Chapter 5 Henstock-Kurzweil Laplace Transformshodhganga.inflibnet.ac.in/bitstream/10603/25466/11/11_chapter_05.pdf · The rst question arises is that whether (or in which conditions)

Chapter 5

Henstock-Kurzweil Laplace

Transform

5.1 Introduction

The Laplace transform of a function f : [0,∞) → R, at s ∈ C, is defined as

the integral [15] ∫ ∞0

e−stf(t) dt. (5.1)

Many authors, in some previous decades, showed their interest in studying

the Laplace transform in a classical sense; by classical meaning Lebesgue

integral and/or Riemann integral on the real line.

In this chapter, we focus on an integral transform that generalizes the clas-

sical Laplace transform. Here the generalization is given by the replacement,

in the definition of Laplace transform, i.e., in (5.1), of Lebesgue integral with

Henstock-Kurzweil (HK) integral. The Henstock-Kurzweil integral is defined

in terms of Riemann sums but with slight modification, yet it includes the

Riemann, improper Riemann, and Lebesgue integrals as special cases. This

integral is equivalent to the Denjoy and Perron integrals.

117

The first question arises is that whether (or in which conditions) the

Laplace transform when treated as HK integral make sense. The sufficient

condition for the existence of Laplace transform is that the function f is

locally integrable on [0,∞), because the product of locally integrable func-

tion and bounded measurable function is locally integrable. Since Henstock-

Kurzweil integral allows nonabsolute convergence, it makes an ideal setting

for the Laplace transform. It was proved [9] that, on a compact interval, the

product of nonabsolute integrable function and a bounded variation function

is still nonabsolute integrable and this remains valid in case of an unbounded

interval [1]. We observed in (5.1) that the kernel function e−st, s ∈ C, is

not bounded variation function on [0,∞) hence we cannot claim (as in the

classical sense) that the HK Laplace transform exists for any HK integrable

function.

The convolution plays an important role in pure and applied mathemat-

ics, approximation theory, differential and integral equations and many other

areas. Laplace transform has the property that it can interact with con-

volutions so we obtain various results on convolution. We give necessary

and sufficient conditions so that the convolution operator as HK integral is

continuous. Finally we solve the problem of inversion by using generalized

differentiation.

The results in this chapter are appeared in [2, 14].

5.2 Existential Conditions

In this section, we tackle the problem of existence of Laplace transform as

HK integral.

According to Zayed [16], the classical sufficient condition for the existence

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of Laplace transform is that the function f is locally integrable on [0,∞), i.e.,

f ∈ Lloc[0,∞). This is because the multiplier for the Lebesgue integrable

functions is bounded measurable function and the function e−st, s ∈ C, is

bounded measurable function. We know that the multipliers for HK inte-

grable functions are precisely the functions of bounded variation [1]. Note

that in (5.1), the function e−st, s ∈ C, is not of bounded variation on [0,∞)

but it is of bounded variation on any compact interval J ⊂ R. Hence we

can not consider (5.1) as HK integral directly. So we cannot get any specific

condition for the existence of Laplace transform as HK integral. However we

do have the following different conditions.

Theorem 5.2.1. Let f : [0,∞) → R be any continuous function such that

F (x) =∫ x

0f , 0 ≤ x <∞, is bounded on [0,∞). Then the Laplace transform

L{f}(s), i.e., L{f}(s) =∫∞

0e−stf(t) dt, Re.s > 0 exists.

Proof. Let 0 < x <∞, s = σ + iω, σ, ω ∈ R.

We have (F (t) e−st)′= −s F (t) e−st + f(t) e−st.

Since e−st is of bounded variation on any compact interval, say [a, b] and F (t)

is bounded on [0,∞). Then F (t) e−st is HK integrable on [a, b].

Therefore −s F (t) e−st is integrable over I = [a, b].

Now ∫ b

a

∣∣∣(e−st)′∣∣∣ dt =

∫ b

a

∣∣−s e−st∣∣ dt=

∫ b

a

√σ2 + ω2 e−σt dt

=

√σ2 + ω2

σ[e−σa − e−σb].

Without loss of generality let us take a = 0. Therefore∫ b

0

∣∣∣(e−st)′∣∣∣ dt =

√σ2 + ω2

σ[1− e−σb] <∞.

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Hence (e−st)′

is absolutely integrable over [a, b].

Now by Hake’s theorem [1], we can write

limb→∞

∫ b

0

∣∣∣(e−st)′∣∣∣ dt = limb→∞

√σ2 + ω2

σ[1− e−σb]

=

√σ2 + ω2

σ, σ > 0

=

√σ2 + ω2

σ, Re.s > 0.

This shows that (e−st)′

is absolutely integrable on [0,∞).

Again limt→∞

e−st = 0, Re.s > 0.

Hence the function e−st : [0,∞)→ R is continuous on [0,∞) with limt→∞

e−st =

0 and (e−st)′

is absolutely integrable over [0,∞) and we have assumed that

f : [0,∞)→ R is continuous function such that F (x) =∫ x

0f , 0 ≤ x <∞, is

bounded on [0,∞).

Therefore by Dedekind’s test [10], we can say that the integral∫ ∞0

e−stf(t) dt exists forRe.s > 0.

Remark 5.2.2. The above result can also be proved by using Du-Bois Rey-

mond’s test [1] as follows:

Take φ(t) = e−st, s = σ + iω, J = [0,∞).

Then clearly φ(t) is differentiable and φ′ ∈ L1(J).

Since F is bounded, we have limt→∞

F (t) e−st = 0, Re.S > 0.

Hence F (x) =∫ x

0f , 0 ≤ x <∞ is bounded on J , e−st is differentiable on J ,

(e−st)′ ∈ L1(J) and F (t) e−st → 0 as t→∞, Re.s > 0.

Therefore by Du-Bois Reymond’s test, f(t)e−st ∈ HK(J).

That is, the Laplace transform of f(t),∫∞

0e−stf(t) dt exists, Re.s > 0.

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Theorem 5.2.3. If the Laplace transform of f(t),∫∞

0e−stf(t) dt exists for

Re.s > 0, then the function f ∈ HKloc.

Proof. Suppose the integral∫∞

0e−stf(t) dt exists for Re.s > 0.

Note that the function e−st, s ∈ C, as a function of real variable t, is not

of bounded variation on [0,∞) but is of bounded variation on any compact

interval I = [a, b].

By Hake’s theorem [1], we can write∫ ∞0

e−stf(t) dt = lima→0b→∞

∫ b

a

e−stf(t) dt.

Since the limits on R.H.S. exist, we can say that the HK integral∫ bae−stf(t)dt

exists, Re.s > 0 for some compact interval [a, b].

Hence the function f(t) is HK integrable necessarily on the compact interval

I = [a, b], that is, f ∈ HKloc.

