396 CHAPTER 15. ISOLATED SINGULARITIES

and the third having θ ∈(

π

2 + arcsin(

cR1−β

), 3π

2 − arcsin(

cR1−β

)). Then,

∫CR

etzF (z)dz =∫ 3π

2 −arcsin(

cR1−β

)π2 +arcsin

(c

R1−β

) e(Rcosθ+iRsinθ)tF(

Reiθ)

Rieiθ dθ+ (15.15)

+∫ π

2 +arcsin(

cR1−β

)π2−arcsin( c

R )e(Rcosθ+iRsinθ)tF

(Reiθ

)Rieiθ dθ

+∫ 3π

2 +arcsin( cR )

3π2 −arcsin

(c

R1−β

) e(Rcosθ+iRsinθ)tF(

Reiθ)

Rieiθ dθ

Consider the last two integrals first. For large |z| , with z ∈ C∗R, the sum of the absolutevalues of these is no more than∣∣∣∣∣

∫ π2 +arcsin

(c

R1−β

)π2−arcsin( c

R )eR(cosθ)t C

RαRdθ

∣∣∣∣∣+∣∣∣∣∣∫ 3π

2 +arcsin( cR )

3π2 −arcsin

(c

R1−β

) eR(cosθ)t CRα

Rdθ

∣∣∣∣∣≤ CeR(cos( π

2−arcsin( cR )))t

(arcsin

( cR1−β

)+ arcsin

( cR

))R1−α

+CeR(cos( 3π2 +arcsin( c

R )))t(

arcsin( c

R1−β

)+ arcsin

( cR

))R1−α

Now from trig. identities, cos(

π

2 − arcsin(θ))= θ ,cos

( 3π

2 + arcsin(θ))= θ , and so

the above reduces to 2Cect(

arcsin(

cR1−β

)+ arcsin

( cR

))R1−α which converges to 0 as

R→ ∞. Recall 0 < β < α . It remains to consider the integral in 15.15. For large |z| ,the absolute value of this integral is no more than

∫ 3π2 −arcsin( c

R )

π2 +arcsin( c

R )eR(cosθ)t C

RαRdθ ≤Cπe

Rt cos(

π2 +arcsin

(c

R1−β

))R1−α =CπR1−α e−ctRβ

which converges to 0 as R→ ∞. ■

Proposition 15.8.4 If Re p < c for all p a pole of F (s) and if F (s) is meromorphic andsatisfies the growth condition 15.14, and if f (t) is defined by the Bromwich integral, thenF (s) is the Laplace transform of f (t) for large s.

Proof: This follows from Lemmas 15.8.2 and 15.8.3. These lemmas say that if F (s)satisfies the growth condition, then we can define a locally Lipschitz function f (t) in termsof that Bromwich integral or equivalently the contour integral. Then consider F̂ (s) theLaplace transform of f (t) for large s. Then Bromwich integral makes F (s) into f (t) bydefinition and by the earlier theory, it makes F̂ (s) into f (t) and so F (s) = F̂ (s). ■

From this proposition, we have the following procedure.

Procedure 15.8.5 Suppose F (s) is a Laplace transform and is meromorphic on Cand satisfies 15.14. (This situation is quite typical) Then to compute the Holder continuousfunction of t, f (t) whose Laplace transform gives F (s) , do the following. Find the sum ofthe residues of eztF (z) for Rez < c where all poles have real part smaller than c.

396 CHAPTER 15. ISOLATED SINGULARITIESand the third having 6 € (5 + arcsin (go ) ; 2 —arcsin (gos )). Then,3B ~aresin(5*[ e°F (z)dz= |Cr ¥ +aresin (5¢$-+aresin( 3%+h—arcsin (#)3a F +aresin( & )hy.3m ~aresin(—° )2 ri-BConsider the last two integrals first. For large |z|, with z € Cp, the sum of the absolutevalues of these is no more than) e(Reos O+iRsin 6)t (Re?) Rie’? d+ (15.15)B)e(Reos 6+iRsin 6)t F (Re’*) Rie!’ d@o(Reos 0-+iRsin @)t pp (Re?) Rie d@aE (#)+aresin( £)eR (cos 6)t © nie[focally ) eRleos Ot © Rd@)| +z .J5 ~arcsin( 8)< Ce R(cos( §—aresin( ¥ )) )t lin (<8) in (9 a+CeR(cos( F +aresin( g)) 1 (aresin (a) + arcsin (5)) Ri-¢3 —aresin(Now from trig. identities, cos (4 —arcsin(@)) = 0,cos (2% +arcsin(@)) = @, and sothe above reduces to 2Ce“ (aresin (5 i ar) +arcsin (4 ) R'—® which converges to 0 asR—-> 0, Recall 0 < B <q. It remains to consider the integral in 15.15. For large |z|,the absolute value of this integral is no more than3 —aresin(§ ) Cis R ef(cos = RdO < cnet ld +aresin( = 5)) pI- a —~CrrR'- a ect RP$+aresin(§)which converges to 0 as R > oo.Proposition 15.8.4 /fRep < c forall p a pole of F (s) and if F (s) is meromorphic andsatisfies the growth condition 15.14, and if f (t) is defined by the Bromwich integral, thenF (s) is the Laplace transform of f (t) for large s.Proof: This follows from Lemmas 15.8.2 and 15.8.3. These lemmas say that if F (s)satisfies the growth condition, then we can define a locally Lipschitz function f (t) in termsof that Bromwich integral or equivalently the contour integral. Then consider F (s) theLaplace transform of f (t) for large s. Then Bromwich integral makes F (s) into f(t) bydefinition and by the earlier theory, it makes F (s) into f (t) and so F (s) = F (s).From this proposition, we have the following procedure.Procedure 15.8.5 Suppose F (s) is a Laplace transform and is meromorphic on Cand satisfies 15.14. (This situation is quite typical) Then to compute the Holder continuousfunction of t, f (t) whose Laplace transform gives F (s) , do the following. Find the sum ofthe residues of e*' F (z) for Rez < c where all poles have real part smaller than c.