漸進行為與拉普拉斯轉換的關係

國立臺灣大學理學院數學所 碩士論文

Department or Graduate Institute of Mathematics College of Science

National Taiwan University Master Thesis

漸進行為與拉普拉斯轉換的關係 Asymptotic behavior on Laplace Transform

蘇哲寬 Jhe-Kuan Su

指導教授﹕陳逸昆 教授

Advisor: Professor I-Ken Chen

中華民國 108 年 7 月

中文摘要

在這篇論文中，我們主要討論拉普拉斯轉換如何影響一個給定函數 的漸進行為。研究發現，一個函數 f(t)在 t 很大的行為會影響 L(f)(s) 在 s 很小的漸進行為。更進一步在某些狀況下我們可以藉由 L(f)(s) 的漸進行為去推得 f(t)在 t 很大的漸進行為。這篇論文主要使用實分

析技巧。我們還將此結果與 Abelian and Tauberian Theorem of

Laplace transform 進行比較。

Abstract

In the present thesis, we concern the relation between the function

and its Laplace transform. More precisely, the asymptotic behavior

when t is large (small) implies asymptotic behavior of its Laplace

transform when s is small (large) and vice versa, especially, for

critical exponent we provide a simple sufficient and necessary

condition. Our approach only involves real analysis. We will also

compare the result with Abelian and Tauberian Theorem of Laplace

transform.

目 錄

口試委員會審定書……… i

中文摘要……… ii

英文摘要………. iii

第一章 Introduction……… 1

第二章 Relation between behaviour of the function and its transformed type ( 1 ) … … … . . 2

第三章 Relation between behaviour of the function and its transformed type ( 2 ) … … … . . 8

第 四 章 Non-informativeness of polynomial term of the transformed f u n c t i o n … … … . . 1 4 第五章 The difference with the previous result ……….…. 15

參考文獻……….…… 19

1 Introduction

Throughout this thesis, we assume that f is a measurable function on R and e^−stf (t) ∈ L₁ for any s > 0.

We define

L(f )(s) ≡ Z ∞

0

f ∗ e^−stdt (1)

and

F (t) ≡ Z t

0

f (u)du. (2)

It was found that the information about the asymptotic behaviour of F near infinity implies some information about the asymptotic behaviour behaviour of L(f) near zero and vice versa, for example the famous Tauberian Theorem for Laplace transform.

According to Feller in theorem of [1], we know that given ρ ≥ 0 if we have

”scaling behaviour” of f near the infinity, i.e.,

t→∞lim F (xt)

F (t) = x^ρ, (3)

then L(f) will have the scaling behaviour near origin, i.e.,

t→0lim

L(f )(xt)

L(f )(t) = x^−ρ (4)

and vice versa.

In [2], if F has some ”logistic” behaviour near infinity, then L(f) has some logistic behaviour near zero and vice versa. In [3], a complex analysis approach is used. With some additional condition, the relation of asymptotic behaviour are characterized.

In the thesis, we will relax the condition and still provide a similar result.

First we found that :

Theorem 1.1.

(a) For integer n ≥ 0, when limt→∞f (t)

tⁿ = 1, then

s→0lim

L(f )(s)

s⁻ⁿ⁻¹ = n!. (5)

(b) If limt→∞

f (t)

1 t

= 1, then lim

s→0

L(f )(s)

− log s = 1. (6)

(c) For integer n < −1, if limt→∞f (t) tⁿ = 1, then

Z ∞ 0

e^−stf (t)dt−Σ⁻ⁿ⁻²_i=0 Z ∞

0

(−1)ⁱtⁱ

i! f (t)dtsⁱ=(−1)⁻ⁿs⁻ⁿ⁻¹log s

(−n − 1)! +o(s⁻ⁿ⁻¹log s (−n − 1)! )

(7) as s → 0

Then we use that result to found that

Theorem 1.2.