Theorem 5.2.4. Let f ∈ HK([0,∞)). Define F (x) =∫∞xf(t) dt. Then

L{f}(s) exists if and only if∫∞

0e−stF (t) dt exists at s ∈ C.

Proof. Let T > 0. The function e−st, s ∈ C, is of bounded variation on a

compact interval [0, T ] and the function f is HK integrable on [0,∞).

Hence e−stf(t), s ∈ C, is HK integrable on [0, T ] and by integrating by parts,

we get ∫ T

0

e−stf(t) dt = e−sTF (T )− F (0) + s

∫ T

0

e−stF (t) dt.

Since the function f is HK integrable on [0,∞), F (0) is finite, say A and by

[9.12(a), [5]], F is continuous function such that limT→∞

F (T ) = 0.

Therefore by Hake’s theorem [1],

limT→∞

∫ T

0

e−stf(t) dt = limT→∞

{e−sTF (T )− F (0) + s

∫ T

0

e−stF (t) dt

}.

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Hence ∫ ∞0

e−stf(t) dt = s

∫ ∞0

e−stF (t) dt− A.

This shows that L{f}(s) exists if and only if∫∞

0e−stF (t) dt exists at s ∈ C,

Re.s > 0.

Theorem 5.2.5. (Uniqueness theorem) Let L{f}(s) =∫∞

0e−stf(t) dt

and L{g}(s) =∫∞

0e−stg(t) dt. If f(t) = g(t) a.e. on [0,∞), then L{f}(s) =

L{g}(s).

5.3 Basic Properties

The usual elementary properties such as linearity, dilation, modulation, trans-

lation are remains same for the HK Laplace transform, the proofs of which

are quite similar to that of classical ones. Also there are some other re-

sults that are analogous to the classical one with some modification in the

hypothesis which we discuss below.

Theorem 5.3.1. (HK Laplace transform of derivative) Let f : [0,∞)→

R be a function which is in ACGδ(R) such that limt→∞

f(t)est

= 0. Then for Re.s >

0 both L{f}(s) and L{f ′}(s) fail to exist or L{f ′}(s) = s L{f}(s)− f(0).

Proof. Let T > 0. Consider the integral J =∫ T

0e−stf

′(t) dt.

This integral is valid since the function e−st, s ∈ C, is bounded variation on

any compact interval [0, T ] and the function f′

is HK integrable on R.

By integrating by parts, we get

J = e−sTf(T )− f(0) + s

∫ T

0

e−stf(t) dt.

By Hake’s theorem [1],

limT→∞

∫ T

0

e−stf′(t) dt = lim

T→∞e−sTf(T )− f(0) + lim

T→∞s

∫ T

0

e−stf(t) dt

122

= s

∫ ∞0

e−stf(t) dt− f(0).

Therefore ∫ ∞0

e−stf′(t) dt = s

∫ ∞0

e−stf(t) dt− f(0).

That is, L{f ′}(s) = sL{f}(s)− f(0).

Remark 5.3.2. The above result can be generalize by imposing the condi-

tions on f as:

For k = 0, 1, 2, . . . , (n− 1), f (k) are in ACGδ(R) and limt→∞

f (k)

est= 0, Re.s > 0.

Then L{f (n)}(s) and L{f}(s) fails to exist or

L{f (n)}(s) = sn L{f}(s)− sn−1 f(0)− sn−2 f′(0)− . . .− f (n−1)(0).

Theorem 5.3.3. (Differentiation of HK Laplace transform) Let f :

[0,∞) → R be a function such that L{f}(s) exists on a compact interval,

say J = [α, β], α, β ∈ R. Define g(t) = t f(t) and assume that g ∈ HK(R+).

Then L{g}(s) exists and L{g}(s) = −(L{f}(s))′ a.e. Here s ∈ R.

Proof. Suppose

L{f}(s) =

∫ ∞0

e−stf(t) dt exists on [α, β], α, β ∈ R.

Define g(t) = t f(t).

We know the necessary and sufficient condition for differentiating under the

HK integral is that∫ ∞t=0

∫ b

s=a

e−st tf(t) ds dt =

∫ b

s=a

∫ ∞t=0

e−st tf(t) dt ds (5.2)

for all [a, b] ⊂ [α, β] [12].

Now ∫ ∞t=0

∫ b

s=a

e−st tf(t) ds dt =

∫ ∞t=0

(e−at − e−bt)f(t) dt

123

= L{f}(a)− L{f}(b)

which exists.

Therefore the double HK integral∫∞t=0

∫ bs=a

e−st tf(t) ds dt exists.

Hence by [lemma 25(a), [13]], the equality in (5.2) holds and we can differ-

entiate under the HK integral. By (5.2), we have∫ b

s=a

∫ ∞t=0

e−st tf(t) dt ds = L{f}(a)− L{f}(b)

= −∫ b

a

(L{f}(s))′ds.

Therefore ∫ b

s=a

∫ ∞t=0

e−st tf(t) dt ds = −∫ b

a

(L{f}(s))′ds

for all [a, b] ⊂ [α, β]. Which implies that∫ ∞0

e−st tf(t) dt = −(L{f}(s))′ a.e. on [α, β].

That is, L{t f(t)}(s) = −(L{f}(s))′ a.e. on [α, β].

Hence the derivative of the HK Laplace transform of f(t) exists a.e. on

[α, β].

Remark 5.3.4. Let G(s) = L{t f(t)}(s) and H(s) = −L{f(t)}(s).

Then (H(s))′

= G(s) a.e. on [α, β]. And, we know that [5], A function

f : [a, b] → R is HK integrable on [a, b] if and only if there exists an ACGδfunction F on [a, b] such that F

′= f a.e. on [a, b].

Again, by fundamental theorem of calculus [10], (H(s))′

is HK integrable,

i.e., every derivative is HK integrable. Hence G(s) is also HK integrable.

Consequently, the function H(s) is ACGδ on [α, β].

Continuing in this way n-times, we can say that if tnf(t) is HK integrable,

then any order of derivative of L{f}(s) exists a.e. So L{f}(s) is analytic a.e.

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Theorem 5.3.5. (Multiplication by 1t) Let f : [0,∞)→ R be a function

such that L{f}(s) exists and assume that f(t)t∈ HK(R+). Then L

{f(t)t

}(s)

exists and

L

{f(t)

t

}(s) =

∫ ∞s

L{f}(u) du.

Here s ∈ R.

Proof. Assume that f(t)t∈ HK(R+).

Since the function e−st is of bounded variation on any compact interval I =

[a, b], we have e−st f(t)t∈ HK([a, b]). That is,

∫ bae−st f(t)

tdt exists.

Without loss of generality, let us take a = 0 so that∫ b

0e−st f(t)

tdt exists.

