We will prove that the improper integral Γ(x) =
Z ∞ 0
e−ttx−1dt
exists for every x > 0. The function Γ(x) is called the Gamma function. Let us recall the comparison test for improper integrals.
Theorem 1.1. (Comparison Test for Improper Integral of Type I) Let f (x), g(x) be two continuous functions on [a, ∞) such that 0 ≤ f (x) ≤ g(x) for all x ≥ a. Then
(1) If Z ∞
a
g(x)dx is convergent, so is Z ∞
a
f (x)dx.
(2) If Z ∞
a
f (x)dx is divergent to infinity, so is Z ∞
a
g(x)dx.
Theorem 1.2. (Limit Comparison Test) Let f (x), g(x) be two nonnegative continuous functions on [a, ∞). Suppose that
x→∞lim f (x)
g(x) = L with L > 0.
Then both Z ∞
a
g(x)dx and Z ∞
a
f (x)dx are convergent or divergent.
Lemma 1.1. For every s > 0, the improper integral Z ∞
0
e−stdt converges.
Proof. Let us compute
Z N 0
e−stdt = e−sM− 1
−s . By definition,
Z ∞ 0
e−stdt = lim
N →∞
Z N 0
e−stdt = lim
N →∞
e−sM − 1
−s = 1
s.
The limit exists and hence the improper integral converges. Now let us study the case when x ≥ 1.
Lemma 1.2. Let n be a natural number. Then
t→∞lim tn−1
e12t = 0.
Proof. By L’ Hospital rule,
t→∞lim tn−1
e12t
= lim
t→∞
(n − 1)tn−2
1 2e12t
. Since tn−1 is a polynomial of degree n − 1, we know
dn
dtntn−1= 0.
Inducitvely, we find
t→∞lim tn−1
e12t = lim
t→∞
0
1 2
n
e12t = 0.
1
By the definition of limit, we choose = 1, there exists M > 0 such that for all t ≥ M
tn−1 e12t
< = 1.
Hence for t ≥ M, 0 ≤ tn−1< e12t. This implies that for t ≥ M, (1.1) 0 ≤ e−ttn−1≤ e−t· e12t= e−12t. Lemma 1.1 implies that
Z ∞ 0
e−12tdt is convergent. (1.1) and Comparison test implies that Z ∞
0
e−ttn−1dt is convergent for every n ∈ N.
Let x ≥ 1 be any real number. Let [x] be the largest integer so that [x] ≤ x < [x] + 1.
Then for t ≥ 0,
(1.2) 0 ≤ e−ttx−1 ≤ e−tt[x].
Since Z ∞
0
e−tt[x]dx is convergent, by comparison test and (1.2), we find Z ∞
0
e−ttx−1dt is convergent.
Now let us study the case when 0 < x < 1. Then we know 1
e12t
≤ tx−1 e12t
≤ t e12t. We know that
t→∞lim t
e12t = lim
t→∞
1 e12t = 0.
By the Sandwich principle,
t→∞lim tx−1
e12t
= 0
for 0 < x < 1. (Of course, this statement is true for all x > 0.) This implies that 0 ≤ e−ttx−1 ≤ e−12t, t ≥ 1.
By comparison test, Z ∞
1
e−ttx−1dt is convergent for 0 < x < 1. Now, we need to verify that Z 1
0
e−ttx−1dt = Z 1
0
et t1−xdt is convergent. Notice that lim
t→0
e−t
t1−x = ∞. Hence the integral Z 1
0
e−ttx−1dt is a type II improper integral.
Theorem 1.3. Let f, g ∈ C(a, b] and 0 ≤ f (x) ≤ g(x) for all a < x ≤ b.
(1) If Z b
a
g(x)dx is convergent, so is Z b
a
f (x)dx.
(2) If Z b
a
f (x)dx is divergent to infinity, so is Z b
a
g(x)dx.
