37.5. THE BROMWICH INTEGRAL 761

Now from trig. identities, cos(

π

2 − arcsin(θ))= θ ,cos

( 3π

2 + arcsin(θ))= θ , and so

the above reduces to

2Cect(

arcsin( c

R1−β

)+ arcsin

( cR

))R1−α

which converges to 0 as R→ ∞. Recall 0 < β < α . It remains to consider the integral in37.9. For large |z| , the absolute value of this integral is no more than∫ 3π

2 −arcsin( cR )

π2 +arcsin( c

R )eR(cosθ)t C

RαRdθ ≤Cπe

Rt cos(

π2 +arcsin

(c

R1−β

))R1−α =CπR1−α e−cRβ

which converges to 0 as R→ ∞. ■

Corollary 37.5.2 Let the contour be as shown and assume 37.8 for meromorphic F (s).Then the above limit in 37.7 exists. Also f (t) , given by the Bromwich integral, is continuouson (0,∞) and its Laplace transform is F (s).

Proof: It only remains to verify continuity. Let R be so large that the above contourη∗R encloses all poles of F . Then for such large R, the contour integrals are not changingbecause all the poles are enclosed. Thus

f (t̂) = limR→∞

12πi

∫ηR

eut̂F (u)du =1

2πi

∫ηR

eut̂F (u)du

| f (t̂)− f (t)| ≤∣∣∣∣ f (t̂)− 1

2πi

∫ηR

eut̂F (u)du∣∣∣∣

+

∣∣∣∣ 12πi

∫ηR

eut̂F (u)du− 12πi

∫ηR

eutF (u)du∣∣∣∣

+

∣∣∣∣ 12πi

∫ηR

eutF (u)du− f (t)∣∣∣∣

=

∣∣∣∣ 12πi

∫ηR

eut̂F (u)du− 12πi

∫ηR

eutF (u)du∣∣∣∣

Since ηR is fixed, it follows that if |t̂− t| is small enough, then | f (t̂)− f (t)| is also small.■

It follows from Lemma 37.5.1 that

f (t) ≡ limR→∞

12πi

∫γ+iR

γ−iReutF (u)du

= limR→∞

(1

2πi

∫γ+iR

γ−iReutF (u)du+

12πi

∫CR

eutF (u)du)

=1

2πi2πi(sum of residues of the poles of eztF (z)

)= sum of residues.

The following procedure shows how the Bromwich integral can be computed to obtainan actual formula for a function. However, the integral itself will make sense and could benumerically computed to solve for the inverse Laplace transform.