By Hake’s theorem [1], limb→∞

∫ b0e−st f(t)

tdt exists.

Hence L{f(t)t

}(s) exists.

Now consider the integrals

I1 =

∫ ∞0

∫ ∞s

e−ut f(t) du dt, I2 =

∫ ∞s

∫ ∞0

e−ut f(t) dt du.

I1 =

∫ ∞0

[e−st − 0]f(t)

tdt

=

∫ ∞0

e−stf(t)

tdt which exists.

Therefore I1 exists on R× R.

Also, note that the function e−st is of bounded variation on any compact

interval J ⊂ R. Therefore∫R VJ(e−st) dt < MJ , for some constant MJ > 0.

Hence by [lemma 25(a), [13]], I2 exists on R× R and I1 = I2 on R× R. So∫ ∞0

e−stf(t)

tdt =

∫ ∞s

∫ ∞0

e−ut f(t) dt du

=

∫ ∞s

L{f}(u) du.

That is, L{f(t)t

}(s) =

∫∞s

L{f}(u) du.

125

Theorem 5.3.6. (HK Laplace transform of integral) Let f : [0,∞)→ R

be HK integrable function such that L{f}(s) exists. Then L{∫ t

0f(u) du

}(s)

exists and L{∫ t

0f(u) du

}(s) = 1

sL{f}(s).

Proof. Define F (t) =∫ t

0f(u) du, t ∈ R+. Then F

′(t) = f(t) a.e. on R+

[9.12(b), [5]].

Since f ∈ HK(R+), the integral∫ ∞0

f(u) du exists on R+.

That is,

limt→∞

∫ t

0

f(u) du exists on R+.

Hence limt→∞

F (t) is bounded on R+.

Now for T > 0, consider∫ T

0

e−st f(t) dt =

∫ T

0

e−st F′(t) dt.

This integral exists since e−st is of bounded variation on [0, T ] and the func-

tion f(t) is HK integrable on R+.

By integrating by parts, we get∫ T

0

e−st f(t) dt = e−sT F (T ) + s

∫ T

0

e−st F (t) dt.

Since limT→∞

F (T ) is bounded on R, set limT→∞

F (T ) = A.

By Hake’s theorem [1], we can write∫ ∞0

e−st f(t) dt = limT→∞

∫ T

0

e−st f(t) dt

= limT→∞

e−sT F (T ) + s limT→∞

∫ T

0

e−st F (t) dt

= s

∫ ∞0

e−st F (t) dt Re.s > 0.

126

Therefore ∫ ∞0

e−st F (t) dt =1

s

∫ ∞0

e−st f(t) dt, Re.s > 0.

That is, ∫ ∞0

e−st∫ t

0

f(u) du dt =1

sL{f}(s), Re.s > 0.

Hence L{∫ t

0f(u) du

}= 1

sL{f}(s), Re.s > 0.

5.4 Convolution

In this section we shall discuss some classical results of Laplace convolution

in the HK integral setting, called HK-convolution. First we give two results

about the existence of HK-convolution.

Theorem 5.4.1. Let f ∈ HK(R+) and g ∈ BV(R+). Then we have f ∗ g

exists on R+.

Proof. By definition [4],

f ∗ g(t) =

∫ t

0

f(t− τ) g(τ) dτ

which exists as HK-integral by multiplier theorem [1].


|f ∗ g(t)| ≤ A‖f‖{

infR+|g|+ V[0,t] g(τ)

}.

Hence f ∗ g exists on R+.

Theorem 5.4.2. Let f ∈ HKloc and g ∈ BV such that the support of g is

in the compact interval I = [a, b]. Then

|f ∗ g(t)| ≤ A‖f‖[t−b,t−a]

{inf[a,b]|g|+ V[a,b] g(τ)

}.

127

Proof. Suppose supp(g) ⊂ [a, b]. Then we have

f ∗ g(t) =

∫ b

a

f(t− τ)g(τ) dτ.

Integrate by parts, we get

|f ∗ g(t)| ≤∣∣∣∣g(τ)

∫ b

a

f(t− τ) dτ

∣∣∣∣+

∣∣∣∣∫ b

a

∫ b

a

f(t− τ) dg(τ)

∣∣∣∣≤ inf

τ∈[a,b]|g(τ)| ·

∣∣∣∣∫ b

a

f(t− τ) dτ

∣∣∣∣+

∣∣∣∣∫ b

a

f(t− τ) dτ

∣∣∣∣ · V[a,b] g(τ)

≤{

infτ∈[a,b]

|g(τ)|+ V[a,b] g(τ)

}sup

[t−b,t−a]

∣∣∣∣∫ t−a

t−bf(u) du

∣∣∣∣ .Therefore

|f ∗ g(t)| ≤ A‖f‖[t−b,t−a]

{inf[a,b]|g|+ V[a,b] g(τ)

}.

The following result shows that the HK-convolution is bounded.

Theorem 5.4.3. Let f ∈ HK(R+) and g ∈ L1(R+) ∩ BV(R+). Then we

have A‖f ∗ g‖ ≤ A‖f‖ ‖g‖1.

Proof. For 0 ≤ a < b ≤ ∞, we have∫ b

a

f ∗ g(t) dt =

∫ b

a

∫ t

0

f(t− τ)g(τ) dτ dt.

Let

I1 =

∫ b

a

∫ t

0

f(t− τ)g(τ) dτ dt and I2 =

∫ t

0

∫ b

a

f(t− τ)g(τ) dt dτ.

Consider

I2 =

∫ t

0

∫ b

a

f(t− τ)g(τ) dt dτ.

Since f ∈ HK(R+), we have∫ baf(t− τ) dt exists and finite.

Let ∫ t−b

t−af(τ) dτ = M.

128

Therefore

|I2| ≤M

∫ t

0

|g(τ)| dτ.

Since g ∈ L1(R+), we have for some finite K,∫ t

0

|g(τ)| dτ = K.

So |I2| ≤MK and hence I2 exists on R+ × R+.

Now as g ∈ L1(R+), we have ‖g‖1 ≤ K1 and also∫R+ V[0,t] g ≤M1.

Therefore by [lemma 25, [13]], I1 = I2 on R+ × R+.

Hence ∣∣∣∣∫ b

a

f ∗ g(t) dt

∣∣∣∣ ≤ sup[t−a,t−b]

∣∣∣∣∫ t−b

t−af(u) du

∣∣∣∣ · ∫ t

0

|g(τ)| dτ

≤ A‖f‖[t−a,t−b] · ‖g‖1.

Therefore

sup[a,b]

∣∣∣∣∫ b

a

f ∗ g(t) dt

∣∣∣∣ ≤ A‖f‖[t−a,t−b] · ‖g‖1.

That is, A‖f ∗ g‖ ≤ A‖f‖ ‖g‖1.