(a) For integer n ≥ 0 when limt→∞ F (t)

tⁿ = 1, then lim

s→0

L(f )(s)

s⁻ⁿ = n! (8)

(b) If limt→∞

F (t)

log t = 1 then

s→0lim

L(f )(s)

− log s = 1 (9)

and vice versa if f is non negative.

(c) For integer n < −1 when, limt→∞

−F (t)+R∞ 0 f (u)du

(n+1)tⁿ⁺¹ = 1, then L(f )(s) − Σ⁻ⁿ⁻²_i=0

Z ∞ 0

(−1)ⁱtⁱ

i! f (t)dtsⁱ =(−1)⁻ⁿs⁻ⁿ⁻¹log s

(−n − 1)! + o(s⁻ⁿ⁻¹log s (−n − 1)! )

(10) as s → 0, and the converse is not true

The organization of the remaining part of the thesis is as follows:

In Section 2, we will discuss how the behaviour of f affect the behaviour of L(f)(s). In Section 3, we will discuss the relation between the behaviour of F(t) and the behaviour of L(f)(s). In Section 4, we will show that the polynomial terms of L(f) tell nothing about of f. In Section 5, we will compare our result with previous result in [1],[2].

2 Relation between behaviour of the function and its transformed type (1)

The proof of Theorem 1.1 is as below.

Proof.

(a) If

f (x) = tⁿ+ h(t) where ^h(t)_tn → 0 as t → ∞, then we have

Z ∞ 0

e^−stf (t)dt

= Z ∞

0

e^−st(tⁿ+ h(t))dt

= n!

sⁿ⁺¹+ Z ∞

0

e^−sth(t)dt.

(11)

So, it suffice to show that:

Z ∞ 0

e^−sth(t)dtsⁿ⁺¹→ 0 (12)

as s → 0 .

Since h(t)t⁻ⁿ→ 0 as t → ∞, we have

∀ε > 0, ∃δ > 0 s.t. |h(t)| < εtⁿ for t ∈ [δ, ∞]

In doing so,

Z ∞ 0

e^−sth(t)dt

= Z ∞

δ

e^−sth(t)dt + Z δ

0

e^−sth(t)dt,

(13)

but

| Z ∞

δ

e^−sth(t)dt|

≤ Z ∞

δ

e^−st|h(t)|dt

≤ Z ∞

δ

e^−stεtⁿdt

≤ n!ε sⁿ⁺¹.

(14)

Plus,

| Rδ

0 e^−sth(t)dt s⁻ⁿ⁻¹ |

=|

Z δ 0

e^−stsⁿ⁺¹h(t)dt|

=sⁿ⁺¹ Z δ

0

e^−st|h(t)|dt

≤sⁿ⁺¹ Z δ

0

|h(t)|dt

(15)

(Note that h(t) is locally integrable.) which approach to 0 as s → 0.

So we have lim sup_s→0|R∞

0 e^−sth(t)| ≤ ε for any ε > 0.

which implies lim sup_s→0|R∞

0 e^−sth(t)| = 0

(b) and (c) For integer n ≥ 1,

if we have f (t) = _t¹n + o(_t¹n) as t → ∞, define

φn(x) := e^−x− Σⁿ⁻²_i=0 ⁽⁻¹⁾_i!^ixi

(−1)ⁿ⁻¹ (n−1)! xⁿ⁻¹

. (16)

For example, φ1(x) = e^−x, φ2(x) = ^e−x_−x⁻¹, φ3(x) = ^e−x−1+x_x2 ...and so on.2

It is not hard to know that this series of functions are all non negative.

Claim: for any n ≥ 1 (1) limx→0φn(x) = 1.

(2) limx→∞xφn(x) = n − 1.

(3) φn(x) is monotone decreasing function.

The first part and second part of the claim is trivial from the definition of φn(x).

So, we only do the third part:

Since the first part and second part of the claim, it suffice to show that all the critical point of this function lie in a decreasing function with initial value 1.