Note that
0 < e−ttx−1 ≤ etx−1. We know that
Z 1 0
tx−1dt = lim
b→0
Z 1 b
tx−1dt
= lim
b→0
tx x
1
b
= lim
b→0
1 x −bx
x
= 1 x. By the comparison test,
Z 1 0
e−ttx−1dt is convergent.
Theorem 1.4. For every x > 0, the improper integral Γ(x) =
Z ∞ 0
e−ttx−1dt is convergent.
Proposition 1.1. For x > 0,
Γ(x + 1) = xΓ(x).
Proof. For every N > 0, using integration by parts, Z N
0
e−ttxdt = −txe−t
N 0 + x
Z N 0
e−ttx−1dt
= −Nxe−N + x Z N
0
e−ttx−1dt.
We have seen that lim
N →∞Nxe−N = lim
N →∞
Nx
eN = 0 for all x > 0. Then Γ(x + 1) = lim
N →∞
Z N
0
e−ttxdt
= lim
N →∞(−Nxe−N+ x Z N
0
e−ttx−1dt)
= x lim
N →∞
Z N 0
e−ttx−1dt
= xΓ(x).
Corollary 1.1. For every n ≥ 0, Γ(n + 1) = n!.
Proof. We know Z ∞
0
e−tdt = 1. Hence Γ(1) = 1. We can prove the formula by induction.
Assume that the statement is true for some nonnegative integer k, i.e. Γ(k + 1) = k!. By the previous proposition,
Γ(k + 2) = Γ((k + 1) + 1) = (k + 1)Γ(k + 1) = (k + 1) · k! = (k + 1)!.
Proposition 1.2. For any s > 0, x > 0, we have Z ∞
0
e−sttx−1dt = Γ(x) sx .
Proof. This formula can be proved by using substitution u = st.
Example 1.1.
Z ∞ 0
e−st(2 − 3t + 5t2)dt = 2 s− 3
s2 +10 s3. This can be proved by the previous proposition.
Here comes a question, if f (t) is a continuous function on [0, ∞) such that Z ∞
0
e−stf (t)dt = 2 s− 3
s2 + 10 s3,
what can you say about f (t)? Let us introduction the notion of Laplace transformation.
2. Laplace Transformation, its inverse transformation, and linear homogeneous o.d.e of constant coefficients
Definition 2.1. Let f (t) be a continuous function on [0, ∞). Suppose Z ∞
0
e−stf (t)dt con- verges for s ≥ a for some a ∈ R. Denote
L(f )(s) = Z ∞
0
e−stf (t)dt called the Laplace transform of f.
Now, let us assume that the Laplace transform of f1, f2 exist on some domain D ⊂ R where f1, f2 are continuous functions on [0, ∞). For any real numbers a1, a2, we know that for any s ∈ D,
L(a1f1+ a2f2)(s) = Z ∞
0
(a1f1(t) + a2f2(t))e−stdt
= lim
M →∞
Z M 0
(a1f1(t) + a2f2(t))e−stdt
= lim
M →∞
a1
Z M 0
f1(t)e−stdt + a2 Z M
0
f2(t)e−stdt
= a1 lim
M →∞
Z M 0
f1(t)e−stdt + a2 lim
M →∞
Z M 0
f2(t)e−stdt
= a1L(f1)(s) + a2L(f2)(s).
Proposition 2.1. Suppose that f1, f2 are two continuous functions on [0, ∞) such that their Laplace transform exist on some domain D ⊂ R for s. Then
L(a1f1+ a2f2)(s) = a1L(f1)(s) + a2L(f2)(s), s ∈ D,
for any a1, a2∈ R. (We say that the Laplace transform is a linear transformation.)
Theorem 2.1. Suppose f1, f2 are two continuous functions on [0, ∞) such that their Laplace transform exist. If L(f1) = L(f2), then f1 = f2.
Hence if g(s) = L(f )(s), we denote
f (t) = L−1(g)(t) called then Laplace inverse transform.
Proposition 2.2. Suppose f (t) is a smooth function on [0, ∞) (f(k)(t) exists for all k ≥ 1.