Now we give some basic elementary properties of HK-convolution.

Theorem 5.4.4. Suppose f ∈ HK(R+) and g ∈ BV(R+). Then the follow-

ing holds:

I) f ∗ g(t) = g ∗ f(t) for all t ∈ R+.

II) (λf) ∗ g = f ∗ (λg) = λ(f ∗ g) for some λ ∈ C.

III) (f ∗ g)x(t) = fx ∗ g(t) = f ∗ gx(t).

IV) h ∗ (f + g)(t) = h ∗ f(t) + h ∗ g(t) for some h ∈ HK(R+).

Proof. The proof is similar to that of the classical case.

129

The associative property of HK-convolution is given in the next theorem.

Theorem 5.4.5. If f ∈ HK(R+), g ∈ BV(R+) and h ∈ L1(R+), then

(f ∗ g) ∗ h(t) = f ∗ (g ∗ h)(t), ∀ t ∈ R+.

Proof. Consider

(f ∗ g) ∗ h(t) =

∫ t

0

∫ τ

0

f(ξ) g(τ − ξ) h(t− τ) dξ dτ.

By changing the order of integration, we get

(f ∗ g) ∗ h(t) =

∫ t

ξ=0

f(ξ)

∫ t−ξ

0

g(u)h(t− u− ξ) du dξ

=

∫ t

ξ=0

f(ξ) g ∗ h(t− ξ) dξ

= f ∗ (g ∗ h)(t).

Now Consider

I1 =

∫ t

τ=ξ

∫ t

ξ=0

f(ξ) g(τ − ξ)h(t− τ) dξ dτ

and

I2 =

∫ t

ξ=0

f(ξ)

∫ t

τ=ξ

g(τ − ξ)h(t− τ) dτ dξ.

Let

J =

∫ t

τ=ξ

g(τ − ξ)h(t− τ) dτ.

Integrate by parts, we get

|J | ≤∣∣∣∣g(t− ξ)

∫ t

τ=ξ

h(t− τ) dτ

∣∣∣∣+

∣∣∣∣∫ t

τ=ξ

∫ u

τ=ξ

h(t− τ) dτ dg(τ − ξ)∣∣∣∣

≤ inf[ξ,t]|g| ·

∣∣∣∣∫ t

τ=ξ

h(t− τ) dτ

∣∣∣∣+

∣∣∣∣∫ t

τ=ξ

h(t− τ) dτ

∣∣∣∣ · V[ξ,t] g(τ − ξ)

≤ inf[ξ,t]|g| ·

∫ t

τ=ξ

|h(t− τ)| dτ +

∫ t

τ=ξ

|h(t− τ)| dτ · V[ξ,t] g(τ − ξ)

≤ inf[ξ,t]|g| · ‖h‖1 + ‖h‖1 · V[ξ,t] g(τ − ξ)

130

≤ ‖h‖1

{inf[ξ,t]|g|+ V[ξ,t] g(τ − ξ)

}.

Therefore

I2 ≤ supR+

∣∣∣∣∫ t

0

f(ξ) dξ

∣∣∣∣ ·{‖h‖1


}}≤ A‖f‖ · ‖h‖1


}.

Hence I2 exists on R+ × R+. And, by [lemma 25, [13]] we are through.

Now we show that the HK-Laplace transform of convolution of two func-

tions is the product of their HK-Laplace transforms.

Theorem 5.4.6. Let f1 ∈ HK(R+), f2 ∈ L1(R+) ∩ BV(R+). Then f ∗ g

exists and if L {f1(t)} and L {f2(t)} exist at s ∈ C, then L {f1 ∗ f2(t)} exists

at s ∈ C and L {f1 ∗ f2(t)} (s) = L {f1(t)} (s) L {f2(t)} (s).

Proof. Consider

L {f1 ∗ f2(t)} (s) =

∫ ∞0

e−stf1 ∗ f2(t) dt

=

∫ ∞0

∫ t

0

e−sτf1(τ)e−s(t−τ)f2(t− τ) dτ dt.

By interchanging the iterated integrals, we have

L {f1 ∗ f2(t)} (s) =

∫ ∞τ=0

e−sτf1(τ)

∫ ∞t=τ

e−s(t−τ)f2(t− τ) dt dτ

=

∫ ∞0

e−sτf1(τ) dτ

∫ ∞0

e−suf2(u) du.

But by hypothesis both the integrals∫ ∞0

e−sτ f1(τ) dτ and

∫ ∞0

e−sτ f2(τ) dτ

exist at s ∈ C.

Therefore L {f1 ∗ f2(t)} exists at s ∈ C and

131

L {f1 ∗ f2(t)} (s) = L {f1(t)} (s) L {f2(t)} (s).

Now it remains to show the justification for interchanging the iterated inte-

grals.

Since the function f2 ∈ L1(R+), ∃ KI such that ‖f2‖1 < KI , where I =

[a, b] ⊂ R+ is any compact interval.

Clearly, V[a,b] e−s(t−τ) f2(t− τ) ≤ e−σu V[t−b,t−a] f2(τ) + |f2(τ)| 2 e−σr for some

reals u and r and so∫ ∞0

V[t−b,t−a]e−sτ f2(τ) dτ ≤ e−σu

∫ ∞0

V[t−b,t−a]f2(τ) dτ+2e−σr∫ ∞

0

|f2(τ)| dτ.

Therefore∫ ∞0

V[t−b,t−a] e−sτ f2(τ) dτ ≤ e−σu

∫ ∞0

V[t−b,t−a] f2(τ) dτ + 2KI e−σr.

Since f2 ∈ BV(R+), there exists a constant MI such that∫ ∞0

V[t−b,t−a]f2(τ) dτ < MI .

Then ∫ ∞0

V[t−b,t−a]e−sτf2(τ) dτ ≤ e−σuMI + 2KI e

−σr.

Thus for each compact interval I = [a, b] ⊂ R+, the integral∫ ∞0

V[t−b,t−a]e−sτf2(τ) dτ

is finite and ‖f2‖1 ≤ KI .

Hence the integral ∫ ∞τ=0

∫ ∞t=τ

e−stf1(τ) f2(t− τ) dt dτ

exists on R+×R+. And, by [lemma 25, [13]], we can interchange the iterated

integrals.

132

Theorem 5.4.7. In addition to the hypothesis of the above theorem if f1 ∗

f2(t) is in HK(R+) and f2 is continuous, then f1 ∗ f2(t) is continuous with

respect to the Alexiewicz norm.

Proof. Choose any δ > 0 such that 0 < δ ≤ 1 (The case for negative δ is

analogous).

Define D(t, δ) = f1 ∗ f2(t+ δ)− f1 ∗ f2(t) = I1 + I2.