If it has a critical point says x and φ⁰_n(x) = 0, then

φ⁰_n(x)

=

(−e^−x− Σⁿ⁻²_i=1 ⁽⁻¹⁾_(i−1)!^ixi−1)(⁽⁻¹⁾_(n−1)!^nxn−1) − (−e^−x− Σⁿ⁻²_i=0 ⁽⁻¹⁾_(i)!^ixi)(⁽⁻¹⁾_(n−2)!^nxn−2) (_(n−1)!¹ )²x²ⁿ⁻²

=(−1)ⁿ(n − 1)!

xⁿ (−e^−x(x + n − 1) + Σⁿ⁻²_i=0 (−1)ⁱxⁱ(n − 1 − i)

i! )

(17) then we have

e^−x= Σⁿ⁻²_i=0 ⁽⁻¹⁾_i!^ixi(n − 1 − i)

x + n − 1 . (18)

Furthermore,

φn(x)

=

Σⁿ⁻²_i=0^{(−1)i xi}_i! (n−1−i)

x+n−1 − Σⁿ⁻²_i=0 ⁽⁻¹⁾_i!^ixi

(−1)ⁿ⁻¹ (n−1)! xⁿ⁻¹

= n − 1 x + n − 1.

(19)

That means all critical point lie in a strictly decreasing sequence function h(x) =_x+n−1ⁿ⁻¹ (note h(0)=1),

but if φ_n(x) is not monotone decreasing i.e. there exist x₁ < x₂ such that φ_n(x₁) < φ_n(x₂)

Then, if x1 6= 0, then there is a maximum point y of φn(x) on (x1, x2], and a minimum point z of φn(x) on [0,y). Note that y is a critical point,

so φn(y) = _y+n−1ⁿ⁻¹ < 1 which imply z 6= 0. z is also a critical point and φn(z) < φn(y), but this contradict the fact that all critical point of φn(x) lie in a strictly decreasing function f (x) = _x+n−1ⁿ⁻¹ .

If x₁= 0, this lead to the contradiction once we notice that f (x) = _x+n−1ⁿ⁻¹ <

1 for any x > 1.

By (1) and (3) of the claim, we have 0 ≤ φ_n ≤ 1 for any x ≥ 0.

By assumption and (2) of claim,we have

∀ε > 0, ∃M (ε) > 0

s.t. ∀t ∈ [M (ε), ∞), we have |f (t) −_t¹n| ≤ ε_t¹n and |φn(t)t| ≤ (n − 1)(1 + ε).

Note that

L(f )(s) − Σⁱ⁼ⁿ⁻²_i=0 ⁽⁻¹⁾_i!^isiR∞

0 tⁱf (t)dt

(−1)ⁿ⁻¹ (n−1)! sⁿ⁻¹

= Z ∞

0

φn(st)tⁿ⁻¹f (t)dt

= Z M (ε)

0

φn(st)tⁿ⁻¹f (t)dt + Z ¹_s

M (ε)

φn(st)tⁿ⁻¹f (t)dt + Z ∞

1 s

φn(st)tⁿ⁻¹f (t)dt

≤ Z M (ε)

0

φ_n(st)tⁿ⁻¹f (t)dt + Z ¹_s

M (ε)

tⁿ⁻¹|f (t)|dt + (n − 1) Z ∞

1 s

(1 + ε)² 1 st²dt

≤ Z M (ε)

0

φn(st)tⁿ⁻¹f (t)dt + Z ¹_s

M (ε)

(1 + ε)1

tdt + (n − 1) Z ∞

1 s

(1 + ε)² 1 st²dt

≤ Z M (ε)

0

φn(st)tⁿ⁻¹f (t)dt + (1 + ε)(log1

s− log M (ε)) + (n − 1) Z ∞

1 s

(1 + ε)² 1 st²dt

≤ Z M (ε)

0

tⁿ⁻¹|f (t)|dt + (1 + ε)(log1

s− log M (ε)) + (n − 1)(1 + ε)².

