Kenneth Kuttler

12.9. THE INVERSION OF LAPLACE TRANSFORMS 295

y ∈ [−1,1] and let it be 0 if |y|> 1. Of course this function has a jump at −1 and 1. Thenfrom 12.16,

limR→∞

2π

∫ R

0sin(tx)

∫ 1

0sin(ty)ydydt = lim

R→∞

2π

∫ R

0sin(tx)

(1t2 (sin t− t cos t)

)dt

Thus

limR→∞

2π

∫ R

0sin(tx)

(1t2 (sin t− t cos t)

)dt =

 x if |x|< 11/2 if |x|= 10 if |x|> 1

It might not be the first thing you would think of. I am not sure whether the integrand iseven in L1 (R) although the above limit does exist.

Now suppose that f (y) = e−|y|. This is an even function. From 12.15

e−|x| = limR→∞

2π

∫ R

0cos(tx)

∫∞

0cos(ty)e−ydydt = lim

R→∞

2π

∫ R

cos(tx)t2 +1

I think this is a pretty amazing formula. Obviously you can make these up all day, amazingformulas which none of the usual tools will allow you to compute.

Definition 12.8.4 Let f ∈ L1 (R) . The Fourier cosine and sine transforms are de-fined respectively as g(x)≡√

2π

∫∞

0f (t)cos(xt)dt,

√2π

∫∞

0f (t)sin(xt)dt

Note that 12.15, 12.16 sort of say that if you take the cosine transform of the cosinetransform, you get the function back, a similar assertion holding for the Fourier sine trans-form.

12.9 The Inversion of Laplace TransformsHow does the Fourier transform relate to the Laplace transform? This is considered next.Recall that from Theorem 10.3.5 if g has exponential growth |g(t)| ≤Ceλ t , then if Re(s)>λ , one can define L g(s) as

L g(s)≡∫

∞

0e−sug(u)du

and also s→L g(s) is differentiable on Re(s) > λ in the sense that if h ∈ C and G(s) ≡L g(s) , then

limh→0

G(s+h)−G(s)h

= G′ (s) =−∫

∞

0ue−sug(u)du

Thus G is analytic and has all derivatives. Then the next theorem shows how to invert theLaplace transform. It is another one of those results which says that you get the mid pointof the jump when you do a certain process. It is like what happens in Fourier series wherethe Fourier series converges to the midpoint of the jump under suitable conditions and likewhat was just shown for the inverse Laplace transform.

The next theorem gives a more specific version of what is contained in Theorem 10.4.8.However, this theorem does assume a Holder continuity condition which is not needed forTheorem 10.4.8. I think that it is usually the case that the needed Holder condition will beavailable.

12.9. THE INVERSION OF LAPLACE TRANSFORMS 295y € [-1,1] and let it be 0 if |y| > 1. Of course this function has a jump at —1 and 1. Thenfrom 12.16,2 /R J 2 Rk 1lim = . . — tim 2 . 7jim = [ sin (tx) [ sin (ty) ydydt jim = [ sin (tx) (3 (sint reost) dtThus> /R 1 xif |x| <1lim — | sin(tx) (3 (sins ~rcosr)) dt=< 1/2if |x|}=1Kove 1 J0 / Oif |x| >1It might not be the first thing you would think of. I am not sure whether the integrand iseven in L! (R) although the above limit does exist.Now suppose that f (y) = e~!!. This is an even function. From 12.1522 pk =“hl |i =| t [ ty) edydt = li = |e jim = hh cos(tx) J cos (ty) e dy jim = hhR cos (tx)dt+1I think this is a pretty amazing formula. Obviously you can make these up all day, amazingformulas which none of the usual tools will allow you to compute.Definition 12.8.4 Ler f €L' (R). The Fourier cosine and sine transforms are de-fined respectively as g(x) =V2 [Flees (xt) dt, 2 [rosin atNote that 12.15, 12.16 sort of say that if you take the cosine transform of the cosinetransform, you get the function back, a similar assertion holding for the Fourier sine trans-form.12.9 The Inversion of Laplace TransformsHow does the Fourier transform relate to the Laplace transform? This is considered next.Recall that from Theorem 10.3.5 if g has exponential growth |g (t)| < Ce’, then if Re (s) >A, one can define fg (s) asL2(s)= [es (u) duand also s + £g(s) is differentiable on Re(s) > A in the sense that if h € C and G(s) =L£g(s), thenlim Glsth)~ G's) (s+h) ~ G(s) =G'(s)= -| ue ““g(u)duh0 h 0Thus G is analytic and has all derivatives. Then the next theorem shows how to invert theLaplace transform. It is another one of those results which says that you get the mid pointof the jump when you do a certain process. It is like what happens in Fourier series wherethe Fourier series converges to the midpoint of the jump under suitable conditions and likewhat was just shown for the inverse Laplace transform.The next theorem gives a more specific version of what is contained in Theorem 10.4.8.However, this theorem does assume a Holder continuity condition which is not needed forTheorem 10.4.8. I think that it is usually the case that the needed Holder condition will beavailable.