Where

I1 =

∫ t

0

f1(τ) [f2(t+ δ − τ)− f2(t− τ)] dτ, I2 =

∫ t+δ

t

f1(τ)f2(t+δ−τ) dτ.

Since f2 ∈ BV(R+), f2 is bounded on R+.

So there exists M2 such that |f2(x)| ≤M2, for all x ∈ R+. Then

|I2| ≤M2

∣∣∣∣∫ t+δ

t

f1(τ) dτ

∣∣∣∣ .But f1 ∈ HK(R+) implies that

limδ→0

∣∣∣∣∫ t+δ

t

f1(τ) dτ

∣∣∣∣ −→ 0.

Therefore |I2| → 0 as δ → 0.

Now consider

I1 =

∫ t

0

f1(τ) [f2(t+ δ − τ)− f2(t− τ)] dτ.


|I1| ≤∣∣∣∣[f2(t+ δ − τ)− f2(t− τ)]

∫ t

0

f1(τ) dτ

∣∣∣∣+

∣∣∣∣∫ t

0

∫ u

0

f1(τ) dτ d(f2(u+ δ − τ)− f2(u− τ))

∣∣∣∣≤ |f2(t+ δ − τ)− f2(t− τ)| · sup

t∈R+

∣∣∣∣∫ t

0

f1(τ) dτ

∣∣∣∣+ sup

u∈R+

∣∣∣∣∫ u

0

f1(τ) dτ

∣∣∣∣ · ∣∣∣∣∫ t

0

d(f2(u+ δ − τ)− f2(u− τ))

∣∣∣∣133

= |f2(t+ δ − τ)− f2(t− τ)| · supt∈R+

∣∣∣∣∫ t

0

f1(τ) dτ

∣∣∣∣+ sup

u∈R+

∣∣∣∣∫ u

0

f1(τ) dτ

∣∣∣∣ · |f2(u+ δ − τ)− f2(u− τ)|.

Therefore

|I1| ≤ |f2(t+ δ − τ)− f2(t− τ)| · supt∈R+

∣∣∣∣∫ t

0

f1(τ) dτ

∣∣∣∣+ sup

u∈R+

∣∣∣∣∫ u

0

f1(τ) dτ

∣∣∣∣ · |f2(u+ δ − τ)− f2(u− τ)| .

Since f2 is continuous on R+, for given ε > 0 there exists η > 0 such that

|I1| ≤ ε, ∀ δ < η.

Therefore |D(t, δ)| ≤ |I1|+ |I2| < ε, ∀ δ < η.

That is, for given ε > 0 there exists η > 0 such that |D(t, δ)| < ε, ∀ δ < η.

Hence f1 ∗ f2 is continuous at t ∈ R+.

Now, let α, β ∈ R+.

Consider∫ β

α

[f1 ∗ f2(t+ δ)− f1 ∗ f2(t)] dt =

∫ α+δ

α

f1 ∗ f2(t) dt−∫ β+δ

β

f1 ∗ f2(t) dt.

Write F1,2(x) =∫ x

0f1 ∗ f2(t) dt. Then

supα, β∈R+

∣∣∣∣∫ β

α

[f1 ∗ f2(t+ δ)− f1 ∗ f2(t)] dt

∣∣∣∣ ≤ supα∈R+

|F1,2(α + δ)− F1,2(α)|

+ supβ∈R+

|F1,2(β + δ)− F1,2(β)| .

Since F1,2 is an indefinite HK integral of f1 ∗ f2 on R+, it is continuous on

R+ [5]. Therefore for given ε > 0 there exists η > 0 such that

|F1,2(ξ + δ)− F1,2(ξ)| < ε

2, ∀ δ < η and for all ξ ∈ R+.

Hence we have

supα, β∈R+

∣∣∣∣∫ β

α

[f1 ∗ f2(t+ δ)− f1 ∗ f2(t)] dt

∣∣∣∣ < ε, ∀ δ < η.

134

Thus for given ε > 0 there exists η > 0 such that A‖f1∗f2(t+δ)−f1∗f2(t)‖ <

ε, ∀ δ < η.

This shows that f1 ∗ f2(t) is continuous at t ∈ R+ with respect to the Alex-

iewicz norm.

The following result concerns with necessary and sufficient conditions

on fn and gn : [0, t] → R so that the HK-convolution as an operator is

continuous. It involves either uniform boundedness or uniform convergence

of the indefinite HK integral of fn.

Theorem 5.4.8. Let {fn} be a sequence of HK-integrable functions, fn :

[0, t] → R, ∀ n such that∫ t

0fn =

∫ t0f as n → ∞ for some HK-integrable

function f : [0, t]→ R. Define Fn (x) =∫ x

0fn, F (x) =

∫ x0f .

Let {gn} be a sequence of uniform bounded variation functions, gn : [0, t]→

R, ∀ n, such that gn → g pointwise on [0, t], where g : [0, t] → R. Then

fn ∗ gn(t)→ f ∗ g(t) as n→∞ if and only if

i) Fn → F uniformly on [0, t] as n→∞.

ii) Fn → F pointwise on [0, t], {Fn} is uniformly bounded and V (gn − g)→

0.

Proof. i) Suppose Fn → F uniformly on [0, t] as n→∞.

By assumption and multiplier theorem [1], fn ∗ gn(t) exists at t ∈ R+ and

fn ∗ gn(t) =

∫ t

0

fn(u) gn(t− u) du and f ∗ g(t) =

∫ t

0

f(u) g(t− u) du.

Consider fn ∗ gn(t)− f ∗ g(t) = I1 + I2,

where

I1 =

∫ t

0

[fn(u)− f(u)] gn(t− u) du, I2 =

∫ t

0

[gn(t− u)− g(t− u)] f(u) du.

135


I1 = gn(t− u)

∫ t

0

[fn(u)− f(u)] du−∫ t

0

∫ y

0

[fn(u)− f(u)] dgn

= gn(t− u)[Fn(t)− F (t)]−∫ t

0

[Fn(u)− F (u)] dgn.

Then

|I1| ≤Max0≤u≤t

|Fn(u)− F (u)|∫ t

0

dgn +M |Fn(t)− F (t)|

−→ 0 as n→∞ (Since Fn → F uniformly).

Hence I1 → 0 as n→∞.

And by integrating by parts, we get

I2 = {gn(t− u)− g(t− u)}∫ t

0

f −∫ t

0

F dgn +

∫ t

0

F dg.

By assumption the first term tends to 0 and since F is continuous and each

gn is of bounded variation with gn → g as n → ∞, so we have∫ t

0F dgn →∫ t

0F dg. Hence I2 → 0 as n→∞. Therefore fn ∗ gn(t)→ f ∗ g(t) as n→∞.

Now for the converse part, suppose that either Fn 9 F on [0, t] or Fn−F → 0

not uniformly on [0, t].

