Kenneth Kuttler

15.8. THE INVERSION OF LAPLACE TRANSFORMS 395

and this is the case we will consider because it is very easy to compute. We assume c > 0.

Lemma 15.8.2 Let the contour be as shown and assume 15.13 for meromorphic F (u)such that F (u) also has only finitely many poles whose real parts are less than η < c,a positive numer, F (u) being analytic if Re(u) ≥ c. Then f (t) , given by the Bromwichintegral, has exponential growth | f (t)| ≤Cect Lipschitz continuous near every point t > 0and its Laplace transform is F (s).

Proof: Since F has only finitely many poles It only remains to verify Lipschitz conti-nuity near a point t > 0. Let R be so large that the above contour γ∗R encloses all poles ofF . Then for such large R, the contour integrals are not changing because all the poles areenclosed. Thus, letting t̂ ∈ (t−δ , t +δ ) , t−δ > 0, it follows that

f (t̂) = limR→∞

12πi

∫γR

eut̂F (u)du =1

2πi

∫γR

eut̂F (u)du

Therefore,

| f (t̂)− f (t)|=∣∣∣∣ 12πi

∫γR

eut̂F (u)du− 12πi

∫γR

eutF (u)du∣∣∣∣

2π

∣∣∣∣∫γR

(eut̂ − eut

)F (u)du

∣∣∣∣= 12π

∣∣∣∣∫γR

(∫ t̂

tseusds

)F (u)du

∣∣∣∣The contour contains no poles and F is continuous, so |F (u)| is bounded by some numberM on γR. Then the above is no more than

≤M1

2π2πR

∣∣∣∣∫ t̂

tseusds

∣∣∣∣= RMt̂ec(t̂+t) |t̂− t| ≤ RM (t +δ )ec(t̂+t) |t̂− t|

The claim of exponential growth follows from observing that the residue at each pole zis of the form eztb1 and |eztb1|< |b1|ect . ■

A sufficient condition for 15.13 is that for all |z| large enough,

|F (z)| ≤ C|z|α

, some α > 0. (15.14)

Note that this assumption implies there are finitely many poles for F (z) because if w is apole, you have limz→w |F (z)|= ∞.

Lemma 15.8.3 Let the contour be as shown and assume the growth condition 15.14.Then the above limit in 15.13 exists for t > 0.

Proof: Assume c≥ 0 as shown and let θ be the angle between the positive x axis and apoint on CR. Let 0 < β < α . Then the contour integral over CR will be broken up into threepieces, two pieces around the y axis

θ ∈[

2− arcsin

( cR

2+ arcsin

( cR1−β

)],[

3π

2− arcsin

( cR1−β

3π

2+ arcsin

( cR

)],

15.8. THE INVERSION OF LAPLACE TRANSFORMS 395and this is the case we will consider because it is very easy to compute. We assume c > 0.Lemma 15.8.2 Let the contour be as shown and assume 15.13 for meromorphic F (u)such that F (u) also has only finitely many poles whose real parts are less than N < c,a positive numer, F (u) being analytic if Re(u) > c. Then f(t), given by the Bromwichintegral, has exponential growth | f (t)| < Ce“ Lipschitz continuous near every point t > 0and its Laplace transform is F (s).Proof: Since F has only finitely many poles It only remains to verify Lipschitz conti-nuity near a point f > 0. Let R be so large that the above contour Vp encloses all poles ofF. Then for such large R, the contour integrals are not changing because all the poles areenclosed. Thus, letting f € (t—6,t+6),t—6 > 0, it follows thatf (@) = lim = / ce" F (u)du= ! / eF (u)duYR YRR00 270i OniTherefore,1. 1_ ce d -— | ut i dIni [e (w)du~ Tri dye | ae[ ee") ronal = 2) ( [eras ronan- 20The contour contains no poles and F is continuous, so |F (w)| is bounded by some numberM on Yp. Then the above is no more than7?| se’ dstThe claim of exponential growth follows from observing that the residue at each pole zis of the form e“b, and |e“b,| < |b, |e". HlA sufficient condition for 15.13 is that for all |z| large enough,IFO —f(O)| =_tOn1 A A A 7 A< M5—2nR — RMieCF) lf —t| < RM (t +6) et) lf—t|CcIFS Ta some > 0. (15.14)Note that this assumption implies there are finitely many poles for F (z) because if w is apole, you have lim,_,,, |F (z)| =.Lemma 15.8.3 Let the contour be as shown and assume the growth condition 15.14.Then the above limit in 15.13 exists for t > 0.Proof: Assume c > 0 as shown and let 0 be the angle between the positive x axis and apoint on Cr. Let 0 < B < a. Then the contour integral over Cr will be broken up into threepieces, two pieces around the y axis@€ E in(5) Fan in( 5 )|5 arcs R)?2 arcs RB):=F _aresin (>) =F aresin (£)2 R!-B/? 2 R/\?