(20) Hence we have

lim sup

s→0

L(f )(s) − Σⁱ⁼ⁿ⁻²_i=0 ⁽⁻¹⁾_i!^isiR∞

0 tⁱf (t)dt

(−1)ⁿ⁻¹sⁿ⁻¹(− log s) ≤ 1. (21) On the other hand, for ^{φ−1n (1−ε)}_s > M (ε),

Z ∞ 0

φn(st)tⁿ⁻¹f (t)dt

=

Z ^{φ−1n (1−ε)}_s

0

φn(st)tⁿ⁻¹f (t)dt + Z ∞

φ−1 n (1−ε)

s

φn(st)tⁿ⁻¹f (t)dt

≥

Z ^{φ−1n (1−ε)}_s

0

φn(st)tⁿ⁻¹f (t)dt

= Z M (ε)

0

φn(st)tⁿ⁻¹f (t)dt + (1 − ε)

Z ^{φ−1n (1−ε)}_s

M (ε)

tⁿ⁻¹f (t)dt

= Z M (ε)

0

tⁿ⁻¹|f (t)|dt + (1 − ε)

Z ^{φ−1n (1−ε)}_s

M (ε)

(1 − ε)1 tdt

= Z M (ε)

0

tⁿ⁻¹|f (t)|dt + (1 − ε)

Z ^{φ−1n (1−ε)}_s

M (ε)

(1 − ε)1 tdt

= Z M (ε)

0

tⁿ⁻¹|f (t)|dt + (1 − ε)² Z

φ−1n (1−ε) s

M (ε)

1 tdt]

= Z M (ε)

0

tⁿ⁻¹|f (t)|dt + (1 − ε)²(log (φ⁻¹_n (1 − ε)

s ) − log M (ε))]

(22)

And we conclude that lim inf

s→0

L(f )(s) − Σⁱ⁼ⁿ⁻²_i=0 ⁽⁻¹⁾_i!^isiR∞

0 tⁱf (t)dt

(−1)ⁿ⁻¹sⁿ⁻¹(− log s) ≥ 1. (23)

Remark 2.1.

The method to prove the part (a) of Theorem also lead to the following result:

(a) If there is a non-negative integer n such that f (x) = tⁿ+ o(tⁿ) as t → 0

then

L(f )(s) = _sn+1^n! + o(_sn+1¹ ) as s → ∞.

Remark 2.2. The converse is not true.

3 Relation between behaviour of the function and its transformed type (2)

Before we prove the theorem 1.2,we introduce a useful lemma.

Lemma 3.1. L(log t)(s) =^{γ−log s}_s ,where γ =R∞

0 e^−ulog udu.

Proof.

L(log t)(s)

= Z ∞

0

e^−stlog tdt

= Z ∞

0

e^−ulogu s 1 sdu

= Z ∞

0

e^−u(log(u) − log s)1 sdu

=1 s(

Z ∞ 0

e^−ulog udu − Z ∞

0

e^−ulog sdu)

=1 s(

Z ∞ 0

e^−ulog udu − log s Z ∞

0

e^−udu)

=1

s(γ − log s).

(24)

Then, we will use this to prove Theorem 1

Proof. Here we will break the proof into three steps.

Step 1: The ”if” part of (a), (b) and (c) Step 2: The ”only if” part of (b)

Step 3: Offering the counter example of ”only if part of (c).

Step 1:

For (a),by Theorem 1.1,

L(f (t))(s) s

=L(

Z t 0

f (x)dx)(s)

= 1

n + 1((n + 1)!

sⁿ⁺² + o( 1 sⁿ⁺²)).

(25)

Therefore

L(f )(s) = n!

sⁿ⁺¹ + o( 1

sⁿ⁺¹). (26)

For (b),

L(f (t))(s) s

=L(

Z t 0

f (x)dx)(s)

=L(log t + g(t))(s) (where g(t)

log t → 0 as t → ∞)

=γ − log s

s + L(g(t))(s)

=γ − log s

s +

Z ∞ 0

e^−stg(t)dt.