Then there is a sequence {xi}i∈Λ in [0, t] on which Fn 9 F uniformly. So,

by Bolzano-Weierstrass theorem [7], we can find a subsequence {yi}i∈Γ of

{xi}i∈Λ, Γ ⊂ Λ, such that Fn(yi) 9 F (yi) for all i and yi → y as i→∞, i.e.,

Fn(yi)− F (yi) 9 0.

Let us assume without loss of generality that 0 < yn ≤ y ≤ t. Let H be

Heaviside step function.

Define

gn(t− u) =

H(u− yn), n ∈ Γ

H(u− y), otherwise

136

and g(t− u) = H(u− y).

Then we have, for n ∈ Γ, fn∗gn(t) = Fn(t)−Fn(yn) and f∗g(t) = F (t)−F (y).

Since Fn(yn) 9 F (yn) as n → ∞, yn → y as n → ∞ and F is continuous,

we have Fn(yn) 9 F (y).

But Fn(t) → F (t) as n → ∞. Therefore fn ∗ gn(t) 9 f ∗ g(t) as n → ∞

which is contradiction. Hence we must have Fn → F uniformly on [0, t].

ii) Suppose Fn → F pointwise on [0, t], Fn is uniformly bounded and V (gn−

g)→ 0.

Consider fn ∗ gn(t)− f ∗ g(t) = I1 + I2,

where

I1 =

∫ t

0

fn(u) [gn(t− u)− g(t− u)] du, I2 =

∫ t

0

[fn(u)− f(u)] g(t− u) du.


I1 = [gn(t− u)− g(t− u)]

∫ t

0

fn −∫ t

0

∫ y

0

fn d(gn − g)

= [gn(t− u)− g(t− u)] Fn(t)−∫ t

0

Fn(y) d(gn − g).

Therefore

|I1| ≤ |gn(t− u)− g(t− u)| |Fn(t)|+ |Fn(y)| V (gn − g).

Since {Fn} is uniformly bounded, we have |Fn| ≤ M , ∀ n and for some

constant M .

Therefore

|I1| ≤ |gn(t− u)− g(t− u)|M +M V (gn − g).

By assumption gn → g as n→∞ and also V (gn − g)→ 0.

Hence I1 → 0 as n→∞.

Again by integrating by parts, we write

I2 = g(t− u) {Fn(t)− F (t)} −∫ t

0

(Fn − F ) dg. (5.3)

137

By assumption Fn(t)→ F (t), so the first term tends to 0. And, since |Fn| ≤

M , Fn → F and g is of bounded variation, by the dominated convergence

theorem for Riemann-Stieltjes integral [6], we have∫ t

0

Fn dg →∫ t

0

F dg as n→∞.

Therefore by (5.3), I2 → 0 as n→∞.

Hence fn ∗ gn(t)→ f ∗ g(t) as n→∞.

Now for the converse part suppose there is c ∈ (a, b) such that Fn(c)−F (c) 9

0 as n→∞.

Let gn(t− x) = g(t− x) = H(x− c).

Then∫ t

0

fn(u) gn(t− u) du = Fn(t)− Fn(c),

∫ t

0

f(u) g(t− u) du = F (t)− F (c).

Since Fn(t)→ F (t) and Fn(c) 9 F (c), we have Fn(t)−Fn(c) 9 F (t)−F (c)

as n→∞. That is,

fn ∗ gn(t) 9 f ∗ g(t) as n→∞

which is contradiction. Hence we must have Fn → F pointwise on [0, t].

Now if {Fn} is not uniformly bounded, then there is a sequence {xi}i∈Λ in

[0, t] on which |Fn| → ∞.

Therefore by Bolzano-Weierstrass theorem [7], we can find a subsequence

{yi}i∈Γ of {xi}i∈Λ, Γ ⊂ Λ, such that Fn(yi) ≥ 1 for all n and Fn(yi) → ∞

and yi → y.

Without loss of generality let us take 0 ≤ yi ≤ y ≤ t.

Define gn(t− u) = H(u−yi)√Fn(yi)

.

Then V (gn) ≤ 1, g = 0 and V (gn − g) = 0 and∫ t

0

fn(u) gn(t− u) du =

∫ t

0

fn(u)H(u− yi)√

Fn(yi)du

138

=

∫ t

yi

fn(u)√Fn(yi)

du

=Fn(t)√Fn(yi)

−√Fn(yi).

Therefore ∫ t

0

fn(u)gn(t− u) du→ −∞ as n→∞

and ∫ t

0

f(u)g(t− u) du = 0.

Hence fn ∗ gn 9 f ∗ g as n→∞ which is contradiction and so we must have

{Fn} is uniformly bounded sequence.

5.5 Inversion

Since in case of Henstock-Kurzweil integral, the integration process and dif-

ferentiation process are inverse of each other. Here we shall establish the

inverse of Henstock-Kurzweil Laplace transform by using Post’s generalized

differentiation method [8]. For this we shall need following lemmas:

Lemma 5.5.1. Let η be such that 0 < η < b − a, and h(x) ∈ C2(a ≤ x ≤

a+ η), h′(a) = 0, h

′′(a) < 0, h(x) is nonincreasing on (a, b]. Then∫ b

a

ek h(x) dx −→ ek h(a)

(−π

2k h′′(a)

) 12

k →∞.

Proof. Since h(x) ∈ C2(a ≤ x ≤ a+ η), h′′(x) is continuous.

Let ε > 0 be such that 0 < ε < −h′′(a).

Choose δ > 0, δ < η such that |h′′(x)− h′′(a)| < ε, a ≤ x ≤ a+ δ.

That is,

h′′(a)− ε < h

′′(x) < h

′′(a) + ε, a ≤ x ≤ a+ δ. (5.4)

139

Consider the integral

Ik =

∫ b

a

ek [h(x)−h(a)] dx.

We write Ik = I′

k + I′′

k ,

where

I′

k =

∫ a+δ

a

ek [h(x)−h(a)] dx, I′′

k =

∫ b

a+δ


Since h is nonincreasing on (a, b], we have h(x) ≤ h(a+ δ) for all x > a+ δ.

Therefore

I′′

k ≤∫ b

a+δ

ek [h(a+δ)−h(a)] dx

and

|I ′′k | ≤ ek [h(a+δ)−h(a)] (b− a− δ).

Hence I′′

k → 0 as k →∞.

Now consider

I′

k =

∫ a+δ

a


By Taylor’s series with remainder, we have

h(x)− h(a) ≈ h′′(ξ)

(x− a)2

2, a ≤ ξ ≤ a+ δ.

Therefore

I′

k =

∫ a+δ

a

ek h′′

(ξ)(x−a)2

2 dx.

So by (5.4), we can write∫ a+δ

a

ek [h′′

(a)−ε] (x−a)22 dx ≤

∫ a+δ

a

ek h′′

(ξ)(x−a)2

2 dx ≤∫ a+δ

a

ek [h′′

(a)+ε](x−a)2

2 dx.

