762 CHAPTER 37. SOME FUNDAMENTAL FUNCTIONS AND TRANSFORMS

PROCEDURE 37.5.3 Suppose F (s) is a Laplace transform and is meromor-phic on C and satisfies 37.8. (This situation is quite typical) Then to compute the functionof t, f (t) whose Laplace transform gives F (s) , do the following. Find the sum of theresidues of eztF (z) for Rez < γ where all the poles of F (z) have real part less than γ. Thisyields the midpoint of the jump of f (t) at each t where f is Holder continuous from theleft and right. (Note there are no jumps by Corollary 37.5.2 so if f is Holder continuous atevery point, then f (t) is recovered.)

Example 37.5.4 Suppose F (s) = s

(s2+1)2 . Find f (t) such that F (s) is the Laplace trans-

form of f (t).

There are two residues of this function, one at i and one at −i. At both of these pointsthe poles are of order two and so we find the residue at i by

res( f , i) = lims→i

dds

(etss(s− i)2

(s2 +1)2

)=−iteit

4

and the residue at −i is

res( f ,−i) = lims→−i

dds

(etss(s+ i)2

(s2 +1)2

)=

ite−it

4

From the above procedure, the function f (t) is the sum of these.

ite−it

4+−iteit

4=

14

it(e−it − eit)

=14

it (cos(t)− isin t− (cos t + isin t))

=12

t sin t

You should verify that this actually works giving L ( f ) = s

(s2+1)2 .

Example 37.5.5 Find f (t) if F (s) , the Laplace transform is e−s/s.

You need to compute the residues of est e−s

s . The function equals

1s

∞

∑k=0

(−1)k (t−1)k sk

k!.

Thus the residue is 1. However, this fails to be the function whose Laplace transform isF (s) . What is wrong? The problem with this is the failure of the estimate on F (s) to holdfor large s. Indeed, if s =−n, you would have en/n but it would need to be less than C/nα

which is not possible. The estimate requires F (s)→ 0 as |s| →∞ and this does not happenhere. You can verify directly that the function which works is u1 (t) which is 0 for t < 1and 1 for t ≥ 1. Thus if the estimate does not hold, the procedure does not necessarily holdeither.