37.2. LAPLACE TRANSFORM 753

Definition 37.2.1 A function φ has exponential growth on [0,∞) if there are positive con-stants λ ,C such that |φ (t)| ≤ Ceλ t for all t. Then for s > λ , one defines the Laplacetransform L φ (s)≡

∫∞

0 φ (t)e−stdt.

Theorem 37.2.2 If s is a complex number and Res > λ where |φ (t)| ≤Ceλ t , and

f (s)≡∫

0e−st

φ (t)dt

then for Res > λ ,

limh→0

f (s+h)− f (s)h

≡ f ′ (s) =∫

0(−t)e−st

φ (t)dt

Thus s→ f (s) is analytic on Res > λ .

Proof: Let Res > λ . s will be complex as will h.∫∞

0

e−(s+h)t − e−st

hφ (t)dt +

∫∞

0te−st

φ (t)dt =∫

0e−st

(e−ht −1

h+ t)

φ (t)dt

Then

e−ht −1h

+ t =1h

(∞

∑k=0

(−1)k hktk−1

)+ t

=

(∞

∑k=1

(−1)k hk−1tk

)+ t

= h∞

∑k=2

(−1)k hk−2tk

Thus ∣∣∣∣(e−ht −1h

+ t)∣∣∣∣≤ |h| t2e|h|

and so ∣∣∣∣∣∫

0

e−(s+h)t − e−st

hφ (t)dt +

∫∞

0te−st

φ (t)dt

∣∣∣∣∣≤∫

0|h| t2e|h|te−Re(s)teλ tdt

which clearly converges to 0 since for all |h| sufficiently small,

e|h|te−Re(s)teλ t ≤ e−(Re(s)−(λ+ε))t

where ε is small enough that Re(s) > λ + ε . Thus the integral is finite for all |h| smallenough and it is multiplied by |h|. ■

This shows that f is analytic on Re(s) > λ . Hence it has all derivatives. In fact, youcan do a similar computation to the above and verify that

f (k) (s) =∫

0(−t)k e−st

φ (t)dt

37.2. LAPLACE TRANSFORM 753Definition 37.2.1 A function ¢ has exponential growth on |0,°°) if there are positive con-stants A,C such that | (t)| < Ce* for all t. Then for s > A, one defines the Laplacetransform £O (s) = fy o(t)e “dt.Theorem 37.2.2 If s is a complex number and Res > A where | (t)| < Ce”, andf(s) = [eo (atthen for Res >A,lim Porn= Le) =f) = | (newowarh>0Thus s — f (s) is analytic on Res > A.Proof: Let Res >. s will be complex as will h.eT (sthyt 00 ew ht _ 1[° t)dt+ [ te “o = a= | e™ +t) (t)dt0 A. 0 hTheneM] 1/< kt= - —1)*nkek—1 ) +1h + h d ) += (yp) +ek=1= hy (-1)'ak 74k=2Thuselt 2 th( 1)| <|niremand so<[ |n|t7e Ihlt 6 —Re(s (s)t At dyew (sthyt[° t)dt+ [ te “do (t)dt0 awhich clearly converges to 0 since for all |h| sufficiently small,elhlt e(s)tp At <e (Re(s)—(A+e))twhere € is small enough that Re(s) > A+. Thus the integral is finite for all |f| smallenough and it is multiplied by |h|. HiThis shows that f is analytic on Re(s) >A. Hence it has all derivatives. In fact, youcan do a similar computation to the above and verify that£9 (5) = [ (oke"o ar