Set f(0)(t) = f (t).) Assume that lim
t→∞f(k)(t)e−st= 0 for all k ≥ 0. Then L(f0)(s) = −f (0) + sL(f )(s).
Proof. Using integration by parts, Z N
0
e−stf0(t)dt = f (t)e−st
N 0 + s
Z N 0
e−stf (t)dt
= f (N )e−sN− f (0) + s Z N
0
e−stf (t)dt.
By assumption, lim
N →∞f (N )e−sN = 0 (consider k = 0.) Hence Z ∞
0
e−stf0(t)dt = lim
N →∞
Z N 0
e−stf0(t)dt
= lim
N →∞
f (N )e−sN − f (0) + s Z N
0
e−stf (t)dt
= −f (0) + s lim
N →∞
Z N 0
e−stf (t)dt
= −f (0) + s Z ∞
0
e−stf (t)dt.
Corollary 2.1. Under the same assumption as above, we have
L(f(n))(s) = snF (s) −X
k=0
sn−k−1f(k). When n = 2, we have L(f00)(s) = s2F (s) − sf0(0) − f (0).
Now let us use Laplace transform to solve ordinary differential equation with constant coefficients.
Example 2.1. Solve for y0+ 2y = 0 with initial condition y(0) = 1.
Take Laplace transform, we obtain
L(y0+ 2y)(s) = L(y0)(s) + 2L(y)(s) = 0.
Let g(s) = L(y)(s). Using the previous proposition,
L(y0)(s) = −y(0) + sL(y)(s) = −1 + sg(s).
Hence (−1 + sg(s)) + 2g(s) = 0. This implies that g(s) = 1
s + 2. Then y = L−1(g)(t) = e−2t.
Example 2.2. Solve for y00+ y = 0 with initial condition y(0) = a and y0(0) = b.
Let g(s) be the Laplace transform of y. Then
L(y00)(s) + g(s) = 0.
On the other hand, L(y00)(s) = s2g(s) − sy(0) − y0(0) = s2g(s) − as − b. This implies that (s2+ 1)g(s) = as + b =⇒ g(s) = a s
s2+ 1+ b 1 s2. We know s
s2+ 1 = L(cos t)(s), and 1
s2+ 1 = L(sin t)(s). Hence y(t) = a cos t + b sin t.
In general, we can solve for the linear (homogeneous) differential equation of constant coefficients through Laplace transformation. A linear homogeneous differential equation of constant coefficients is a differential equation
(2.1) any(n)+ an−1y(n−1)+ · · · + a1y0+ a0y = 0, where a0, · · · , an∈ R. If an6= 0, we say that the equation is of order n.
Let us study the case when n = 2 :
(2.2) ay00+ by0+ cy = 0.
Here a 6= 0. Taking the Laplace transformation of the equation, we have a(s2F (s) − sy0(0) − y(0)) + b(sF (s) − y(0)) + cF (s), where F (s) = L(y)(s). Therefore F (s) is a rational function in s and given by
F (s) = As + B as2+ bs + c.
where A = ay0(0) and B = ay0(0) + by(0). We can use partial fraction expansion to decom- pose F (s).
Definition 2.2. The polynomial χ(s) = as2+ bs + c is called the characteristic polynomial of (2.2).
We assume that a = 1 and let D = b2 − 4c be the discriminant of the characteristic polynomial.
case 1: If D > 0, χ(s) has two distinct real roots. We denote the root of χ(s) by λ1 and λ2. Thus we may write
F (s) = C1
1 s − λ1
+ C2
1 s − λ2
for some C1, C2∈ R. We know that
L(eat)(s) = 1 s − a.
Using the inverse transform and the linearity of L−1, we see that y(t) = L−1(F )(t)
= C1L−1
1
s − λ1
+ C2L−1
1
s − λ2
= C1eλ1t+ C2eλ2t. Hence y is a linear combination of eλ1t and eλ2t.
case 2: If D = 0, χ(s) has one repeated root, called λ. Then we write F (s) = C1
1
s − λ + C2
1 (s − λ)2. Recall that
L(tneλt)(s) = Z ∞
0
e−sttnetλdt = Γ(n + 1)
(s − λ)n+1 = n!