That is, ∫ a+δ

a

ek [h′′

(a)−ε] (x−a)22 dx ≤ I

′

k ≤∫ a+δ

a

ek [h′′

(a)+ε](x−a)2

2 dx. (5.5)

Consider

J =

∫ a+δ

a

ek [h′′

(a)−ε] (x−a)22 dx.

140

After substituting x− a = u, we get

J =

∫ δ

0

ek2

[h′′

(a)−ε] u2 du

=1√

−k2

(h′′(a)− ε)

∫ δ√−k2

(h′′ (a)−ε)

0

e−u2

du

−→ 1√−k2

(h′′(a)− ε)

∫ ∞0

e−u2

du as k →∞

=1√

−k2

(h′′(a)− ε)

√π

2as k →∞.

Therefore∫ a+δ

a

ek [h′′

(a)−ε] (x−a)22 dx =

(π

−2k (h′′(a)− ε)

) 12

as k →∞.

Similarly,∫ a+δ

a

ek [h′′

(a)+ε](x−a)2

2 dx =

(π

−2k (h′′(a) + ε)

) 12

as k →∞.

Therefore (5.5) becomes(π

−2k (h′′(a)− ε)

) 12

≤ I′

k ≤(

π

−2k (h′′(a) + ε)

) 12

.

Since ε > 0 is arbitrary, we have(π

−2k h′′(a)

) 12

≤ I′

k ≤(

π

−2k h′′(a)

) 12

.

Hence

I′

k −→(

π

−2k h′′(a)

) 12

as k →∞.

So ∫ b

a

ek [h(x)−h(a)] dx −→(

π

−2k h′′(a)

) 12

as k →∞.

That is, ∫ b

a


(π

−2k h′′(a)

) 12

as k →∞.

141

Lemma 5.5.2. Let η be such that 0 < η < b − a, and h(x) ∈ C2(b − η ≤

x ≤ b), h′(b) = 0, h

′′(b) < 0, h(x) is nonincreasing on [a, b). Then∫ b

a

ek h(x) dx −→ ek h(b)

(π

−2k h′′(b)

) 12

as k →∞.

Proof. The proof is similar to that of previous lemma.

Lemma 5.5.3. Let η be such that 0 < η < b − a, and h(x) ∈ C2(a ≤

x ≤ a + η), h′(a) = 0, h

′′(a) < 0, h(x) is nonincreasing on (a, b]. Suppose

φ(x) ∈ HK([a, b]). Then∫ b

a

φ(x) ek h(x) dx −→ φ(a) ek h(a)

(π

−2k h′′(a)

) 12

as k →∞.

Proof. Since h(x) ∈ C2(a ≤ x ≤ a+ η), we have h′′(x) is continuous.

Let ε > 0 be such that 0 < ε < −h′′(a).

Choose δ > 0, δ < η such that |h′′(x)− h′′(a)| < ε, a ≤ x ≤ a+ δ.

That is,

h′′(a)− ε < h

′′(x) < h

′′(a) + ε, a ≤ x ≤ a+ δ.

Consider the integral

Ik =

∫ b

a

[φ(x)− φ(a)] ek [h(x)−h(a)] dx.

We show that Ik → 0 as k →∞.

Let

gk(x) =

ek [h(x)−h(a)], x ∈ [a, b]

0, x = a.

We claim that gk → 0 as k →∞.

Since h is nonincreasing on (a, b], we have h(x)− h(a) < 0 for all x ∈ (a, b].

Therefore gk(x) = ek [h(x)−h(a)] −→ 0 as k →∞.

Set g(x) = 0 for all x ∈ [a, b]. Then

gk(x) = ek [h(x)−h(a)] −→ 0 = g(x) for all x ∈ [a, b], as k →∞.

142

Since φ(x) ∈ HK([a, b]), the integral∫ b

a

[φ(x)− φ(a)] dx

exists.

Observe that the sequence {gk(x)} is of uniform bounded variation on [a, b]

with gk(x)→ 0 as k →∞.

Then by [Cor.3.2, [11]], we have∫ b

a

[φ(x)− φ(a)] ek [h(x)−h(a)] dx −→∫ b

a

[φ(x)− φ(a)] · 0 dx as k →∞

−→ 0 as k →∞.

Therefore ∫ b

a

φ(x) ek [h(x)−h(a)] dx = φ(a)

∫ b

a


That is, ∫ b

a

φ(x) ek h(x) dx = φ(a)

∫ b

a

ek h(x) dx.

But by lemma (5.5.1), we have∫ b

a


(−π

2k h′′(a)

) 12

as k →∞.

Therefore∫ b

a

φ(x) ek h(x) dx −→ φ(a) ek h(a)

(−π

2k h′′(a)

) 12

as k →∞.

Lemma 5.5.4. Let η be such that 0 < η < b − a, and h(x) ∈ C2(b − η ≤

x ≤ b), h′(b) = 0, h

′′(b) < 0, h(x) is nonincreasing on [a, b) and suppose

φ(x) ∈ HK([a, b]). Then∫ b

a

φ(x) ek h(x) dx −→ φ(b) ek h(b)

(−π

2k h′′(b)

) 12

as k →∞.

143

Proof. The proof is similar to that of the previous lemma.

Now we are ready to obtain the inversion theorem for Henstock-Kurzweil

Laplace transform.

Theorem 5.5.5. (Inversion Theorem) If f(t) ∈ HK(R+) and if the HK-

Laplace transform of f(t), L{f}(s) exists, then

limk→∞

1

k!

(k

t

)k+1 ∫ ∞0

e−kut uk f(u) du = f(t).

Proof. The inversion theorem follows immediately, if we can prove the fol-

lowing:

I)

limk→∞

1

k!

(k

t

)k+1 ∫ ∞t

e−kut uk f(u) du =

f(t)

2(5.6)

II)

limk→∞

1

k!

(k

t

)k+1 ∫ t

0

e−kut uk f(u) du =

f(t)

2. (5.7)

I) We have the relation∫ b

a

f(x) ek φ(x) dx −→ f(a) ek φ(a)

(−π

2k φ′′(a)

) 12

as k →∞.

Take a = t and φ(x) = ln(x)− xt

for all x ∈ [t,∞). Then we have∫ b

t

f(x) ek [ln(x)−xt

] dx −→ f(t) ek [ln(t)−1]

(−π

2k (−1t2

)

) 12

as k →∞.

Therefore∫ b

t

f(x) xk e−kxt dx −→ f(t) tk e−k

(π t2

2k

) 12

as k →∞.

That is, ∫ b

t

f(x) xk e−kxt dx −→ f(t) tk+1 e−k

( π2k

) 12

as k →∞.