(27)

Note that for any ε > 0 there is real number M (ε) > 0 s.t. |g(t)| ≤ ε log t whenever t > M (ε).

Therefore,

γ − log s

s +

Z ∞ 0

e^−stg(t)dt

=γ − log s

s +

Z M (ε) 0

e^−stg(t)dt + Z ∞

M (ε)

e^−stg(t)dt

≤γ − log s

s +

Z M (ε) 0

e^−stg(t)dt + Z ∞

M (ε)

e^−stε log tdt

≤(1 + ε)γ − log s

s +

Z M (ε) 0

g(t)dt.

(28)

That means for any ε > 0 lim sup

s→0

L(f (t))(s)

− log s ≤ 1 + ε. (29)

So we have

lim sup

s→0

L(f (t))(s)

− log s ≤ 1. (30)

Similarly,

lim sup

s→0

L(f (t))(s)

− log s ≥ 1 (31)

and the result follows.

For (c), the method is similar to (b)

Step 2:

Note that for any 0 < M < 1,

L(f )(s)

= Z ∞

0

e^−stf (t)dt

≥

Z −¹_slog M 0

e^−stf (t)dt

≥

Z −¹_slog M 0

M f (t)dt

≥M

Z −¹_slog M 0

f (t)dt

=M F (−1

slog M ).

(32)

Since

lim sup

s→0

L(f )(s)

− log s ≤ 1, (33)

we have

lim sup

s→0

F (−¹_slog M ) log¹_s ≤ 1

M. (34)

Hence,

lim sup

s→0

F (¹_s)

log_{−s log M}¹ ≤ 1

M. (35)

So,

lim sup

s→0

F (¹_s) log¹_s ≤ 1

M (36)

for any 0 < M < 1.

Then, we have

lim sup

s→0

F (¹_s)

log¹_s ≤ 1. (37)

On the other hand,

we suppose lim inf_t→∞^{F (t)}_{log t} < 1,

then there exist ti% ∞ such that F (ti) ≤ (1 − ε) log ti.

Let si:= t

−_{1− ε}^1−ε

2

i .

For convenience, we set γ := ₁₋^1−εε 2

. Then,

L(f )(sj)

= Z ∞

0

e^−sjtf (t)dt

= Z ∞

0

e⁻

t tγ

jf (t)dt

= Z t_j

0

e⁻

t tγ

jf (t)dt + Z ∞

tj

e⁻

t tγ

jf (t)dt.

(38)

Set G(t) ≡ (1 + δ) log t for t ≥ tj. Note that by (39)

∃M > 0s.t.∀t ≥ M,_{− log t}^{F (t)} ≤ 1 + δ.

For ti> M , by integration by part, we have Z ∞

t_j

e⁻

t tγ

j f (t)dt

= − e⁻

tj tγ

jF (tj) + s Z ∞

tj

e⁻

t tγ

jF (t)dt

≤ − e⁻

tj tγ

jF (tj) + s Z ∞

t_j

e⁻

t tγ

jG(t)dt

≤ − e⁻

tj tγ

jF (t_j) + e⁻

t tγ

jG(t_j) + Z ∞

t_j

e⁻

t tγ

jdG

≤e⁻

tj tγ

jG(tj) + Z ∞

tj

e⁻

t tγ

jdG

≤(1 + δ)e⁻

tj tγ

j log tj+ Z ∞

t_j

e⁻

t tγ

j dG.