(s − λ)n+1. Thus we obtain
y(t) = L−1(F )(t)
= C1L−1
1 s − λ
+ C2L−1
1
(s − λ)2
= C1eλt+ C2teλt
= (C1+ C2t)eλt.
case 3: If D < 0, by completing the square, we write χ(s) = (s − α)2+ β2. We write
F (s) = As
(s − α)2+ β2 + B
(s − α)2+ β2 = C1 s − α
(s − α)2+ β2 + C2 β
(s − α)2+ β2.
Notice that
L(eαtcos βt)(s) = s − α
(s − α)2+ β2, L(eαtsin βt) = β
(s − α)2+ β2. Thus
y(t) = L−1(F )(t)
= C1L−1
s − α (s − α)2+ β2
+ C2L−1
β
(s − α)2+ β2
= C1eαtcos βt + C2eαtsin βt
= eαt(C1cos βt + C2sin βt).
Remark. Let V be the subset of C2[0, ∞) consisting of y such that (2.2) holds. The above observation implies that V is in fact a two dimensional real vector (sub)space (of C2[0, ∞)).
This method can be applied to (2.1) for any n ∈ N.
Definition 2.3. The characteristic polynomial of (2.1) is defined to be χ(s) = ansn+ · · · + a1s + a0.
Taking the Laplace transformation of (2.1) and using Corollary 2.1, we find that F (s) = P (s)
χ(s)
for some polynomial P (s) ∈ R[s]. Thus F (s) is a rational function and we can solve for y using the partial fraction for F (s). Thus we reduce our problems to the cases when χ(s) = (s − a)nor χ(s) = ((s − α)2+ β2)m. When χ(s) = (s − a)n, we write
F (s) = A1
s − a + A2
(s − a)2 + · · · + An (s − a)n
= C0 1
s − a + C1 1!
(s − a)2 + · · · + Cn−1(n − 1)!
(s − a)n,
where Ci−1= Ai/(i − 1)! for i = 1, · · · , n. By taking the inverse transform, we obtain y(t) = (C0+ C1t + · · · + Cn−1tn−1)eat.
When χ(s) = ((x − α)2+ β2)m, we write F (s) = A1s + B1
(x − α)2+ β2 + · · · + Ams + Bm
((s − α)2+ β2)m
= C1(s − a) + D1β
(x − α)2+ β2 + · · · + Cm(s − a) + Dmβ ((s − α)2+ β2)m .
Here Ci, Dj are constants. By finding the inverse transform of (Ci(s − a) + Diβ)/(((s − α)2+ β2)i), we obtain y.
3. Beta Function In this section, x, y are positive real numbers. Let us define
B(x, y) = Γ(x)Γ(y) Γ(x + y).
The function B(x, y) is called the Gamma function. It follows immediately from the defi- nition that
B(x, y) = B(y, x), ∀x, y > 0.
Using the property of Gamma function: Γ(x + 1) = xΓ(x) for x > 0, we obtain:
Proposition 3.1. Suppose p, q > 0. Then (1) B(p, q) = B(p + 1, q) + B(p, q + 1).
(2) B(p, q + 1) = q
pB(p + 1, q) = q
p + qB(p, q).
Theorem 3.1. For x, y > 0,
B(x, y) = Z ∞
0
tx−1 (1 + t)x+ydt.
Let us postpone the proof of this equation.
Let us consider the substitution t = tan2θ with 0 ≤ θ ≤ π/2. Then dt = 2 tan θ sec2θdθ.