144

Therefore as k →∞, we have

1

k!

(k

t

)k+1 ∫ b

t

f(x) xk e−kxt dx =

1

k!

(k

t

)k+1

f(t) tk+1 e−k( π

2k

) 12

=

(k

e

)kk

k!f(t)

( π2k

) 12.

The Stirling’s formula is given by(ne

)n √2πn→ n!, using this we can write

1

k!

(k

t

)k+1 ∫ b

t


k√2πk

f(t)( π

2k

) 12

as k →∞

=f(t)

2as k →∞.

This is true for every b ∈ R, b > t. Therefore by Hake’s theorem [1], we have

limk→∞

1

k!

(k

t

)k+1 ∫ ∞t


f(t)

2.

II) Again we have the relation∫ b

a

f(x) ek φ(x) dx −→ f(b) ek φ(b)

(−π

2 k φ′′(b)

) 12

as k →∞.

Take b = t and φ(x) = ln(x)− xt, for all x ∈ (0, t]. Then we get

limk→∞

1

k!

(k

t

)k+1 ∫ t

a


f(t)

2.

By Hake’s theorem [1], we have

limk→∞

1

k!

(k

t

)k+1 ∫ t

0


f(t)

2.

Therefore from (5.6) and (5.7), we can write

limk→∞

1

k!

(k

t

)k+1 ∫ t

0

f(x) xk e−kxt dx = f(t).

145

5.6 Examples

Here we shall give some examples of HK-Laplace transformable functions

whose classical Laplace transform do not exist. However we could not able

to find the exact form of the HK-Laplace transform of these functions.

Example 5.6.1. Consider the function g(t) = e−st sin tln(t+2)

, s ∈ C.

Let f(t) = e−st sin t and φ(t) = 1ln(t+2)

.

Observe that the function φ(t) is differentiable on R+ and φ′ ∈ L1(R+) and

if F (t) =∫ t

0e−su sinu du, Re.s > 0, then |F (t)| ≤ M for all t ∈ (0,∞),

Re.s > 0 for some constant M > 0.

Now

F (t) φ(t) =

{e−st(s sin t− cos t)

s2 + 1+

1

s2 + 1

}· 1

ln(t+ 2).

Then

limt→∞

F (t) φ(t) = limt→∞

{e−st(s sin t− cos t)

s2 + 1+

1

s2 + 1

}· 1

ln(t+ 2)

−→ 0.

Therefore limt→∞

F (t) φ(t) exists.

Hence by Du-Bois-Reymond’s test [1], we have e−st sin tln(t+2)

∈ HK(R+), Re.s > 0.

That is, ∫ ∞0

e−st sin t

ln(t+ 2)dt exists, Re.s > 0.

Hence the HK-Laplace transform of the function g(t) = sin tln(t+2)

exists, Re.s >

0.

Example 5.6.2. Consider the function f(t) = cos t√t

.

We have

F (t) =

∫ t

0

cosu√udu = 2

∫ √t0

cosu2 du.

146

Let φ(t) = e−st, s ∈ C. Then φ′(t) = −s e−st ∈ L1(R+), Re.s > 0.

Now

F (t)φ(t) = 2 e−st∫ √t

0

cosu2 du.

Then

limt→∞

F (t)φ(t) = 2 limt→∞

e−st∫ √t

0

cosu2 du

= 2 limt→∞

e−st · limt→∞

∫ √t0

cosu2 du

= 0, Re.s > 0.

Therefore by Du-Bois-Reymond’s test [1], we have f(t)φ(t) ∈ HK(R+),

Re.s > 0.

Hence the HK-Laplace transform of the function f(t) = cos t√t

exists for Re.s >

0.

Example 5.6.3. Let φ(t) = e−σt

t, t ∈ R+, σ > 0.

This function is continuous on R+ and

φ(t+ 1) =1

(t+ 1) eσ(t+1)<

1

t eσt= φ(t) for all t ∈ R+

and also limt→∞

φ(t) = limt→∞

1t eσt

= 0, σ > 0.

Therefore φ(t) is decreasing to zero as t→∞.

Observe that the integral ∫ ∞0

e−σt

tdt, σ > 0

diverges.

Now let n0 ∈ N be such that 0 < (1 + 4n0) π4ω

.

For t ∈ R, we have | sinωt| ≥ 1√2

if and only if ωt ∈ ∪∞n=n0[(1 + 4n)π

4, (3 +

4n)π4].

147

If n ∈ N, then (3 + 4n) π4< (1 + n) π.∫ (1+n)π

0

e−σt

t| sin t| dt =

n∑j=n0

∫ (3+4j)π4

(1+4j)π4

e−σt

t| sinωt| dt

≥n∑

j=n0

∫ (3+4j)π4

(1+4j)π4

e−σt

t

1√2dt

≥ 1√2

n∑j=n0

∫ (3+4j)π4

(1+4j)π4

1

(3 + 4j) π4eσ(3+4j)π

4

dt

=1√2

n∑j=n0

4

(3 + 4j) π es(3+4j)π4

· π2

≥ π

2√

2

n∑j=n0

1

(1 + j) π eσ(1+j)π.

On the other hand, we have∫ (1+n)π

0

e−σt

tdt =

∫ n0π

0

e−σt

tdt+

∫ (1+n)π

n0π

e−σt

tdt

=

∫ n0π

0

e−σt

tdt+

n∑j=n0

∫ (1+j)π

jπ

e−σt

tdt

≤∫ n0π

0

e−σt

tdt+

n∑j=n0

∫ (1+j)π

jπ

e−σjπ

jπdt

=

∫ n0π

0

e−σt

tdt+

n∑j=n0

e−σjπ

jπ· π.

But since e−σt

t∈ HK(R+), we have

∫∞0

e−σt

tdt =∞.

Therefore∑∞

j=n0

e−σjπ

jπ=∞.

So e−σt

tsinωt /∈ L1(R+), σ > 0.

Similarly, e−σt

tcosωt /∈ L1(R+), σ > 0.

Now let f(t) = sinωt, φ(t) = e−σt

t, t ∈ R.

Observe that φ(t) is monotone function and φ(t)→ 0 as t→∞.

And∣∣∣∫ t0 sinωu du

∣∣∣ ≤ 2|ω| , ω 6= 0 for all t ∈ R+. That is, F is bounded.

Therefore by Chartier-Dirichlet’s test [1], we have e−σt

tsinωt ∈ HK(R+).

148

Similarly, e−σt

tcosωt ∈ HK(R+).

Hence e−σt

t{cosωt− i sinωt} ∈ HK(R+).

That is,∫∞

0e−σt

te−iωt dt exists, σ > 0.

Thus∫∞

0e−st

tdt exists, Re.s > 0 as a HK integral, where s = σ + iω.

149

References

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150

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151

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