(39)

Therefore Z tj

0

e⁻

t tγ

j f (t)dt + Z ∞

t_j

e⁻

t tγ

jf (t)dt

≤ Z t_j

0

f (t)dt + (1 + δ)e⁻

tj tγ

j log tj+ Z ∞

tj

e⁻

t tγ

j dG

≤(1 − ε) log tj+ (1 + δ)e⁻

tj tγ

j log tj+ Z ∞

t_j

e⁻

t tγ

jdG

≤ − (1 −ε

2) log s_j−(1 + δ) γ e⁻

tj tγ

j log s_j+ Z ∞

t_j

e⁻

t tγ

j(1 + δ)1 tdt

= − (1 −ε

2) log sj−(1 + δ) γ e⁻

tj tγ

j log sj+ (1 + δ)[log tje^−sjtj+ Z ∞

sjtj

log ue^−udu − e^−sjtjlog sj]

= − (1 −ε

2) log sj−(1 + δ) γ e^−s

γ−1 γ

j log sj+ (1 + δ)[−1

γlog sje^−s

γ−1 γ

j +

Z ∞ s

γ−1 γ j

log ue^−udu − e^−s

γ−1 γ

j log sj].

(40) Since s

γ−1 γ

j → ∞, we have lim inf

s→0

L(f )(s)

− log s < 1 −ε

2. (41)

which contradict the assumption.

Hence lim inft→∞

F (t) log t ≥ 1.

Combining (36), we finish the step 2.

Step 3:

For n < −1,

set f := Σ^∞_i=0δt_i(tⁿ⁺¹_i − tⁿ⁺¹_i+1), then

Z ∞ 0

φ_−n(st)t⁻ⁿ⁻¹f dt

= Z ∞

0

φ_−n(st)t⁻ⁿ⁻¹Σ^∞_i=0δt_i(tⁿ⁺¹_i − tⁿ⁺¹_i+1)dt

=Σ^∞_i=0φ_−n(sti)t⁻ⁿ⁻¹_i (tⁿ⁺¹_i − tⁿ⁺¹_i+1)

≤Σ^∞_i=0φ_−n(st_i)t⁻ⁿ⁻¹_i (tⁿ⁺¹_i )

=Σ^∞_i=0φ_−n(sti).

(42)

We assume that t1 ≥ 2, ti+1 ≥ (i + 1)ti (it is easily follow once we choose the subsequence of ti ).

Set sj := _t¹

j+1 (here j is large enough such that φ−n(x) < (1 + ε)(¹_x) for x ≥ (j + 2),

then using property of φ, we have Z ∞

0

φ_−n( t tj+1

)t⁻ⁿ⁻¹f dt

≤Σ^∞_i=0φ−n( t_i tj+1

)

=Σ^j+1_i=0φ_−n( t_i tj+1

) + Σ^∞_i=j+2φ_−n( t_i tj+1

)

≤Σ^j+1_i=0φ_−n( ti

t_j+1) + Σ^∞_i=j+2φ_−n(

i

Y

k=j+2

k)

≤Σ^j+1_i=0φ_−n( ti

t_j+1) + Σ^∞_i=j+2((−n − 1)(1 + ε) 1 Qi

k=j+2k)

=Σ^j+1_i=0φ_−n( ti

t_j+1) + (−n − 1)(1 + ε)Σ^∞_i=j+2( 1 Qi

k=j+2k)

≤Σ^j+1_i=0φ_−n( ti

t_j+1) + (−n − 1)(1 + ε)Σ^∞_i=j+2( 1 (j + 2)^i−j−1)

≤Σ^j+1_i=0φ_−n( t_i tj+1

) + (−n − 1)(1 + ε)( 1 j + 1)

≤(j + 2) + (−n − 1)(1 + ε)( 1 j + 1)

< < log t_j+1

= − log sj.

(43)

Given s ∈ (s_j+1, s_j), by monotonicity of φ, we know thatR∞

0 φ_−n(st)t⁻ⁿ⁻¹f dt is monotonically deceasing on s.

Hence,

Z ∞ 0

φ_−n(st)t⁻ⁿ⁻¹f dt

≤ Z ∞

0

φ_−n(sj+1t)t⁻ⁿ⁻¹f dt

≤(j + 3) + (−n − 1)(1 + ε)( 1 j + 2)

< < log tj+1

= − log s_j

≤ − log s.

(44)

Let g(t) := (1 + t)ⁿ+ Σ^∞_i=0δt_i(tⁿ⁺¹_i − tⁿ⁺¹_i+1).

