(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

We will prove that the improper integral Γ(x) =

Z ∞

0

e^−tt^x−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)dxis 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 withL >0.

Then both Z ∞

a

g(x)dx and Z ∞

a

f(x)dxare 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 whenx≥1.

Lemma 1.2. Let nbe a natural number. Then

t→∞lim tⁿ⁻¹

e^12t = 0.

Proof. By L’ Hospital rule,

t→∞lim tⁿ⁻¹

e^12t

= lim

t→∞

(n−1)tⁿ⁻²

1 2e^12t

. Since tⁿ⁻¹ is a polynomial of degreen−1,we know

dⁿ

dtⁿtⁿ⁻¹ = 0.

Inducitvely, we find

t→∞lim tⁿ⁻¹

e^12t = lim

t→∞

0

1 2

n

e^12t = 0.

1

By the definition of limit, we choose = 1,there existsM >0 such that for all t≥M

tⁿ⁻¹ e^12t

< = 1.

Hence for t≥M, 0≤tⁿ⁻¹< e^12t.This implies that fort≥M, (1.1) 0≤e^−ttⁿ⁻¹ ≤e^−t·e^12t=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 everyn∈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^−tt^x−1 ≤e^−tt^[x]. Since

Z ∞

0

e^−tt^[x]dx is convergent, by comparison test and (1.2), we find Z ∞

0

e^−tt^x−1dt is convergent.

Now let us study the case when 0< x <1.Then we know 1

e^12t

≤ t^x−1 e^12t

≤ t e^12t. We know that

t→∞lim t

e^12t = lim

t→∞

1 e^12t = 0.

By the Sandwich principle,

t→∞lim t^x−1

e^12t

= 0

for 0< x <1.(Of course, this statement is true for all x >0.) This implies that 0≤e^−tt^x−1 ≤e^−12t, t≥1.

By comparison test, Z ∞

1

e^−tt^x−1dtis convergent for 0< x <1.Now, we need to verify that Z 1

0

e^−tt^x−1dt= Z 1

0

e^t t^1−xdt is convergent. Notice that lim

t→0

e^−t

t^1−x = ∞. Hence the integral Z 1

0

e^−tt^x−1dt is a type II improper integral.

Theorem 1.3. Letf, g∈C(a, b] and 0≤f(x)≤g(x) for all a < x≤b.

(1) If Z b

a

g(x)dxis 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^−tt^x−1 ≤et^x−1. We know that

Z 1 0

t^x−1dt= lim

b→0

Z 1 b

t^x−1dt

= lim

b→0

t^x x

1 b

= lim

b→0

1 x −b^x

x

= 1 x. By the comparison test,

Z 1 0

e^−tt^x−1dt is convergent.

Theorem 1.4. For everyx >0,the improper integral Γ(x) =

Z ∞

0

e^−tt^x−1dt is convergent.

Proposition 1.1. For x >0,

Γ(x+ 1) =xΓ(x).

Proof. For everyN >0,using integration by parts, Z N

0

e^−tt^xdt= −t^xe^−t

N 0 +x

Z N 0

e^−tt^x−1dt

=−N^xe^−N +x Z N

0

e^−tt^x−1dt.

We have seen that lim

N→∞N^xe^−N = lim

N→∞

N^x

e^N = 0 for all x >0.Then Γ(x+ 1) = lim

N→∞

Z _N

0

e^−tt^xdt

= lim

N→∞(−N^xe^−N+x Z N

0

e^−tt^x−1dt)

=x lim

N→∞

Z N 0

e^−tt^x−1dt

=xΓ(x).

Corollary 1.1. For everyn≥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^−stt^x−1dt= Γ(x) s^x .

Proof. This formula can be proved by using substitutionu=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, iff(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 fors≥afor somea∈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 wheref1, 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 domainD⊂Rfors. 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),thenf₁ =f₂.

Hence if g(s) =L(f)(s),we denote

f(t) =L⁻¹(g)(t) called then Laplace inverse transform.

Proposition 2.2. Supposef(t) is a smooth function on [0,∞) (f^(k)(t) exists for allk≥1.

Setf⁽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 (considerk= 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

s^n−k−1f^(k). When n= 2,we haveL(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.

Letg(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 thatg(s) = 1

s+ 2.Then y=L⁻¹(g)(t) =e^−2t.

Example 2.2. Solve for y⁰⁰+y = 0 with initial conditiony(0) =aand y⁰(0) =b.

Letg(s) be the Laplace transform ofy. 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(cost)(s),and 1

s²+ 1 =L(sint)(s).Hencey(t) =acost+bsint.

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, wherea0,· · · , an∈R.Ifan6= 0,we say that the equation is of ordern.