Using sec2θ = tan2θ + 1, Beta function B(x, y) can be rewritten as B(x, y) =
Z π
2
0
(tan2θ)x−1
(1 + tan2θ)x+y · 2 tan θ sec2θdθ
= 2 Z π
2
0
tan2x−2θ
sec2x+2yθ · tan θ sec2θdθ
= 2 Z π
2
0
tan2x−1θ · 1
sec2x+2y−2θdθ
= 2 Z π
2
0
sin θ cos θ
2x−1
· cos2x+2y−2θdθ
= 2 Z π
2
0
sin2x−1θ cos2y−1θdθ.
Using B(y, x) = B(x, y), we also obtain B(x, y) = 2
Z π
2
0
cos2x−1θ sin2y−1θdθ.
This is the second form of Beta function. Consider substitution t = sin2θ for 0 ≤ θ ≤ π/2.
Then dt = 2 sin θ cos θdθ, we obtain the third form of the Beta function:
B(x, y) = Z 1
0
tx−1(1 − t)y−1dt.
We conclude that
Theorem 3.2. The Beta function has the following forms:
(1) B(x, y) = Z ∞
0
tx−1 (1 + t)x+ydt,
(2) B(x, y) = 2 Z π2
0
cos2x−1θ sin2y−1θdθ, (3) B(x, y) =
Z 1 0
tx−1(1 − t)y−1dt.
Example 3.1. Compute Z ∞
0
e−x2dx.
Let us make a change of variable t = x2. Then dt = 2xdx and thus dx = t−12dt 2 .The integral can be rewritten as
Z ∞ 0
e−x2dx = 1 2
Z ∞ 0
e−tt−12dt = Γ 12 2 . Now, we only need to compute Γ(1/2). Using Beta function,
B 1 2,1
2
= Γ 122
Γ 12 +12 = Γ
1 2
2
On the other hand, B 1
2,1 2
= 2 Z π
2
0
cos2·12−1θ sin2·12−1θdθ = 2 Z π
2
0
dθ = π.
Hence Γ(1/2) =√
π. We obtain that Z ∞
0
e−x2dx =
√π 2 . Example 3.2. Compute
Z 2π 0
sin4θdθ.
We know Z 2π
0
sin4θdθ = 4 Z π
2
0
sin4θdθ = 2 · 2 Z π
2
0
sin2·52−1θ cos2·12 θdθ = 2B 5 2,1
2
. We compute
B 5 2,1
2
= Γ 52 Γ 12 Γ(3) . Using Γ(x + 1) = xΓ(x), we obtain
Γ 5 2
= 3 2 ·1
2· Γ 1 2
= 3 4
√π.
Hence B 5 2,1
2
=
3 4
√π ·√ π
2! = 3
8π. Thus Z 2π
0
sin4θdθ = 3 4π.
Now, let us go back to the proof of Theorem 3.1.
Γ(x)Γ(y) =
Z ∞ 0
e−ttx−1dt
Z ∞ 0
e−ssx−1ds
= Z ∞
0
Z ∞ 0
e−(t+s)tx−1sy−1dtds.
Let us compute Z ∞
0
e−(t+s)tx−1dt. Consider t = su, we rewrite Z ∞
0
e−(t+s)tx−1dt = sx Z ∞
0
e−(u+1)sux−1du.
Hence the integral becomes Γ(x)Γ(y) =
Z ∞ 0
Z ∞ 0
e−(u+1)ssx+y−1ds
ux−1du
= Z ∞
0
Γ(x + y)
(u + 1)x+y · ux−1du
= Γ(x + y) Z ∞
0
ux−1 (1 + u)x+ydu.
Here we use Proposition 1.2. This shows that Γ(x)Γ(y)
Γ(x + y) = Z ∞
0
ux−1 (1 + u)x+ydu.
Example 3.3. Compute Z 3
1
(x − 1)10(x − 3)3dx.
Let t = (x − 1)/2. Then the integral becomes Z 3
1
(x − 1)10(x − 3)3dx = Z 1
0
(2t)10(2t − 2)32dt = −214 Z 1
0
t10(1 − t)3dt.
We obtain
Z 3 1
(x − 1)10(x − 3)3dx = −214B(11, 4) = −214·10!3!
14! .