It is obvious thatRt

0g(u)du 6= tⁿ+ o(tⁿ), but

L(f )(s) − Σ⁻ⁿ⁻²_i=0 R∞ 0

(−1)ⁱtⁱ

i! f (t)dtsⁱ= ⁽⁻¹⁾⁻ⁿ_(−n−1)!^{a−ns−n−1log s}+ o(^s−n−1_(−n−1)!^{log s}).

Then the result is followed.

4 Non-informativeness of polynomial term of the transformed function

Theorem 4.1. For any Ci∈ R, i = 1, 2, ..., m there exist f with compact support such that

Z ∞ 0

e^−stf (t)dt = C₀+ C₁s + C₂s²... + C_ms^m+ o(s^m). (45) Proof. Note that

Z ∞ 0

e^−st1_[a,b]dt = e^−as− e^−bs s

=Σ^∞_i=0(−1)ⁱ⁻¹(aⁱ⁺¹− bⁱ⁺¹) (i + 1)! sⁱ

=Σ^m_i=0(−1)ⁱ⁻¹(aⁱ⁺¹− bⁱ⁺¹)

(i + 1)! sⁱ+ O(s^m+1).

(46)

By choosing fj := 1_[0,j+1], for j = 0, ..., m, we have Z ∞

0

e^−stf_jdt = Σ^m_i=0(−1)ⁱ⁻¹(−(j + 1)ⁱ⁺¹)

(i + 1)! sⁱ+ O(s^m+1). (47)

So it suffice to show that the matrix A ≡















1 1

2 1

3

1 ... ^m+1₁

−1 2!

−4 2!

−9

2! ... ^−(m+1)_2! ²

1 3!

8 3!

27

3! ... ^(m+1)_3! ³ ...

(−1)^m (m+1)!

(−1)^m2^m+1 (m+1)!

(−1)^m3^m+1

(m+1)! ... ^(−1)m_(m+1)!^(m+1)m+1













 has non zero determinant.

But, the is equal to say the matrix(by elementary matrix operation)













1 2 3 ... m + 1

1² 2² 3² ... (m + 1)²

1³ 2³ 3³ ... (m + 1)³

...

1^(m+1) 2^(m+1) 3^(m+1) ... (m + 1)^(m+1)











 has zero determinant and it is a known fact.

5 The difference with the previous result

We will show the difference between asymptotic behaviour and scaling be- haviour.

The following theorem is prove in [1]:

Proposition 5.1. Let u(x) be a measurable function on [0, ∞) We define U (t) :=Rt

0u(dx) L(u)(s) :=R∞

0 e^−stdU for s > 0

Then, for any ρ > 0, the following two statements are equivalent:

(1)

t→∞lim

L(u)(^s_t) L(u)(¹_t) = 1

λ^ρ (48)

for any s > 0 (2)

t→0lim U (ts)

U (t) = s^ρ (49)

for any s > 0

Which is the difference of the condition between this result and the result in chapter 1?

It is obvious that if we have f (t) = Ctⁿ+ o(tⁿ) as t → ∞ for some C ∈ R, then limt→∞ f (xt)

t , but the converse is false even when n=1.

Here we provide example.

Let g(t) := t(1 + ₁₀¹ sin log log t), it is clear that g(t) 6= Ctⁿ+ o(tⁿ) for any C ∈ R.

However,

g(xt) g(t)

=xt(1 +₁₀¹ sin log log xt) t(1 +₁₀¹ sin log log t)

=x(1 + ₁₀¹ sin log log xt) (1 +₁₀¹ sin log log t)

=x(1 +

1

10sin log log xt −₁₀¹ sin log log t (1 + ₁₀¹ sin log log t) )

=x(1 + 1 10

sin log log xt − sin log log t (1 +₁₀¹ sin log log t) )

→x

(50)

as t → ∞.

