Kenneth Kuttler

296 CHAPTER 12. SERIES AND TRANSFORMS

Theorem 12.9.1 Let g be a piecewise continuous function defined on (0,∞) whichhas exponential growth |g(t)| ≤Ceλ t for some real λ and g(t) is Holder continuous fromthe right and left as in 10.8 and 10.9. For Re(s)> λ

L g(s)≡∫

∞

0e−sug(u)du

Then for any γ > λ ,

limR→∞

12π

∫ R

−Re(γ+iy)tL g(γ + iy)dy =

g(t+)+g(t−)2

(12.17)

Proof: This follows from plugging in the formula for the Laplace transform of g andthen using the above. Thus

12π

∫ R

−Re(γ+iy)tL g(γ + iy)dy =

12π

∫ R

−Re(γ+iy)t

∫∞

−∞

e−(γ+iy)ug(u)dudy

2π

∫ R

−Reγteiyt

∫∞

−∞

e−(γ+iy)ug(u)dudy = eγt 12π

∫ R

−Reiyt∫

∞

−∞

e−iyue−γug(u)dudy

Now apply Theorem 12.8.2 to conclude that

limR→∞

12π

∫ R

−Re(γ+iy)tL g(γ + iy)dy = eγt lim

R→∞

12π

∫ R

−Reiyt∫

∞

−∞

e−iyue−γug(u)dudy

= eγt g(t+)e−γt++g(t−)e−γt−

g(t+)+g(t−)2

In particular, this shows that if L g(s) = L h(s) for all s large enough, both g,h havingexponential growth, then f ,g must be equal except for jumps and in fact, at any pointwhere they are both Holder continuous from right and left, the mid point of their jumps isthe same. That integral is called the Bromwich integral.

This answers the question raised earlier about whether the Laplace transform methodeven makes sense to use because it shows that if two functions have the same Laplacetransform, then they are the same function except at jumps where the midpoint of the jumpscoincide.

Using the method of residues, one can actually compute the inverse Laplace transformusing this Bromwich integral. I will give an easy example, leaving out the technical detailsrelative to estimates. These are in my book on my web site Calculus of Real and ComplexVariables.

Example 12.9.2 Find the inverse Laplace transform of 11+s2 .

The idea is that this is (hopefully) of the form L ( f (t)) and your task is to find f (t) . Todo this, you use the following contour and write the above Bromwich integral as a contourintegral. In this example, let γ > 0. You should verify that∫ R

−Re(γ+iy)tL g(γ + iy)dy =−i

∫γ∗R

ezt 11+ z2 dz−

(−i∫

ezt 11+ z2 dz

)

296 CHAPTER 12. SERIES AND TRANSFORMSTheorem 12.9.1 Lez g be a piecewise continuous function defined on (0,°°) whichhas exponential growth |g (t)| < Ce*! for some real A and g(t) is Holder continuous fromthe right and left as in 10.8 and 10.9. For Re(s) >ALg(s)= [ets (u) du0Then for any y> i,_ g(t+) +8(t-)lim x, (r+iy) "Po (y+iy)dy 5jim = (12.17)Proof: This follows from plugging in the formula for the Laplace transform of g andthen using the above. Thus1Qn=f. ene [|e “(FN g (y )dudy =e" — Lf em [et “e ™ 9 (u) dudyNow apply Theorem 12.8.2 to conclude that1 R . eo .[i Pe (yt iy)dy = li, elroy [ e (7+9)"9 (u) dudy1 ael(ttit — elt —_ eit “NgHn Oe 20 [, La(y+iy)dy=e am. 20 =I, [ie ve (u) dudyengittye M +elirjeM _ gtt) tlt) 52 2 ,In particular, this shows that if 2g (s) = Lh(s) for all s large enough, both g,h havingexponential growth, then f,g must be equal except for jumps and in fact, at any pointwhere they are both Holder continuous from right and left, the mid point of their jumps isthe same. That integral is called the Bromwich integral.This answers the question raised earlier about whether the Laplace transform methodeven makes sense to use because it shows that if two functions have the same Laplacetransform, then they are the same function except at jumps where the midpoint of the jumpscoincide.Using the method of residues, one can actually compute the inverse Laplace transformusing this Bromwich integral. I will give an easy example, leaving out the technical detailsrelative to estimates. These are in my book on my web site Calculus of Real and ComplexVariables.Example 12.9.2 Find the inverse Laplace transform of ThThe idea is that this is (hopefully) of the form Y (f (t)) and your task is to find f (+) . Todo this, you use the following contour and write the above Bromwich integral as a contourintegral. In this example, let y > 0. You should verify thatRi 1 1vmstat (4.27834[ie Bly iy)dy Ty 142 ‘Jeg? 142%