32.51) directly
Equating the last two expressions gives
L [ t x−1] = Γ(x)
s x (32.54)
Γ(x)
s x
= s Γ(x + 1)
x s x+1 (32.55)
which after cancellation gives us the fundamental property of the Gamma
function:
Γ(x + 1) = xΓ(x) (32.56)
Now observe from (32.57) that
From (32.56)
Γ(1) =
∫ ∞
0
u 1−1 e −u du =
∫ ∞
0
e −u du = 1 (32.57)
Γ(2) = 1 · Γ(1) = 1 · 1 = 1 = 1! (32.58)
Γ(3) = 2 · Γ(2) = 2 · 1 = 2 = 2! (32.59)
Γ(4) = 3 · Γ(3) = 3 · 2 = 6 = 3! (32.60)
Γ(5) = 4 · Γ(4) = 4 · 6 = 24 = 4! (32.61)
Γ(6) = 5 · Γ(5) = 5 · 4! = 5! (32.62)
. (32.63)
Γ(n) = (n − 1)! for n ∈ Z + (32.64)
hence for n an integer
L[t n ] = n!
s n+1 (32.65)