Let us study the case whenn= 2 :

(2.2) ay⁰⁰+by⁰+cy= 0.

Here a6= 0.Taking the Laplace transformation of the equation, we have a(s²F(s)−sy⁰(0)−y(0)) +b(sF(s)−y(0)) +cF(s), whereF(s) =L(y)(s).ThereforeF(s) is a rational function insand given by

F(s) = As+B as²+bs+c.

whereA=ay⁰(0) andB =ay⁰(0) +by(0).We can use partial fraction expansion to decom- pose F(s).

Definition 2.2. The polynomialχ(s) =as²+bs+cis 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: IfD >0, χ(s) has two distinct real roots. We denote the root ofχ(s) byλ1 and λ₂.Thus we may write

F(s) =C1

1 s−λ1

+C2

1 s−λ2

for someC1, C2∈R.We know that

L(e^at)(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

=C1e^λ1t+C2e^λ2t. Hence y is a linear combination ofe^λ1t and e^λ2t.

case 2: IfD= 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(tⁿe^λt)(s) = Z ∞

0

e^−sttⁿe^tλdt= 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

=C1e^λt+C2te^λt

= (C1+C2t)e^λt.

case 3: IfD <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(e^αtcosβt)(s) = s−α

(s−α)²+β², L(e^αtsinβt) = 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

=C₁e^αtcosβt+C₂e^αtsinβt

=e^αt(C₁cosβt+C₂sinβt).

Remark. LetV be the subset ofC²[0,∞) consisting ofysuch that (2.2) holds. The above observation implies thatV is in fact a two dimensional real vector (sub)space (ofC²[0,∞)).

This method can be applied to (2.1) for anyn∈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)ⁿorχ(s) = ((s−α)²+β²)^m.When χ(s) = (s−a)ⁿ,we write

F(s) = A₁

s−a+ A₂

(s−a)² +· · ·+ A_n (s−a)ⁿ

=C₀ 1

s−a+C₁ 1!

(s−a)² +· · ·+Cn−1

(n−1)!

(s−a)ⁿ,

whereCi−1 =Ai/(i−1)! fori= 1,· · · , n.By taking the inverse transform, we obtain y(t) = (C0+C1t+· · ·+Cn−1tⁿ⁻¹)e^at.

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 obtainy.

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. Forx, y >0,

B(x, y) = Z ∞

0

t^x−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. Thendt= 2 tanθsec²θdθ.

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 tanθsec²θdθ

= 2 Z ^π

2

0

tan^2x−2θ

sec^2x+2yθ ·tanθsec²θdθ

= 2 Z ^π

2

0

tan^2x−1θ· 1

sec^2x+2y−2θdθ

= 2 Z ^π

2

0

sinθ cosθ

2x−1

·cos^2x+2y−2θdθ

= 2 Z ^π

2

0

sin^2x−1θcos^2y−1θdθ.

UsingB(y, x) =B(x, y),we also obtain B(x, y) = 2

Z ^π

2

0

cos^2x−1θsin^2y−1θdθ.

This is the second form of Beta function. Consider substitution t= sin²θfor 0≤θ≤π/2.

Then dt= 2 sinθcosθdθ, we obtain the third form of the Beta function:

B(x, y) = Z 1

0

t^x−1(1−t)^y−1dt.

We conclude that

Theorem 3.2. The Beta function has the following forms:

(1) B(x, y) = Z ∞

0

t^x−1 (1 +t)^x+ydt,

(2) B(x, y) = 2 Z ^π₂

0

cos^2x−1θsin^2y−1θdθ, (3) B(x, y) =

Z 1 0

t^x−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

cos^2·12−1θsin^2·12−1θdθ= 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

sin^2·52−1θcos^2·12 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^−tt^x−1dt

Z ∞

0

e^−ss^x−1ds

= Z ∞

0

Z ∞

0

e^−(t+s)t^x−1s^y−1dtds.

Let us compute Z ∞

0

e^−(t+s)t^x−1dt.Considert=su,we rewrite Z ∞

0

e^−(t+s)t^x−1dt=s^x Z ∞

0

e^−(u+1)su^x−1du.

Hence the integral becomes Γ(x)Γ(y) =

Z ∞

0

Z ∞

0

e^−(u+1)ss^x+y−1ds

u^x−1du

= Z ∞

0

Γ(x+y)

(u+ 1)^x+y ·u^x−1du

= Γ(x+y) Z ∞

0

u^x−1 (1 +u)^x+ydu.

Here we use Proposition 1.2. This shows that Γ(x)Γ(y)

Γ(x+y) = Z ∞

0

u^x−1 (1 +u)^x+ydu.

Example 3.3. Compute Z 3

1

(x−1)¹⁰(x−3)³dx.

Lett= (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! .