(note that | sin log log xt − sin log log t| ≤ | log log xt − log log t|

≤ | log^{log xt}_{log t}| → 0 )

The following theorem is prove in [2]:

Proposition 5.2. For any increasing and right continuous function F : R⁺→ R with F (0⁺) = 0, let L(dF ) ≡R∞

0⁺e^−tsdF (s) then the following are equivalent:

(a) for every positive x and y (y 6= 1)

t→∞lim

F (tx) − F (t)

F (ty) − F (t) = log x

log y (51)

(b)

lim

t→0

L(dF )(tx) − L(dF )(t)

L(dF )(ty) − L(dF )(t)= log x

log y. (52)

Here we will show Theorem 5.2 and Theorem 1.2 won’t imply each other.

For instance, set g(t) := log t(1 + ε sin log log log t) and h(t) := log t +

√log t sin t.

Clearly, lim_{t→∞log t}^g(t) doesn’t exist.

However, g(xt) − g(t) g(et) − g(t)

=log xt(1 + ε sin log log log xt) − log t(1 + ε sin log log log t) log et(1 + ε sin log log log et) − log t(1 + ε sin log log log t)

=log x(1 + ε sin log log log xt) + log t(1 + ε sin log log log xt) − log t(1 + ε sin log log log t) (1 + ε sin log log log et) + log t(1 + ε sin log log log et) − log t(1 + ε sin log log log t)

=log x(1 + ε sin log log log xt) + log t(ε(sin log log log xt − sin log log log t)) (1 + ε sin log log log et) + log t(ε(sin log log log et − sin log log log t))) .

(53) Note that (Here without losing generosity we assume x > 1.)

| log t(ε(sin log log log xt − sin log log log t))|

≤| log t(ε(log log log xt − log log log t))|

≤|(1 − x)t log t(ε 1 log log t

1 log t

1 t)|

≤|(1 − x)(ε 1

log log t)| → 0

(54)

as t → ∞.

Hence

t→∞lim

log x(1 + ε sin log log log xt) + log t(ε(sin log log log xt − sin log log log t)) (1 + ε sin log log log et) + log t(ε(sin log log log et − sin log log log t)))

= lim

t→∞

log x(1 + ε sin log log log xt) (1 + ε sin log log log et)

= log x.

(55) It is obvious that limt→∞ h(t)

log t = 1.

However,

h(2et) − h(t) h(et) − h(t)

=log 2e +√

log 2et sin 2et −√

log t sin t 1 +√

log et sin et −√

log t sin t .

(56)

If t = ^πn_e , then it

=log 2e +√

log 2et sin 2et −√

log t sin t 1 +√

log et sin et −√

log t sin t

=log 2e −plog^πn_e sin^πn_e 1 −plog^πn_e sin^πn_e .

(57)

This doesn’t approach to log x when n → ∞.

Also it is shown in P.282 of [3] that:

(Here we simply the result by setting s₀ = 0 and consider s as complex number.)

If L(f )(s) = Σ^∞_n=0ansⁿ⁻¹+ log sΣ^∞_n=0bnsⁿ and the following additional assumption holds.

(1) L(f )(s) is analytic for Re(s) ≥ −δ, δ > 0 , except at s=0

(2) L(f )(s) → 0 uniformly as Im(s) → ∞ for −δ ≤ Re(s) ≤ γ for some γ (3)Rκ+i∞

κ−i∞ |L(f )(s)|ds < ∞ for −δ ≤ κ ≤ γ, Then we have

f = a₀− Σ^∞_n=0(−1)ⁿb_nn!t⁻ⁿ⁻¹+ o(t⁻ⁿ⁻¹) (58) as t → ∞.

References

[1] Feller, William. An introduction to probability theory and its applications.

Vol. II. Second edition, 1971,Pages 443

[2] Laurens de Haan. An Abel-Tauber Theorem for Laplace Transforms Journal of the London Mathematical Society, Volume s2-13, Issue 3, 1 August 1976, Pages 537

[3] H.S. Carslaw, J.C. Jaeger. Operation method in applied Mathematics,1947, Pages 282